1. Extension-Contraction and the Spectrum of a Polynomial Ring

For subsets A, B of a ring R, we define

AB = {a_{1}b_{1} + … + a_{n}b_{n} | n ∈ ℕ, a_{i} ∈ A, b_{i} ∈ B}.

For a subset S of a commutative ring R with 1, the ideal iof R generated by the set S is SR.

∗ ∗ ∗ Extension and Contraction of Ideals ∗ ∗ ∗

If R ⊆ S are commutative rings then every ideal I of R defines an ideal of S by expansion: I^{e} is the ideal of S generated by I. If S is a commutative ring with 1 then I^{e} = IS. Every ideal J of S defines an ideal of R by contraction: J^{e} = J ∩ R.

More generally, if f: R → S is a ring-homomorphism, then every I ⊴ R defines an ideal I^{c} of S: the ideal of S generated by f(I). If S is a commutative ring with 1 then I^{c} = f(I)S. Every ideal J of S defines an ideal of R via contraction J^{c} = f^{−1}(J) of R.

(When f is understood, I^{e} is often written as IS and J^{e} as J ∩ R by abuse of notation even if f is not injective.)

Contraction preserves the property of being a prime ideal (or a primary ideal); extension, in general, does not.

Clearly, for ideals A, B of R and C, D of S, A ⊆ B ⇒ A^{c} ⊆ B^{c} and C ⊆ D ⇒ C^{e} ⊆ D^{e}; and A^{ce} ⊆ A and D^{ec} ⊇ D.

Therefore, for every ring homomorphism f: R → S, extension and contraction give a bijective correspondence between ideals of R of the form J^{c} = J ∩ R (J an ideal of S) and ideals of S of the form I^{e} = IS (I an ideal of R). In two noteable cases, all ideals of S are of the form I^{e}:

• Let H be an ideal of R and π:R→R/H, π(r)=r+H, the canonical projection. Then extension and contraction induce a bijection between all ideals of R/H and the ideals of R containing H. In both directions of this bijection the properties of being a prime ideal, a primary ideal, or a G-ideal are preserved. Note that for a subset S of R/H π^{−1}(S) = ⋃_{r+H∈S} r + H.

• Let S be a multiplicative subset of R, R_{S} the ring of quotients with denomiators in S and f: R → R_{S} the canonical map f(r) = r/1 (if R has a unit element, or f(r) = rs/s otherwise). Recall that f is injective whenever S contains no zero-divisors. Then extension and contraction induce a bijection between all ideals of R_{S} and those ideals of R that are disjoint from S. Most importantly, if S = R ⊆ P, P a prime ideal, then the bijection is between all ideals of R_{P} and the ideals of R contained in P. Recall that for I an ideal of R, the rings R[x]/I[x] and (R/I)[x] are canonically

isomorphic via a_{0} +a_{1}x+…+a_{n}x^{n} +I[x] ↦ (a_{0} +I)+(a_{1} +I)x+…+(a_{n} +I)x^{n}.

∗ ∗ ∗ Spectrum of a Polynomial Ring ∗ ∗ ∗

Let R be a polynomial ring and P a prime ideal of R. Then every prime ideal Q of R[x] with Q ∩ R ⊇ P contains P [x] and is therefore of the form π^{−1}(Q̄), where π: R[x] → R[x]/P [x] = (R/P)[x] is the canonical projection and Q̄ is a prime ideal of (R/P)[x]. Those Q among them with Q ∩ R = P correspond to those Q̄ for which Q̄ ∩ (R/P) = (0 + P).

Let D be a domain with quotient field K. Then every prime ideal Q of D[x] with Q ∩ D = (0) is of the form Q̄ ∩ D[x], where Q̄ is a prime ideal of K[x] (and thus Q̄ = (f), where f is an irreducible polynomial of K[x].

To summarize: to determine for a given prime ideal P of R all prime ideals Q of R[x] with Q ∩ R = P: consider the polynomial ring K_{P} [x], where K_{P} is the quotient field of R/P, i.e. K_{P} = (R/P \{0+P })^{−1}R/P for each monic irreducible f ∈ K_{P} [x], consider all polynomials of R/P [x] that are divisible by f in K_{P} [x], i.e. all products fg with g ∈ K_{P} [x], such that fg ∈ R/P [x]. Then the collection of all polynomials of D[x] whose residue class (under the projection of coefficients from R to R/P is of the form fg (for some g ∈ K_{P} [x]) is a prime ideal of D[x] and all prime ideals Q of D[x] with Q ∩ R = P are of this form.

2. Modules

2.1 Definition. Let R be a commutative ring. An Abelian group (A, +) together with a “scalar” multiplication ·: R × A → A by elements of R is called an Rmodule, if the following conditions hold for all r, s ∈ R and a, b ∈ A:

(r + s)a = ra + sa (rs)a = r(sa) r(a + b) = ra + rb In addition, we will always assume that modules are unitary, which means

that multiplication by 1 ∈ R is the identity map on A, i.e., for all a ∈ A, 1_{R}a = a.

2.2 Definition. Given an R-module A, the annihilator of a subset B of A is Ann_{R}(B) = {r ∈ R | ∀b ∈ B rb = 0}.

Clearly, Ann_{R}(B) is an ideal of R for any B. Of particular importance is the annihilator of the whole module. If Ann_{R}(A) = I for an R-module A then A can be regarded as an R/I-module by defining (r + I)a:= ra.

2.3 Definition. An R-module A is called faithful if Ann_{R}(A) = (0).

2.4 Definition. A (possibly non-commutative) ring A is called an R-algebra, if (A, +) is an R-module and the multiplication of the ring A interacts peacefully with the scalar multiplication as follows: for all r ∈ R and a, b ∈ A

r(ab) = (ra)b = a(rb).

Examples of R-algebras include the polynomial ring R[x] and the ring M_{n}(R) of n × n matrices with entries in R. More generally, every ring T of which R is a subring is an R-algebra (by restriction of the ring-multiplication on T to ·: R × T → T).

Because of rb = (r1_{A})b, an R-algebra A is faithful if and only if Ann(1_{A}) = (0), and therefore every faithful R-algebra A admits an embedding of R by r ↦ r1_{A}. If we call R̃ the copy of R thus embedded in A, then, again by rb = (r1_{A})b, the scalar multiplication of the R-algebra A is just the restriction of the ring multiplication of A to R̃ × A.

3. Cayley-Hamilton and McCoy Theorems

A little excursion into linear algebra.

For R a commutative ring, let M_{n}(R) be the R-algebra of n × n matrices with entries in R. We denote the identity matrix by I.

3.1 Lemma. Let R be a commutative ring. For A = (a_{ij}) ∈ M_{n}(R), define the adjoint of A as adj(A) = B = (b_{ij}) ∈ M_{n}(R) with b_{ij} = (−1)^{i+j} det(A_{j}^{i}), where A_{j}^{i} ∈ M_{n−1}(R) results from A by removing the i-th column and the j-th row.

Then

AB = BA = (det A)I.

Proof. Routine verification. □

The adjoint of a matrix can be used to give a short proof of Cayley-Hamilton’s theorem. It uses the natural R-algebra homomorphism between M_{n}(R[x]) and M_{n}(R)[x], which maps the matrix whose (i, j)-th entry is ∑_{k} a_{k}^{(i,j)}x^{k} ∈ R[x], to the polynomial ∑_{k} A_{k}x^{k}, where A_{k} ∈ M_{n}(R) is the matrix whose (i, j)-th entry is a_{k}^{(i,j)}.

If S is a non-commutative ring, such as M_{n}(R), and we want to substitute a ring element s for the variable in f(x) = ∑ a_{k}x^{k} ∈ S[x], we must specify whether substitution happens on the right or on the left of the coefficients, i.e., if f(s) means ∑ a_{k}s^{k} (right substitution) or ∑ s^{k}a_{k} (left substitution).

In any case, it is important to remember that substitution (whether right or left) is in general not a homomorphism: if f, g ∈ S[x] and s ∈ S, it is in general not true that (f · g)(s) = f(s) · g(s) (unless s commutes with the coefficients of f and g). One thing that does work similarly to polynomials over commutative rings is the correspondence between zeros and linear factors:

3.2 Lemma. Let S be a (possibly non-commutative) ring and f = ∑ a_{k}x^{k} ∈ S[x]. For s ∈ S define f(s) = ∑ a_{k}s^{k}. Then f(s) = 0 if and only if f(x) = g(x)(x − s) for some g ∈ S[x].

Proof. If f(s) = 0 then f(x) = f(x)−f(s) = ∑_{k} a_{k}x^{k} −∑_{k} a_{k}s^{k} = ∑_{k} a_{k}(x^{k} −s^{k}) = ∑_{k} a_{k}(∑_{j=0}^{k−1} s^{j}x^{k−j−1})(x − s) = g(x)(x − s).

If, on the other hand, f(x) = g(x)(x − s) with g(x) = ∑_{k=0}^{n} b_{k}x^{k} then f(x) = ∑_{k=0}^{n} b_{k}x^{k+1} − ∑_{k=0}^{n} b_{k}sx^{k} = ∑_{k=0}^{n+1}(b_{k−1} − b_{k}s)x^{k} and right substitution yields f(s) = 0. □

3.3 Theorem (Cayley-Hamilton). Let C be a n × n matrix with entries in a commutative ring R, and χ(x) = det(xI − C) its characteristic polynomial. Then χ(C) = 0.

Proof. Let B ∈ M_{n}(R[x]) be the adjoint of xI − C then

B · (xI − C) = χ(x)I.

Applying the R-algebra isomorphism M_{n}(R[x]) ≃ M_{n}(R)[x], we get

B(x)(x − C) = χ(x),

where B(x) ∈ M_{n}(R)[x] is the polynomial corresponding to the matrix B ∈ M_{n}(R[x]). Now χ(x) has a factor x − C in M_{n}(R)[x], and therefore χ(C) = 0. □

With the argument we used to prove Cayley-Hamilton (3.3), one can actually show a stronger result, characterizing the ideal of those polynomials in R[x], which have a given matrix C ∈ M_{n}(R) as a zero:

3.4 Theorem (McCoy). Let C be a n × n matrix with entries in a commutative ring R. For 1 ≤ k ≤ n let J_{k}(xI − C) be the ideal of R[x] generated by the k × k minors of xI −C. Then the ideal of all polynomials f ∈ R[x] with f(C) = 0 is equal to the ideal quotient (J_{n}(xI −C) : J_{n−1}(xI −C)) = (χ_{C}(x)R[x] :_{R[x]} J_{n−1}(xI −C)).

Proof. Since the generators of J_{n−1}(xI − C), the (n − 1) × (n − 1)-minors of (xI − C), are (up to sign) the entries of adj(xI − C)),

g(x) ∈ (χ(x)R[x] :_{R[x]} J_{n−1}(xI − C))

is equivalent to

g(x)adj(xI − C) ∈ χ(x)M_{n}(R[x]).

Multiplying by (xI − C), which, being monic, is certainly not a zero-divisor in M_{n}(R)[x], we again get an equivalent statement:

g(x)χ(x)I ∈ χ(x)M_{n}(R[x])(xI − C).

We may cancel the scalar χ(x) ∈ R[x] (a monic polynomial - not a zero-divisor in R[x]) to arrive at the equivalent statement

g(x)I ∈ M_{n}(R[x])(xI − C),

and apply the isomorphism between M_{n}(R[x]) and M_{n}(R)[x] to get

g(x) ∈ M_{n}(R)[x](x − C),

which is (by 3.2) equivalent to C being a zero of g(x). □

From Cayley-Hamilton theorem for matrices it is possible to derive a version of Cayley-Hamilton theorem for linear operators, by associating matrices to linear operators:

Suppose M is an R-module generated by m_{1}, …, m_{n}. Then every b ∈ M is representable as b = b_{1}m_{1} + … + b_{n}m_{n} with (in general non-unique) b_{1}, …, b_{n} in R. Let φ ∈ End_{R}(M). We say that a matrix C ∈ M_{n}(R) represents φ with respect to the system of generators m_{1}, …, m_{n}, if for every b ∈ M, b = b_{1}m_{1} + … + b_{n}m_{n} implies φ(b) = b′_{1}m_{1} + … + b′_{n}m_{n} with C[b_{1}, …, b_{n}] = [b′_{1}, …, b′_{n}]. (We use square brackets to denote column vectors: [b_{1}, …, b_{n}] means the transpose of (b_{1}, …, b_{n}). It is easy to see that C represents φ if and only if the k-th column of C is a coordinate vector of φ(m_{k}), i.e., if φ(m_{k}) = c_{1k}m_{1} + c_{2k}m_{2} + … + c_{nk}m_{n}.

If the generators m_{1}, …, m_{n} are not R-linearly independent, then the matrix associated to a linear operator φ ∈ End_{R}(M) is non-unique, and not every matrix in M_{n}(R) represents a linear operator. If C represents φ and D represents ψ, however, then C + D and CD represent φ + ψ and φ ∘ ψ, respectively, and, for r ∈ R, rC represents rφ, so the set E ⊆ M_{n}(R) of matrices representing R-endomorphisms of M is a R-subalgebra of M_{n}(R), and there is a surjective R-algebra homomorphism π: E → End_{R}(M), mapping every matrix in E to the endomorphism it represents. In particular, for every matrix C represening some φ∈End_{R}(M), restriction of π to R[C] gives a surjective R-algebra homomorphism π: R[C] → R[φ], mapping f(C) ∈ R[C] to f(φ) ∈ R[φ] for every f ∈ R[x].

3.5 Lemma. Let M be an R-module generated by n elements and φ ∈ End_{R}(M). If C ∈ M_{n}(R) is a matrix associated to φ (with respect to a system of generators m_{1}, …, m_{n} ∈ M) and χ(x) the characteristic polynomial of C then χ(φ) = 0.

Proof. By Cayley-Hamilton (3.3), χ(C) = 0, where χ = det(xI − C) is the characteristic polynomial of C. Applying the R-algebra homomorphism π: R[C]→ R[φ] mentioned above, we see that 0 = π(χ(C)) = χ(φ). □

3.6 Corollary. Let M be an R-module generated by n elements and φ∈End_{R}(M), such that φ(M) ⊆ JM for some ideal J ⊴ R. Then φ is a zero of a monic polynomial in R[x] of degree n, f(x) = x^{n} + a_{n−1}x^{n−1} + … + a_{1}x + a_{0} with a_{k} ∈ J^{n−k} for 0 ≤ k < n.

Proof. If m_{1}, …, m_{n} generate M as a R-module and φ(M) ⊆ JM then there exist c_{ij} ∈ J with φ(m_{k}) = c_{1k}m_{1} +c_{2k}m_{2} +…+c_{nk}m_{n}. Now χ(x) = det(xI −C), the characteristic polynomial of C = (c_{ij}) is a monic polynomial of degree n with χ(φ) = 0 and the property that the coefficient of x^{k} is in J^{n−k}. □

4. Integral elements

4.1 Definition. Let R be a commutative ring, T an R-algebra and u ∈ T. Then u is called integral over R if and only if there exists a monic polynomial f ∈ R[x] with f(u) = 0.

4.2 Definition. A ring extension R ⊆ T is called integral if every element of T is integral over R.

4.3 Theorem. Let R be a commutative ring, T an R-algebra and u ∈ T. Then the following are equivalent:

(1) u is integral over R.

(2) R[u] is finitely generated as an R-module.

(3) R[u] ⊆ A ⊆ T for some ring A that is finitely generated as an R-module.

(4) There exists a faithful R[u]-module that is finitely generated as an R module.

(5) R[u] is integral over R.

Proof. (1 ⇒ 2) R[u] is generated as an R-module by the powers of u. Since u is integral over R, there exists an n ∈ ℕ and r_{0}, …, r_{n−1} ∈ R such that u^{n} = r_{n−1}u^{n−1} + … + r_{1}u + r_{0}, and by induction, every power of u is expressible as an R-linear combination of 1, u, …, u^{n−1}. (2 ⇒ 3) Take A = R[u]. (3 ⇒ 4) Take A as the faithful R[u] module. (4 ⇒ 5) Let M be a faithful R[u]-module, generated by m_{1}, …, m_{n} as an Rmodule, and let w ∈ R[u]. Then φ_{w}(m) = wm (scalar multiplication by w) is an R-module endomorphism of M, φ_{w} ∈ End_{R}(M). Let C = (c_{ij}) in M_{n}(R) be a matrix with with φ_{w}(m_{k}) = c_{1k}m_{1} + c_{2k}m_{2} + … + c_{nk}m_{n} then C represents φ_{w} ∈ End_{R}(M) with respect to the system of generators m_{1}, …, m_{n}.

By Cayley-Hamilton (3.5), χ(φ_{w}) = 0, where χ(x) = det(xI_{n} − C) is the characteristic polynomial of C. 0 = χ(φ_{w}) = φ_{χ(w)}, means that scalar multiplication by χ(w) is the 0-mapping on M, in other words, χ(w) ∈ Ann_{R[w]}(M). Since M is a faithful R[u] module, χ(w) = 0 follows. □

4.4 Definition. Let R be a commutative ring, I an ideal of R, T an R-algebra and u ∈ T. Then u is called integral over I if and only if there exists a monic polynomial f ∈ R[x] whose coefficients (apart from the leading coefficient) are all in I with f(u) = 0.

4.5 Theorem. Let R be a commutative ring, I an ideal of R, T an R-algebra and u ∈ T. Then the following are equivalent:

(1) u is integral over I.

(2) R[u] is finitely generated as an R-module and u ∈ \sqrt{IR[u]} .

(3) R[u] ⊆ A ⊆ T for some ring A that is finitely generated as an R-module

and u ∈ \sqrt{IA} .

(4) There exists a faithful R[u]-module M that is finitely generated as an R module and for some k ∈ ℕ, u^{k}M ⊆ IM.

Proof. (1 ⇒ 2) There exists an n ∈ ℕ and r_{0}, …, r_{n−1} ∈ I such that u^{n} = r_{n−1}u^{n−1} + … + r_{1}u + r_{0}. Therefore u ∈ \sqrt{IR[u]} and R[u] is generated as an R-module by 1, u, …, u^{n−1}. (2 ⇒ 3) Take A = R[u]. (3 ⇒ 4) Take A as the faithful R[u] module. (4 ⇒ 1) Let M be a faithful R[u]-module, generated by m_{1}, …, m_{n} as an Rmodule. Let w=u^{k} such that wM ⊆IM. Then φ_{w}(m)=wm (scalar multiplication by w) is an R-module endomorphism of M. Let C = (c_{ij}) in M_{n}(R) be a matrix with with φ_{w}(m_{k}) = c_{1k}m_{1} + c_{2k}m_{2} + … + c_{nk}m_{n} and c_{ij} ∈ I for all i, j. Then C represents φ_{w} ∈ End_{R}(M) with respect to the system of generators m_{1}, …, m_{n}.

By Cayley-Hamilton (3.5), χ(φ_{w}) = 0, where χ = det(xI_{n} − C) is the characteristic polynomial of C, whose coefficients (apart from the leading coefficient) are all in I. 0 = χ(φ_{w}) = φ_{χ(w)}, means that scalar multiplication by χ(w) is the 0-mapping on M, in other words, χ(w) ∈ Ann_{R[w]}(M). Since M is a faithful R[u] module, χ(w) = 0 follows. As w = u^{k} we have also found a monic polynomial satisfied by u whose coefficients (except for the leading coefficient) are in I. □

If R ⊆ T are commutative rings and T is finitely generated as an R-module, then T is also finitely generated over R as a ring, but the converse does not hold in general, viz. the polynomial ring R[x]. Integrality provides the answer to the question when the converse does hold. Note the analogy to field extensions, where finitely generated by algebraic elements equals finite-dimensional.

4.6 Proposition. Let R be a commutative ring and T a commutative R-algebra. Then the following are equivalent:

(1) T is finitely generated as an R-module.

(2) T is finitely generated over R a a ring and integral over R.

(3) T is finitely generated by integral elements over R as a ring.

Proof. (1 ⇒ 2) by 4.3. (2 ⇒ 3) holds a fortiori. To see (3 ⇒ 1), assume T is generated as a ring over R by t_{1}, …, t_{m}. Then T is generated as an R-module by monomials t_{1}^{k_{1}}t_{2}^{k_{2}} … t_{m}^{k_{m}}. If each t_{i} is integral over R then for each i there exists an exponent n_{i} such that every power of t_{i} is expressible as an R-linear combination of 1, t, …, t^{n_{i}}, and therefore T is generated as an R-module by monomials t_{1}^{k_{1}}t_{2}^{k_{2}} … t_{m}^{k_{m}} with 0 ≤ k_{i} ≤ n_{i}. □

4.7 Corollary. Let R be a commutative ring, T an R-algebra and a, b ∈ T with ab = ba. If both R[a] and R[b] are finitely generated as R-modules then so is R[a, b]. In particular, if a and b are commuting integral elements over R, then a − b and ab are integral over R, too.

4.8 Corollary. Let R ⊆ T be commutative rings. Then the set of all elements of T integral over R is a ring.

Commutativity is essential here. If the integral elements a and b do not commute, neither their sum nor their product needs to be integral over R.

4.9 Definition. Let R ⊆ T be commutative rings. The ring

R′ = {t ∈ T | t is integral over R}

is called the integral closure of R in T. R is called integrally closed in T if R′ = R. “The” integral closure of a domain R is the integral closure in its quotient field. R is called integrally closed if R is integrally closed in its quotient field.

4.10 Proposition. Let R be a commutative ring, R ⊆ S an integral ring extension and u integral over S. Then u is integral over R.

Proof. Suppose u^{n} + a_{n−1}u^{n−1} + … + a_{0} = 0 with a_{0}, …, a_{n−1} ∈ S. Then R[a_{0}, …, a_{n−1}] is finitely generated integral over R. Let g_{1}, …, g_{m} be generators of R[a_{0}, …, a_{n−1}] as an R-module. Then R[a_{0}, …, a_{n−1}, u] is generated as an R-module by the elements g_{i}u^{k} for 1 ≤ i ≤ m and 1 ≤ k ≤ n − 1. □

4.11 Corollary (Transitivity of integral extensions). Let R ⊆ S ⊆ T be commutative rings. If T is integral over S and S is integral over T then T is integral over R.

4.12 Corollary (Integral closure is integrally closed). Let R ⊆ T be commutative rings and R′ the integral closure of R in T. Then R′ is integrally closed in T.

4.13 Theorem. Let R be an integrally closed domain with quotient field K, L a field containing K, and u ∈ L algebraic over K. Then u is integral over R iff the minimal polynomial of u over K is in R[x].

Proof. Assume u integral over R. Let f be the minimal polynomial of u over K and L̄ its splitting field over L. Over L̄, f(x) = (x − u)(x − u_{1}) … (x − u_{m}). Since u and the u_{i} are roots of the same irreducible polynomial on K[x] there exists for each i an isomorphism φ: K[u] → K[u_{i}] with φ(u) = u_{i} and φ(k) = k for all k ∈ K. If u is a root of some monic g ∈ R[x], then 0 = φ(0) = φ(g(u)) = g(u_{i}) shows u_{i} to be a root of the same monic g ∈ R[x]. All u_{i} are integral over R. Therefore the coefficients of f (as sums of products of the roots) are also integral over R. At the same time, the coefficients are in K, therefore, R being integrally closed, the coefficients of f are in R. □

4.14 Lemma. Let K ⊆ F be an algebraic extension of fields and R ⊆ K a domain integrally closed in K.

(i) u ∈ F is integral over R if and only if its minimal polynomial over K is in R[x].

(ii) Let P be a prime ideal of R. u ∈ F is integral over P if all coefficients of its minimal polynomial over K (apart from the leading coefficient) are in P.

Proof. Ad (i). Let f ∈ K[x] be the minimal polynomial of u over K and g ∈ R[x] with g(u) = 0. Then f divides g in K[x]. Let u_{1}, …, u_{n} be the roots of f in its splitting field K̄ over F, then all u_{i} are also roots of g and therefore integral over R. Let R′ be the integral closure of R in K̄. The coefficients of f (as elementary symmetric polynomials in the u_{i} are in R′. As they are also in K, they are in R′ ∩ K = R.

Ad (ii). Let f, g as before, and all coefficients (except leading coefficient) of g in P. The roots of f are integral over P, and therefore in \sqrt{P R′} . The coefficients of f are, as elementary symmetric polynomials in the u_{i}, in the subring generated by the u_{i} in K̄, which is contained in \sqrt{P R′} . As they are also in K, the coefficients

of f are in K ∩ \sqrt{P R′} = \sqrt{P} = P.

□

4.15 Corollary. Let R be a domain with quotient field K and u ∈ L algebraic over K. Then u is integral over R iff the coefficients of the minimal polynomial of u over K are integral over R.

4.16 Lemma. Let R ⊆ T be commutative rings and u ∈ T an invertible element. Then u^{−1} is integral over R if and only if u^{−1} ∈ R[u].

Proof. An equation showing u^{−1} to be integral over R, such as

(u^{−1})^{n} = a_{n−1}(u^{−1})^{n−1} + … + a_{1}(u^{−1}) + a_{0},

can be multiplied by u^{n−1} to show u^{−1} to be in R[u]:

u^{−1} = a_{n−1} + … + a_{1}u^{n−2} + a_{0}u^{n−1}.

Conversely, an equation showing u^{−1} ∈ R[u], can be multiplied by an appropriate power of u^{−1} to show u^{−1} integral over R. □

4.17 Proposition. Let R ⊆ L be an integral extension of domains. Then R is a field if and only if L is a field.

Proof. If R is a field then L is an algebraic extension of a field and therefore a field. If L is a field and u is a non-zero element of R then u has an inverse u^{−1} in L, which by the previous lemma is in R[u] = R. Therefore R is a field. □

4.18 Definition. Ler R ⊆ T be commutative rings. We introduce names for a few properties that the ring extension R ⊆ T may or may not satisfy:

Lying over. For every prime ideal P ⊴ R there exists a prime ideal P̃ ⊴ T with P̃ ∩ R = P. Going up. If P ⊆ Q are prime ideals of R and P̃ a prime ideal of T with P̃ ∩ R = P then there exists a prime ideal Q̃ of T with P̃ ⊆ Q̃ and Q̃ ∩ R = Q. Going down. If P ⊆ Q are prime ideals of R and Q̃ a prime ideal of T with Q̃ ∩ R = Q then there exists a prime ideal P̃ of T with P̃ ⊆ Q̃ and P̃ ∩ R = P. Incomparability. For any two prime ideals P̃ and Q̃ of T with P̃ ∩R = Q̃∩R neither P̃ ⊆ Q̃ nor Q̃ ⊆ P̃ holds.

Let R ⊆ T be commutative rings, P a prime ideal of R and S = R \ P. If Q is an ideal of T, then Q ∩ R ⊆ P if and only if Q ∩ S = ∅. Since S is a multiplicative subset of T containing (0), we know there exist ideals of T not intersecting S, and among them ideals maximal with this property, which are necessarily prime. The question is now whether an ideal maximal among those not intersecting S actually intersects R in P and vice versa. It turns out that the properties “incomparability” and “going up” can be characterized in this way.

4.19 Proposition. Let R ⊆ T be commutative rings. Then the following are equivalent:

(1) R ⊆ T satisfies “going up”.

(2) For every prime ideal P of R, if Q is an ideal of T maximal with respect to the property Q ∩ (R \ P) = ∅ then Q ∩ R = P.

Since ideals maximal with respect to avoiding a multiplicative set S with 0 ∉ S are guaranteed to exist and are prime, we get:

4.20 Corollary. “Going up” implies “lying over”.

4.21 Proposition. Let R ⊆ T be commutative rings. Then the following are equivalent:

(1) R ⊆ T satisfies “incomparability”.

(2) For every prime ideal P of R, if Q is a prime ideal of T with Q ∩ R = P then Q is maximal with respect to the property Q ∩ (R \ P) = ∅.

4.22 Theorem. Let R ⊆ T be an integral extension of rings. Then “incomparability” and “going up” (and therefore “lying over”) hold.

Proof. To show “going up”, we assume P is a prime ideal of R and P̃ an ideal of T maximal with respect to the property P̃ ∩(R\P) = ∅; we must show P̃ ∩R = P. Suppose otherwise and choose u ∈ P \ P̃. Then, by maximality of P̃, there exists an s ∈ (R \ P) ∩ (P̃ + T u), s = q + tu with q ∈ P̃ and t ∈ T. t is integral over R, i.e.,

t^{n} + c_{n−1}t^{n−1} + … + c_{0} = 0

with coeffiecients c_{i} ∈ R. Multiplying by u^{n}, we get

(tu)^{n} + c_{n−1}u(tu)^{n−1} + … + u^{n}c_{0} = 0.

Since tu + q = s, tu ≡ s mod P̃, and therefore

s^{n} + c_{n−1}us^{n−1} + … + u^{n}c_{0} ≡ 0 (mod P̃).

The left side of this congruence is in R, so it is actually in P̃ ∩ R ⊆ P. Since u ∈ P we get s^{n} ∈ P and therefore s ∈ P, a contradiction.

To show “incomparability”, we assume P̃ ∩ R = P and must show that P̃ is maximal with respect to not intersecting S = R \ P. Suppose there exists Q̃ with P̃ ⊂ Q̃ and Q̃ ∩ S = ∅. Choose u ∈ Q̃ \ P̃. u is integral over R. So in particular there exists a monic polynomial f in R[x] with f(u) ∈ P̃. Let f(x) = x^{n} + ∑_{i=0}^{n−1} a_{i}x^{i} be of minimal degree with this property. Clearly, deg f = n ≥ 1. Since u^{n} + a_{n−1}u^{n−1} + … + a_{0} ∈ P̃ ⊆ Q̃ and u ∈ Q̃, it follows that a_{0} ∈ Q̃ ∩ R = P. Now u(u^{n−1} + a_{n−1}u^{n−2} + … + a_{1}) ∈ P̃, but no factor is in P̃, a contradiction. □

4.23 Theorem. If R ⊆ T is an integral extension of domains and R is integrally closed then “going down” holds.

Proof. Suppose P_{0} ⊆ P_{1} prime ideals of R and Q_{1} a prime ideal of T with Q_{1} ∩ R = P_{1}. Let S_{0} = R \ P_{0}, S_{1} = T \ Q_{1}, and S = S_{0}S_{1} = {rs | r ∈ S_{0}, s ∈ S_{1}}. S is a multiplicative set containing S_{0} and S_{1}. We will show P_{0}T ∩ S =. Once we have shown this we are done, because then there exists an ideal Q_{0} of T maximal with respect to P_{0}T ⊆ Q_{0} and Q_{0} ∩ S =, which is prime, and does what we want.

So, suppose r ∈ S_{0} and s ∈ S_{1} with rs ∈ P_{0}T. Let f = x^{n} +a_{n−1}x^{n−1} +…+a_{0} be the minimal polynomial of rs over K. As rs is integral over P_{0}, the coefficients a_{0}, …, a_{n−1} are in P_{0}. Let

g(x) = x^{n} + \frac{a_{n−1}}{r} x^{n−1} + \frac{a_{n−2}}{r^{2}} x^{n−2} + … + \frac{a_{0}}{r^{n}}

then g(s) = 0 and g must be the minimal polynomial of s over K (or else we could construct a polynomial in K[x] of degree less than n satisfied by rs). Therefore b_{k} = a_{k}/r^{n−k} ∈ R for 0 ≤ k < n. Now r^{n−k}b_{k} = a_{k} ∈ P_{0} and r ∉ P_{0} implies b_{k} ∈ P_{0} for 0 ≤ k < n and s is therefore integral over P_{0}. So s ∈ \sqrt{P_{0}T} ⊆ Q_{1}, a contradiction. □

The length of a chain of ideals I_{0} ⊇ I_{1} ⊇ I_{2} ⊇ … is defined as the number of proper inclusions occurring in the chain.

4.24 Definition. Let P be a prime ideal in a ring R. The height of P is the maximal length of a chain of prime ideals descending from P: ht(P) = n ∈ ℕ_{0} if n is maximal such that there exists a chain of prime ideals P ⊃ P_{1} ⊃ … ⊃ P_{n}, and ht(P) = ∞ if there are chains of prime ideals of arbitrary length descending from P.)

4.25 Definition. The Krull dimension dim(R) of a ring R is the supremum of the heights of the prime ideals of R.

Note that a zero-dimensional commutative ring is either a field or else a ring with zero-divisors in which every prime ideal is maximal; and that an integral domain is one-dimensional if and only if every prime ideal other than (0) is maximal.

4.26 Theorem. Let R ⊆ T be commutative rings.

(1) If “incomparability” holds and P is a prime ideal of R then for every prime ideal Q of T with Q ∩ R = P we have ht(Q) ≤ ht(P).

(2) If “going up” holds and P is a prime ideal of R of finite height then there exists a prime ideal Q of R with Q ∩ R = P and ht(Q) ≥ ht(P).

4.27 Corollary. If both “incomparability” and “going up” hold for R ⊆ T (in particular, if R ⊆ T is an integral extension) then for every prime ideal P of R there exists a prime ideal Q of R with Q ∩ R = P and ht(Q) = ht(P).

Proof. Ad 1) Every chain of prime ideals descending from Q contracts to a chain of the same length descending from P, hence ht(P) ≥ ht(Q). Ad 2) “Lying over” and repeated applications of “going up” allow us to construct for every prime ideal chain descending from P a chain of prime ideals in T of the same length descending from some Q with Q ∩ R = P. □

4.28 Definition. The co-height of a prime ideal P of a commutative ring R is the maximal length of a chain of prime ideals ascending from P (in other words, co-ht(P) = dim(R/P).)

4.29 Theorem. Let R ⊆ T be commutative rings. If “going up” and “incomparability” hold (in particular, if R ⊆ T is an integral extension) then for every prime ideal P of R and prime ideal Q of T with Q∩R = P we have co-ht(Q) = co-ht(P).

Proof. Easy. □

4.30 Theorem. Let S be a multiplicative subset of a domain R. If R is integrally closed then so is R_{S}.

Proof. Let K be the quotient field of R and R_{S}, u ∈ K integral over R_{S}; to show u ∈ R_{S}.

u^{n} + \frac{a_{n−1}}{s_{n−1}} u^{n−1} + … + \frac{a_{0}}{s_{0}} = 0 a_{i} ∈ R, s_{i} ∈ S

By multiplication with s = s_{0} … s_{n−1} we get

su^{n} + r_{n−1}a_{n−1}u^{n−1} + … + r_{0}a_{0} = 0 r_{i}, a_{i} ∈ R.

Multiplication with s^{n−1} shows su to be integral over R; therefore su ∈ R, and u ∈ R_{S}. □

4.31 Theorem. Let R_{α}, α ∈ A be a collection of integral domains all contained in a field K. If every R_{α} is integrally closed (in its quotient field) then so is ⋂_{α∈A} R_{α}. The same holds for “integrally closed in K”.

Proof. Easy. □

Since every domain R satisfies R = ⋂_{P prime} R_{P} = ⋂_{Mmaximal} R_{M}, we see:

4.32 Corollary. Let R be an integral domain. Then the following are equivalent: (1) R is integrally closed.

(2) For every prime ideal P of R, R_{P} is integrally closed.

(3) For every maximal ideal M of R, R_{M} is integrally closed.

A commutative ring R is called normal if for every prime ideal P of R the localization R_{P} is an integrally closed domain. We have seen that the concepts of normal and integrally closed coincide in the case of an integral domain.

4.33 Definition. Let R be a commutative ring, T an R-algebra and u ∈ T. Then u is called almost integral over R if and only if there exists a finitely generated R-submodule B of T with R[u] ⊆ B.

Comparing the definition of “almost integral” with the preceding theorem, we see that

(i) “integral” implies “almost integral” (ii) for Noetherian R, the notions of “integral” and “almost integral” coincide.

(For Noetherian R, every R-submodule of a finitely generated R-module is finitely generated, such that “almost integral” implies condition (1) of 4.3.)

4.34 Lemma. If R is a domain with quotient field K, then u ∈ K is almost integral over R if and only if R[u] is a fractional ideal of R, i.e., if there exists a non-zero d ∈ R such that dw ∈ R for every w ∈ R[u], or equivalently, such that du^{n} ∈ R for all n ∈ ℕ:

Proof. If u is almost intgral and R[u] ⊆ B, where b_{1}, …, b_{m} generate B as an R-module then there exists a non-zero common denominator d with db_{i} ∈ R (1 ≤ i ≤ m) and hence dR[u] ⊆ dB ⊆ R. Conversely, if there exists d ∈ R, d ≠ 0, with dR[u] ⊆ R then R[u] ⊆ d^{−1}R, and d^{−1}R is finitely generated (even cyclic) as an R-module. □

4.35 Definition. A domain D with quotient field K is called completely integrally closed if every element of K that is almost integral over D lies in D.

As we have seen, for Noetherian domains the notions of “integrally closed” and “completely integrally closed” coincide.

5. Rings of Quotients and Localization

A multiplicatively closed subset of a ring is a set S ≠ ∅ such that st ∈ S whenever s, t ∈ S. A multiplicatively closed set S is saturated, if S contains all divisors of every s ∈ S. The saturation of a multiplicatively closed set S is S̄ = {t ∈ R | ∃r ∈ R rt ∈ S}.

5.1 Definition. If S is a multiplicatively closed subset of a commutative ring R and 0 ∉ S then the ring of fractions of R with denominators in S, (R_{S}, +, ·) and the canonical homomorphism f_{S}: R → R_{S} are defined as follows:

R_{S} is the set of equivalence classes on S × R under the equivalence relation

(s, r) ∼ (s′, r′) :⟺ ∃t ∈ S tsr′ = ts′r.

The equivalence class of (s, r) is denoted by \frac{r}{s}. Addition and multiplication are defined by

\frac{r}{s} + \frac{r′}{s′} = \frac{s′r + sr′}{ss′}  and  \frac{r}{s} · \frac{r′}{s′} = \frac{rr′}{ss′}

and the canonical homomorphism f_{S}: R → R_{S} by f_{S}(r) = \frac{rs}{s} (for any s ∈ S).

A lot of facts have to be checked before the above definition can be called valid, for instance that ∼ is really an equivalence relation, that addition and multiplication are well-defined (i.e. the resulting equivalence class depends only on the equivalence classes of the arguments and not on the numerator and denominator of the fractions representing them), that R_{S} satisfies the ring axioms, and that f_{S} is well-defined and a ring homomorphism. These verifications are easy, if tedious. In doing them, one inevitably encounters the following properties of R_{S}:

5.2 Remark:

(1) We do not exclude any interesting cases by stipulating 0 ∉ S. If 0 ∈ S then the construction would yield R_{S} = {0}.

(2) For all r ∈ R and s, t ∈ S \frac{r}{s} = \frac{rt}{st}.

(3) R_{S} is a commutative ring with zero \frac{0}{s} and identity \frac{s}{s} (for any s ∈ S).

(4) \frac{r}{s} = \frac{r′}{s′} iff there exists a t ∈ S with trs′ = tr′s.

(5) If S contains no zero-divisors, then \frac{r}{s} = \frac{r′}{s′} iff rs′ = r′s.

(6) \frac{r}{s} = 0 iff there exists a t ∈ S with rt = 0.

(7) If S contains no zero-divisors, then \frac{r}{s} = 0 iff r = 0.

(8) \frac{r}{s} is a unit iff there exists a t ∈ S with rt ∈ S.

(9) If S is saturated, then \frac{r}{s} is a unit iff r ∈ S.

(10) For every s ∈ S f_{S}(s) is a unit in R_{S}.

(11) Kerf_{S} = {r ∈ R | ∃t ∈ S rt = 0}.

(12) f_{S} is injective iff S contains no zero-divisor.

5.3 Lemma. Let S be a multiplicatively closed subset of R and S̄ its saturation. Then φ: R_{S} → S̄^{−1}R given by φ(\frac{r}{s}) = \frac{r}{s} is an isomorphism of rings.

Proof. φ is well-defined and a ringhomomorphism; to see injectivity, suppose \frac{r}{s} ∈ Kerφ, then rt = 0 for some t ∈ S̄. t divides some t′ ∈ S, such that t′r = 0 and therefore \frac{r}{s} = 0 in R_{S}. For surjectivity, consider \frac{r}{t} with r ∈ R and t ∈ S̄. For some s ∈ S and w ∈ R we have s = tw (and thus w ∈ S̄) and \frac{r}{t} = \frac{rw}{tw} = \frac{rw}{s} = φ(\frac{rw}{s}). □

5.4 Proposition. All rings of quotients of R are of the form form R_{S} with S = R \ ⋃_{P ∈P} P for a set P of prime ideals of R.

Proof. By the previous lemma, every ring of quotients is of the form R_{S} with S saturated. By Lemma 10.3, every saturated multiplicative set is the complement of a union of prime ideals and vice versa. □

The importance of the construction lies in the relationship between ideals of R and ideals of R_{S}. We already know that for every ideal I of R, f_{S}(I)R_{S} is an ideal of R_{S} - the extension of I into R_{S}. We denote this ideal by I_{S}.

5.5 Definition. Names and notation for examples of rings of fractions:

• If S is the set of all non-zero-divisors of R, then R_{S} is called the total ring of fractions of R; a special case of this is the quotient field of an integral domain, which is R_{S} with S = R \ {0}.

• If S = R \ P for a prime ideal P of R, then R_{S} is denoted R_{P} and called the localization of R at P. The ideal I_{S} of R_{P} is denoted I_{P} or IR_{P}.

5.6 Remark: The apparent ambiguity of notation in the preceding definition – R_{S} the ring of fractions with denominators in S, and R_{P} the ring of fractions with denominators in the complement R \ P – we resolve by the following convention: if 0 ∉ S then R_{S} denotes the ring of fractions of R with denominators in S, but if 0 ∈ S, e.g., if S is a union of prime ideals, then R_{S} denotes the ring of fractions of R with denominators in R \ S.

5.7 Lemma. Let I be an ideal of R and I_{S} = f_{S}(I)R_{S} the extension of I to I_{S}. Then I_{S} consists of precisely those elements of R_{S} that admit a representation as a fraction with numerator in I:

I_{S} = \{\frac{i}{s} ∈ R_{S} | i ∈ I\}.

Proof. (⊆) is obvious. For (⊇), let i ∈ I, s ∈ S. Then

\frac{i}{s} = \frac{is^{2}}{s^{3}} = \frac{is}{s} \frac{s}{s^{2}} = f_{S}(i)\frac{s}{s^{2}} ∈ f_{S}(I)R_{S}.

□

Note, however, that \frac{r}{s} ∈ I_{S} does not imply r ∈ I, except in special cases such as:

5.8 Lemma. If P is a primary ideal of R with P ∩ S = ∅ then \frac{r}{s} ∈ P_{S} implies r ∈ P.

Proof. Let \frac{r}{s} ∈ P_{S} with P primary. By 5.7, there exist p ∈ P and s′ ∈ S with \frac{r}{s} = \frac{p}{s′}, and ts′r = tsp for some t ∈ S. Since ts′ ∈ S, no power of ts′ is in P. Therefore r ∈ P. □

Likewise, for every ideal J of R_{S}, f_{S}^{−1}(J) is an ideal of R - the contraction of J to R. (If f_{S} is injective then f_{S}^{−1}(J) = J ∩R.) Also, if J is a prime [or primary] ideal then so is f_{S}^{−1}(J).

5.9 Lemma. Let J be an ideal of R_{S} and I and ideal of R.

The inverse image of J under f_{S} consists precisely of all numerators that occur in fraction representations of elements of J:

f_{S}^{−1}(J) = \{r ∈ R | ∃s ∈ S \frac{r}{s} ∈ J\}.

Proof. (⊆) Let r ∈ R with f_{S}(r) ∈ J. Then \frac{s}{s^{2}} f_{S}(r) = \frac{s}{s^{2}} \frac{sr}{s} = \frac{r}{s} ∈ J. For (⊇), consider r ∈ R with \frac{r}{s} ∈ J. Then f_{S}(r) = \frac{sr}{s} = \frac{r}{s} \frac{s^{2}}{s} ∈ J. □

5.10 Theorem.

(a) For every ideal J of R_{S}, (f_{S}^{−1}(J))_{S} = J.

(b) Every ideal of R_{S} is of the form I_{S} = \{\frac{i}{s} ∈ R_{S} | i ∈ I\} for some ideal I of R.

(c) I_{S} = R_{S} iff I ∩ S ≠ ∅.

(d) A bijection between those prime ideals P of R with S ∩P = ∅ and all prime ideals of R_{S} is given by P ↦ P_{S} (for P a prime ideal of R); its inverse is Q ↦ f_{S}^{−1}(Q) (for Q a prime ideal of R_{S}).

(e) Like (d), with “prime” replaced by “primary”.

5.11 Corollary. In terms of extension and contraction by f_{S}: R → R_{S}, we have J^{ce} = J for every ideal J of R_{S} and P^{ec} = P for every primary ideal P of R with P ∩ S = ∅.

5.12 Corollary. Let P be a prime ideal of R. Then every ideal of R_{P} is of the form I_{P} = \{\frac{i}{s} ∈ R_{P} | i ∈ I\} for some ideal I of R with I ⊆ P. Via Q ↦ Q_{P} the prime ideals of R_{P} correspond bijectively to the prime ideals Q of R with Q ⊆ P. In particular, R_{P} is a local ring with maximal ideal P_{P}.

5.13 Definition. A ring containing exactly one maximal ideal is called a local ring. (Note, however, that many authors call such rings quasi-local, reserving the name local for Noetherian rings with a unique maximal ideal.)

5.14 Exercise. A ring is local if and only if the set of non-units forms an ideal.

5.15 Theorem (localization commutes with quotients). Let S be a multiplicative subset of R, I an ideal of R, and π: R → R/I the canonical projection then

(R/I)_{((S+I)/I)} ≃ R_{S}/I_{S}.

5.16 Easy exercise. The following is a well-defined ring-isomorphism:

\frac{(r + I)}{(s + I)} ↦ \frac{r}{s} + I_{S}

5.17 Corollary. For every prime ideal P of R

(R/P)_{(0)} ≃ R_{P} /P_{P} .

5.18 Definition. If P is a prime ideal of R, we have seen that the quotient field of the integral domain R/P is the same as the residue field of the maximal ideal P_{P} of R_{P}. This field is called the residue field of P.

∗ ∗ ∗ Symbolic powers ∗ ∗ ∗

5.19 Definition. Let P be a prime ideal of R. The n-th symbolic power of P is defined by P^{(n)} = R ∩ P^{n}R_{P}.

The n-th symbolic power of P is the contraction back to R of the extension of P^{n} into R_{P} and thus contains P^{n}.

5.20 Proposition. Let P be a prime ideal of R and n ∈ ℕ.

(1) P^{(n)} = (P_{P})^{n} ∩ R

(2) P^{(n)} is P -primary.

(3) If Q is a primary ideal then P^{n} ⊆ Q ⟹ P^{(n)} ⊆ Q.

(4) P^{(n)} = P^{n} ⟺ P^{n} is primary.

(5) If M is a maximal ideal of R then M^{(n)} = M^{n} for all n ∈ ℕ. Proof. (2) Since P_{P} is a maximal ideal of R_{P}, (P_{P})^{n} is P_{P} -primary. By 10.18, P^{(n)} is P -primary. (4) follows from (2) and (3); (5) follows from (4). □

6. Valuation rings, Prüfer rings, Dedekind domains.

6.1 Definition. A commutative ring R is called a valuation ring if for every two elements a, b ∈ R either a | b or b | a holds.

The definition of valuation ring is equivalent to saying that the principal ideals of R are totally ordered with respect to inclusion. In fact, more is true.

6.2 Lemma. Let R be a valuation ring. Then the ideals of R are totally ordered with respect to inclusion. In particular, R is a local ring.

Proof. Let I, J be ideals of R and suppose I ⊈ J. Let i ∈ I \ J. Then for all j ∈ J, j |/i, and therefore i | j. We have shown J ⊆ iR ⊆ I. □

Although the above definition of valuation ring is phrased for commutative rings in general, we will only be interested in valuation rings that are integral domains.

6.3 Remark: Let V be an integral domain contained in a field K. It is easy to see that V is a valuation ring with quotient field K if and only if for every u ∈ K either u ∈ V or u^{−1} ∈ V .

6.4 Remark: If V is a valuation ring with quotient field K and R a ring with V ⊆ R ⊆ K then R is also a valuation ring. This is an easy consequence of the preceeding remark.

The unique maximal ideal of a valuation domain V then consists of all elements u in the quotient field K with u^{−1} ∉ V . Moreover, if M is this unique maximal ideal of V then for every u ∈ K exactly one of the alternatives u ∈ M or u^{−1} ∈ V holds.

∗ ∗ ∗ Valuations ∗ ∗ ∗

6.5 Definition. Let (G, +) be an Abelian group and ≤ an order relation compatible with the group operation (i.e., g ≤ g′ and h ≤ h′ implies g + h ≤ g′ + h′). Then (G, +, ≤) is called an ordered group. If, moreover, ≤ is a total ordering of G (i.e., for all g, h ∈ G, either g ≤ h or h ≤ g holds) then (G, +, ≤) is called a totally ordered group.

Notation: we will use g < h for “g ≤ h and g ≠ h” and g ≥ h for h ≤ g.

6.6 Definition. Let K be a field and (G, +) a totally ordered group. v: K^{∗} → G (where K^{∗} = K \ {0}) is called a valuation on K if

(1) v(ab) = v(a) + v(b) and

(2) v(a + b) ≥ min(v(a), v(b)).

The subgroup Γ_{v} = Im(v) of G is called the valuation group of v. It is customary to extend the definition of v to all of K by adding an infinity element to G and setting v(0) = ∞. With the conventions ∞ + ∞ = ∞, g + ∞ = ∞ and g ≤ ∞ for all g ∈ G, (1) and (2) still hold and

(3) v(a) = ∞ ⟺ a = 0

Some easy consequences of the definition of a valuation v:

(i) v(1) = 0 (ii) v(a^{−1}) = −v(a) (iii) v(−a) = v(a) (iv) v(a − b) ≥ min(v(a), v(b)) (iv) v(a) ≠ v(b) ⟹ v(a + b) = min(v(a), v(b)). If v is a valuation on K then R_{v} = {k ∈ K | v(k) ≥ 0} is a local ring with

maximal ideal M_{v} = {k ∈ K | v(k) > 0}. (Clearly, the units of R_{v} are exactly those field elements with both k and k^{−1} in R_{v}, which are the elements of R_{v} with v(k) = 0.) R_{v} is a valuation ring, since for all a, b ∈ R_{v} either v(a) ≤ v(b) or v(b) ≤ v(a) and v(a) ≤ v(b) means b/a ∈ R_{v} and therefore a | b in R_{v}. We will show that every valuation domain arises from a valuation in this way.

6.7 Lemma. Let R be a valuation domain with quotient field K then the R-submodules of K are totally ordered by inclusion

Proof. Let I, J ⊆ K be R-modules and I ⊈ J. Let i ∈ I \ J. Then for all j ≠ 0 in J either i/j ∈ R or j/i ∈ R, but i/j ∈ R is impossible, since it would imply i = j(i/j) ∈ J. Therefore for all j ≠ 0 in J we have j/i ∈ R, which implies j = i(j/i) ∈ iR ⊆ I. Thus I ⊈ J implies J ⊆ iR ⊆ I. □

6.8 Proposition. Let R be a valuation domain with quotient field K. Then G = {zR | z ∈ K, z ≠ 0} is totally ordered by

yR ≤ zR ⟺ yR ⊇ zR

and

yR ⋆ zR := yzR

defines an Abelian group structure on G compatible with the ordering. Moreover,

v: K^{∗} → G v(z) = zR

is a valuation on K with R = {k ∈ K | v(k) ≥ 0}.

Proof. Easy. □

∗ ∗ ∗ Valuation rings and integral closure ∗ ∗ ∗

6.9 Definition. A ring is called Bézout ring if every finitely generated ideal is principal.

6.10 Remark: It is clear that every valuation ring is a Bézout ring. In particular, every valuation ring is a GCD-ring. Therefore every valuation domain is integrally closed. (This is of course easy to show directly.)

We have seen that valuation domains are integrally closed. It follows that arbitrary intersections of valuation domains contained in a field are integrally closed. We shall see that conversely, every integral closure of a domain is an intersection of valuation domains.

Before showing this, however, we note in passing that valuation rings are characterized among local rings by the Bézout ring property.

6.11 Lemma. A local ring is a valuation ring if and only if it is a Bézout ring.

Proof. Every valuation ring is a Bézout ring. Conversely, let R be a local Bézout ring and a, b ∈ R. We must show a | b or b | a. By dividing a and b by their greatest common divisor, we may assume gcd(a, b) = 1 and therefore Ra+Rb = R. This shows that a and b are not both in the unique maximal ideal of R, hence one of them is a unit and therefore divides the other. □

6.12 Lemma. Let R ⊆ T be commutative rings, I an ideal of R and u ∈ T a unit. If IR ≠ R then IA ≠ A for either A = R[u] or for A = R[u^{−1}].

Proof. Suppose otherwise. Then there exist a_{i} and b_{i} ∈ I such that

(∗)  1 = a_{0} + a_{1}u + … + a_{n}u^{n}
(∗∗)  1 = b_{0} + b_{1}u^{−1} + … + b_{m}u^{−m}

where n, m ∈ ℕ are minimal and w.l.o.g. n ≥ m.

Multiplication of (∗∗) with u^{n} gives

(1 − b_{0})u^{n} = −b_{1}u^{n−1} − … − b_{m}u^{n−m}.

By multiplying (∗) with (1 − b_{0}) and substituting for (1 − b_{0})u^{n} from the above formula we can express 1 as an I-linear combination of powers u^{k} with 0≤k≤n−1, contradicting the minimality of n. □

6.13 Lemma. Let R ⊆ K be an integral domain contained in a field and I an ideal of R. If I ≠ R then there exists a valuation domain V with quotient field K such that R ⊆ V and IV ≠ V .

Proof. Let S be the set of pairs (R_{α}, I_{α}) where R_{α} is a ring with ideal I_{α} ≠ R_{α} and R ⊆ R_{α} as well as I ⊆ I_{α}. We order S by (R_{α}, I_{α}) ≤ (R_{β}, I_{β}) if and only if both R_{α} ⊆ R_{β} and I_{α} ⊆ I_{β}. By Zorn’s Lemma, S has a maximal element (V, J). We know IV ≠ V , since I ⊆ J. Now let u be an element of K, we must show u ∈ V or u^{−1} ∈ V . Suppose not. Since by a previous lemma the extension of J is a proper ideal of either V [u] or V [u^{−1}], either (V [u], JV [u]) or (V [u^{−1}], JV [u^{−1}]) is an element of S strictly greater than (V, J), a contradiction. □

6.14 Theorem. Let R be an integral domain with quotient field K and R′ the integral closure of R. Then R′ = ⋂_{R⊆V ⊆K} V , where V ranges over all valuation domains between R and K.

Proof. ⋂_{R⊆V ⊆K} V contains R, is integrally closed (as an intersection of integrally closed domains) and therefore contains R′. Conversely, let z ∈ ⋂_{R⊆V ⊆K} V ; we must show z is integral over R. Suppose not. Let z = u^{−1}. Then u^{−1} ∉ R′ implies u^{−1} ∉ R[u]. Since u is not a unit in R[u], uR[u] ≠ R[u] and there exists a valuation domain V containing R[u] with uV ≠ V . This contradicts z = u^{−1} ∈ V . □

∗ ∗ ∗ Invertible ideals and Prüfer rings ∗ ∗ ∗

6.15 Definition. Let R be an integral domain with quotient field K and I an R-submodule of K.

• I is called a fractional ideal of R if there exists a non-zero r ∈ R such that rI ⊆ R.

• The inverse of I is defined as I^{−1} = {z ∈ K | zI ⊆ R}.

• I is called invertible if I^{−1}I = R. (Note that I^{−1}I ⊆ R is automatic.)

In the context of invertible ideals, fractional ideals are usually just called ideals, “fractional” being understood. No confusion should result.

6.16 Remark:

1) Every finitely generated R-submodule of K is a fractional ideal. (We can multiply I by a common denominator of the generators.)

2) Every non-zero principal fractional ideal is invertible. (The inverse of Ra is Ra^{−1} and RaRa^{−1} = R.)

6.17 Proposition. Let I be an R-submodule of K. If I is invertible then I is finitely generated.

Proof. I^{−1}I = R, so there exist a_{i} ∈ I^{−1} and b_{i} ∈ I with ∑_{i−1}^{n} a_{i}b_{i} = 1. We claim the b_{i} generate I. Let c ∈ I. c = 1c = ∑_{i−1}^{n} a_{i}cb_{i} =. Now a_{i} ∈ I^{−1} implies a_{i}c ∈ R and we have c = ∑_{i−1}^{n} r_{i}b_{i} with r_{i} ∈ R. □

6.18 Definition. A commutative ring R with only finitely many maximal ideals is called semi-local.

6.19 Proposition. Let R be a semi-local domain. Then every invertible ideal is principal.

Proof. Let M_{1}, …, M_{n} be the maximal ideals of R. By II^{−1} = R, there exist a_{i} ∈ I and b_{i} ∈ I^{−1} such that a_{i}b_{i} ∉ M_{i}. By the Chinese Remainder Theorem, there exists u_{i} ∈ R with u_{i} ∉ M_{i} and u_{i} ∈ ⋂_{j≠i} M_{j}. Let v = ∑_{i=1}^{n} u_{i}b_{i}. Then vI ⊆ R is an ideal of R and not contained in any M_{i}, since va_{i} ∉ M_{i}. Therefore vI = R. This shows I = v^{−1}R. I is principal. □

To be able to consider localizations of fractional ideals we need to extend the construction of rings of fractions to modules:

6.20 Definition. Let R be a commutative ring, S ⊆ R a multiplicative set and M an R-module. The R_{S} -module M_{S} is defined as follows: Its elements are the equivalence classes on S × M w.r.t. the equivalence relation

(s, m) ∼ (s′, m′) :⟺ ∃t ∈ S t(sm′ − s′m) = 0.

The equivalence class of (s, m) is denoted by \frac{m}{s}. Addition and scalar multiplication are defined by:

\frac{m}{s} + \frac{m′}{s′} = \frac{s′m + sm′}{ss′}  and  \frac{r}{s} · \frac{m}{s′} = \frac{rm}{ss′}

If S = R \ P for a prime ideal P of R then M_{S} is denoted M_{P} and called the localization of M at P.

6.21 Remark:

(i) The zero-element of M_{S} is \frac{0}{s} for any s ∈ S, and \frac{m}{s} = 0 iff there exists a t ∈ S with tm = 0.

(ii) The map f: M → M_{S} given by f(m) = \frac{sm}{s} (or f(m) = \frac{m}{1} , if 1 ∈ S) is an R-module homomorphism with Kerf = {m ∈ M | ∃s ∈ S sm = 0}.

6.22 Lemma. If f: M → N is an R-module homomorphism, then

f_{S}: M_{S} → N_{S} f_{S}(\frac{m}{s}) := \frac{f(m)}{s}

is an R_{S} -module homomorphism. Also, (i) S^{−1}id_{M} = id_{M_{S}} (ii) f_{S} ∘ g_{S} = S^{−1}(f ∘ g) (iii) Kerf_{S} = S^{−1}Kerf; Imf_{S} = S^{−1}Imf

(Furthermore, S^{−1} is an exact functor, i.e. it maps exact sequences of Rmodules to exact sequences of R_{S} -modules.)

6.23 Proposition. Let I be an invertible ideal of a domain R, S ⊆ R a multiplicative subset of R. Then I_{S} is an invertible ideal of R_{S}.

Proof. Easy. □

6.24 Theorem. Let I be a finitely generated ideal of an integral domain R. Then I is invertible if and only if I_{M} is principal for every maximal ideal M of R.

Proof. If I is invertible, we have seen that I_{M} is invertible and hence principal for every maximal ideal M of R. Conversely, assume every I_{M} is principal. Suppose I is not invertible then II^{−1} ⊆ M for some maximal ideal M of R. I_{M} is principal, and we may choose a generator in I: I_{M} = iR_{M}, with i ∈ I. Let I = Ri_{1}+…+Ri_{n}. Then there exist r_{k} ∈ R and s_{k} ∈ R \ M with i_{k} = \frac{r_{k}}{s_{k}} i. Set s = s_{1} … s_{n}. Then si^{−1}i_{k} ∈ R for every generator i_{k} of I, and therefore si^{−1} ∈ I^{−1}. This implies s = si^{−1}i ∈ I^{−1}I ⊆ M, a contradiction to s ∈ R \ M. □

6.25 Theorem. Let R be an integral domain. Then the following are equivalent. (1) Every finitely generated non-zero ideal of R is invertible.

(2) R_{P} is a valuation domain for every prime ideal P of R.

(3) R_{M} is a valuation domain for every maximal ideal M of R.

6.26 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceding theorem is called Prüfer domain.

Proof. (1 ⇒ 2) We show that R_{P} is Bézout. Let \frac{r_{1}}{s_{1}} , …, \frac{r_{n}}{s_{n}} ∈ R_{P}, r_{i} ∈ R, s_{i} ∈ R \ P. Since R is Prüfer, I = r_{1}R + … r_{n}R is invertible, and therefore I_{P} = \frac{r_{1}}{s_{1}} R_{P} +…+ \frac{r_{n}}{s_{n}} R_{P} is invertible. Since R_{P} is local, invertible implies principal. (2 ⇒ 3) Obvious. (3 ⇒ 1) If I is a finitely generated ideal of R then I_{M} is principal (as a finitely generated ideal in the Bézout domain R_{M}) for all maximal ideals M of R. Therefore I is invertible. □

6.27 Theorem. Let R be a Prüfer domain with quotient field K. If V is a valuation ring with R ⊆ V ⊆ K then V = R_{P} for some prime ideal P of R.

Proof. Let M be the maximal ideal of V and P = M ∩ R. For any s ∈ R \ P, s ∉ M implies s^{−1} ∈ V . Therefore R_{P} ⊆ V .

Suppose V ⊈ R_{P}. Let v ∈ V \ R_{P}. Since R_{P} is a valuation domain, v^{−1} ∈ R_{P}. This means v = \frac{s}{r} with r ∈ R, s ∈ R \ P. We must have r ∈ P, since otherwise v ∈ R_{P}. Now s = rv ∈ P V ⊆ M, a contradiction. □

∗ ∗ ∗ Discrete valuation rings and Dedekind domains ∗ ∗ ∗

6.28 Theorem. Let R be a local integral domain with maximal ideal M, and not a field. The following are equivalent:

(1) R is Noetherian and integrally closed and dim R = 1.

(2) R is Noetherian and M is principal.

(3) R is a unique factorization domain with just one (up to units) irreducible element.

(4) R is a principal ideal domain.

Proof. (1 ⇒ 2) Let m ∈ M with m ≠ 0 and mR ≠ M and consider the ring T =R/mR. mR not being prime, T is not a domain. The set Z(T) of zero-divisors of T is a union of prime ideals by 10.4, therefore Z(T) equals the unique prime ideal of T: Z(T) = M/mR. By 8.7 (universal zero-divisor) applied to the ideal M/mR of T, there exists a ∈ R \ mR such that aM ⊆ mR.

Now \frac{a}{m} ∈ M^{−1} \ R and \frac{a}{m}M is an ideal of R. Suppose \frac{a}{m}M ≠ R. Then \frac{a}{m}M ⊆ M and therfore R[\frac{a}{m}]M ⊆ M. Now M is a faithful R[\frac{a}{m}]-module and finitely generated as an R-module, which means \frac{a}{m} is integral over R and therefore in R; a contradiction. Therefore \frac{a}{m}M = R and M = \frac{m}{a}R is principal. (2 ⇒ 3) The ascending chain condition implies that every non-zero element is a product of irreducible elements (the ascending chain condition for principal ideals suffices for that). Also, every irreducible element is prime, and there is (up to multiplication by units) only one irreducible element. (Since the unique maximal ideal M of R is principal, it is the only principal ideal maximal among proper principal ideals, and therefore the only ideal generated by an irreducible ideal. So there is up to multiplication by units only one irreducible element and it generates M. Now M is also a prime ideal, wherfore the unique (up to units) irreducible element is also prime). Together this implies a unique factorization domain with just one irreducible element. (3 ⇒ 4) and (4 ⇒ 1) are easy. □

6.29 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceeding theorem is called discrete valuation domain (DVR), or, more precisely discrete valuation domain of rank 1.

6.30 Proposition. A domain is a DVR if and only if it is the valuation ring of a valuation with value group isomorphic to (ℤ, +).

Proof. Exercise. □

6.31 Theorem. Let R be an integral domain. Then the following are equivalent (1) Every non-zero ideal of R is invertible.

(2) R is Noetherian and integrally closed and dim R ≤ 1.

(3) R is either a field or a Noetherian domain s.t. for every maximal ideal M of R, R_{M} is a DVR.

6.32 Definition. A domain satisfying one (and hence all) of the equivalent conditions in the preceding theorem is called a Dedekind ring.

Proof. (1 ⇒ 3) Invertible implies finitely generated, so R is Noetherian. Let M be a maximal ideal of R. Every ideal of R_{M} is of the form I_{M} for an invertible ideal I of R and hence is principal. R_{M} is a local principal ideal domain and therefore a DVR. (3 ⇒ 1) For every non-zero ideal I of R, every localization I_{M} at a maximal ideal is principal, since R_{M} is a principal ideal domain. (3 ⇒ 2) R is integrally closed as the intersection of its localizations R_{M}, which are integrally closed as valuations rings. The Krull dimension of R is the supremum of dim R_{M}, where M ranges through the maximal ideals of R and therefore dim R ≤ 1. (2 ⇒ 3) If dim R ≤ 1 and R is not a field, then dim(R) = 1. For every maximal ideal M of R, the localization R_{M} is local, Noetherian, integrally closed and one-dimensional and therefore a DVR by the previous Theorem. □

6.33 Theorem. Let D be a domain. If every non-zero ideal of D is invertible, then every non-zero ideal is expressible as a product of prime ideals.

Proof. Invertible implies finitely generated, so D is Noetherian. Suppose there exist non-zero ideals not expressible as products of prime ideals. The set of such ideals has a maximal element, I. This I is not R (since R is an empty product of prime ideals); nor is I itself prime, so there exists a prime ideal P ⊃ I properly containing I. We have I ⊆ IP^{−1} ⊆ R. If IP^{−1} = I, then (since P is invertible) multiplication by P gives I = IP and by Nakayama’s Lemma, there is an element a ∈ R with aI = 0 and a ≡ 1 mod P, which is impossible in a domain. Therefore IP^{−1} properly contains I and, by maximality of I, is expressible as a product of prime ideals IP^{−1} = P_{1} … P_{m}. Using P P^{−1} = R again, we get I = P_{1} … P_{m}P, a contradiction. □

6.34 Corollary. In a Dedekind ring, every non-zero ideal is uniquely expressible as a product of prime ideals.

Proof. By 6.31, every non-zero ideal in a Dedekind ring is invertible, so 6.33 implies existence of a factorization. If P_{1} … P_{n} = Q_{1} … Q_{m} then P_{1}, being prime, contains one of the Q_{i}, say Q_{1} ⊆ P_{1}. Since D is at most 1-dimensional, it follows that Q_{1} = P_{1} and we can cancel invertible ideals and get P_{2} … P_{n} = Q_{2} … Q_{m}. Inductively we see that the factorizations are the same (up to order). □

6.35 Remark: Prime factorization of ideals characterizes Dedekind rings, even without the assumption of uniqueness: a domain in which every ideal is a product of prime ideals is a Dedekind domain.

∗ ∗ ∗ rank 1 valuation rings ∗ ∗ ∗

6.36 Definition. An ordered group (Γ, +) satisfies the Archimedean axiom if for any a, b ∈ Γ with a > 0 there exists n ∈ ℕ with na ≥ b.

6.37 Proposition. An ordered group satisfies the Archimedean axiom if and only if it is order-isomorphic to a subgroup of (ℝ, +).

Proof. Every subgroup of (ℝ, +) satisfies the Archimedean axiom. Now let (Γ, +) be an ordered group satisfying the Archimedean axiom and fix an element a > 0 of Γ. We will construct an order-preserving injective map f: Γ → ℝ by defining a Dedekind cut for every b ∈ Γ. First consider only elements b > 0 and define

S_{b} = \{\frac{m}{n} ∈ ℚ | m ∈ ℤ, n ∈ ℕ, ma ≤ nb\}.

Note that \frac{m}{n} ∈ S_{b} and \frac{m̃}{ñ} = \frac{m}{n} (with n, ñ > 0) implies that m̃a ≤ ñb. Also, by the Archimedean axiom, S_{b} ≠ ℚ and S_{b} ≠ ∅.

If \frac{m}{n} ∈ S_{b} then every \frac{m′}{n′} ≤ \frac{m}{n} in ℚ is also in S_{b}. (By choosing a common denominator, we may assume n′ = n, which implies m′ ≤ m, and then m′a ≤ ma ≤ nb = n′b.) S_{b} thus defines a unique real number sup S_{b} and we set f(b) = sup S_{b}. We extend the definition of f to all of Γ by seting f(0) = 0 and f(b) = −f(−b) for b < 0. It is then not hard to check that f is an order-preserving group homomorphism and injective. □

6.38 Proposition. Let v be a valuation on a field K with valuation ring R_{v} and valuation group Γ_{v} ≠ 0. Then Γ_{v} satisfies the Archimedean axiom if and only if dim R_{v} = 1.

Proof. Assume the Archimedean axiom. Γ_{v} ≠ 0 implies that R_{v} is not a field. Suppose dim R > 1. Then there exists a prime ideal P with (0) ⊂ P ⊂ M_{v}. Let a ∈ M_{v} \P and b ∈ P \(0) and set α = v(a), β = v(b). Since α > 0 there exists n ∈ ℕ with nα ≥ β. This means v(a^{n}) ≥ v(b), i.e., \frac{a^{n}}{b} ∈ R_{v} and therfore a^{n} ∈ bR_{v} ⊆ P. We conclude that a ∈ P, a contradiction.

Conversely, assume dim R = 1. R is a one-dimensional local ring, so M_{v} is its only non-zero prime ideal and therefore the radical of every proper non-zero ideal. In particular, for every non-zero a ∈ M_{v} and any b ∈ R_{v} there exists n ∈ ℕ such that a^{n} ∈ bR_{v}, which implies that for every α > 0 in Γ and every β ≥ 0 in Γ there exists n ∈ ℕ with nα = γ + β for some γ ≥ 0, or in other words, nα ≥ β. □

∗ ∗ ∗ Quick and dirty Dedekind ring properties ∗ ∗ ∗

6.39 Lemma. In a Noetherian ring, every non-zero ideal contains a product of non-zero prime ideals.

Proof. Suppose M maximal among the counterexamples. Then M is not prime, so there exist r, s with rs ∈ M but r ∉M and s ∉ M. Since M is striclty contained in M + rR and M + sR, these ideals each contain a product of prime ideals, and so does (M + rR)(M + sR), which is contained in M, a contradiction. □

6.40 Lemma. Let R be a Noetherian ring with dim R = 1 and quotient field K, and (0) ⫋ I ⫋ R a proper ideal of R. Then there exists γ ∈ K \ R with γI ⊆ R.

Proof. Let a ≠ 0 in I then there exist non-zero prime ideals P, P_{1}, …, P_{r} with P ⊇ I ⊇ aR ⊇ P_{1} · . . . · P_{r}, where r is minimal with the property that aR contains a product of r prime ideals. P being prime, it contains one of the P_{i}, say P_{1} ⊆ P and therefore P_{1} = P. So we have

P ⊇ I ⊇ aR ⊇ P · P_{2} · . . . · P_{r}.

Pick b ∈ P_{2} · . . . · P_{r} \ aR (possible by minimality of r) and set γ = b/a. Then γ ∉ R, and for every i ∈ I, iγ = (ib)/a is in R, because ib ∈ P · P_{2} · . . . · P_{r} ⊆ aR. □ 6.41 Theorem. In a Dedekind ring, every non-zero ideal is invertible.

Proof. Let I ≠ (0) and suppose I^{−1}I ⫋ R. Let γ ∈ K \ R with γI^{−1}I ⊆ R, then γI^{−1} ⊆ I^{−1}. This makes I^{−1} a faithful R[γ]-module which is finitely generated as R-module, showing γ ∈ K \ R to be integral over R - a contradiction to R being integrally closed. □

6.42 Corollary. In a Dedekind ring, every non-zero ideal is a product of prime ideals, and this prime factorization is unique.

Proof. Existence first: R is an empty product of prime ideals. Let I ≠ R be a non-zero ideal of R then I ⊆ P_{0} for some prime ideal P_{0}. If I = P_{0} we are done. If I ⊂ P_{0}, then, P_{0} being invertible, I = (IP_{0}^{−1})P_{0} and we can iterate the process of finding a prime factor with I_{1} = (IP_{0}^{−1}). This process terminates with I_{n} = IP_{0}^{−1} … P_{n−1}^{−1} = P_{n} a prime ideal (and therefore I = P_{0}^{−1} … P_{n}) because otherwise we would have an infinte ascending chain of ideals I ⊂ I_{1} ⊂ I_{2} ⊂ …, where I_{k} = IP_{0}^{−1} … P_{k−1}.

Uniqueness: Suppose P_{1} … P_{n} = Q_{1} … Q_{m}, all P_{i} and Q_{i} prime. Induction on min(n, m). If P_{1} … P_{n} = Q_{1} then, Q_{1} being prime, there exists an index i with P_{i} ⊆ Q_{1}, w.l.o.g. P_{1} ⊆ Q_{1}. As P_{1} is maximal, we have P_{1} = Q_{1}. We see that n = 1, because multiplication with P_{1}^{−1} gives P_{2} … P_{n} = R, in particular R ⊆ P_{i} for i ≥ 2, therefore prime ideals with indices i ≥ 2 do not exist.

If min(n, m) > 1 and P_{1} … P_{n} = Q_{1} … Q_{m}, then again P_{1} … P_{n} ⊆ Q_{1} implies that for some i P_{i} = Q_{1}, w.l.o.g. P_{1} = Q_{1}, and we may cancel P_{1} = Q_{1} by multiplying both sides with Q_{1}^{−1}, and get P_{2} … P_{n} = Q_{2} … Q_{m}. By induction hypothesis n = m and P_{2}, …, P_{n} and Q_{2}, …, Q_{n} are (up to order) the same list of prime ideals. □

7. Noetherian Rings.

7.1 Proposition. Let R be a commutative ring. Then the following are equivalent: (1) Every ideal of R is finitely generated.

(2) Every ascending chain of ideals of R is of finite length.

(3) Every non-empty set of ideals of R has a maximal element.

Proof. (1 ⇒ 2) and (2 ⇒ 3) are easy. (3 ⇒ 1) Let I be an ideal of R and S the set of finitely generated ideals contained in I. S (being non-empty because it contains (0)) has a maximal element J. Suppose J ≠ I and pick i ∈ I \ J. Then iR + J ⊆ I is finitely generated and strictly contains J, a contradiction. □

7.2 Definition. A commutative ring satisfying one and hence all of the equivalent conditions of the preceeding theorem is called Noetherian.

Condition (2) above is called “the ascending chain condition” (ACC) (for ideals). Similar conditions for objects other than ideals, or for special kinds of ideals, often play a rôle. For instance, a generalization of Noetherian domains is given by requiring the ascending chain condition only for divisorial ideals; the resulting domains are called Mori domains.

7.3 Definition. Let R be a commutative ring, M an R-module. M is called Noetherian if every R submodule of M is finitely generated.

Since the ideals of a ring R are precisely the R-submodules of R, a ring R is Noetherian (as a ring) if and only if it is a Noetherian R-module.

7.4 Remark: As in Proposition 7.1, the following are equivalent:

(1) M is a Noetherian R-module

(2) Every ascending chain of R-submodules of M is of finite length.

(3) Every non-empty set of R-submodules of M contains a maximal element.

7.5 Lemma. Let M be am R-module and L a submodule of M. Then M is Noetherian if and only if both L and M/L are Noetherian.

Proof. If M is Noetherian, then L is Noetherian, since every submodule of L is s submodule of M. Also, M/L is Noetherian, since every submodule of M/L is of the form π(N), where N is a submodule of M and π: M → M/L the canonical projection, and π(N) is generated by the images under π of a set of generators of N.

Conversely, suppose L and M/L are Noetherian. Given a submodule N of M, let a n_{1}, … n_{s} be generators of N ∩ L and m_{1} + L, …, m_{t} + L generators of (N + L)/L with m_{1}, …, m_{t} ∈ N. Then N is generated by n_{1}, … n_{s}, m_{1}, …, m_{t}. □

7.6 Proposition. For every Noetherian ring R, every finitely generated R-module is Noetherian.

Proof. We only need to show that every free R-module of finite rank is Noetherian, since every R-module generated by n elements is a homomorphic image of the free R-module of rank n.

Induction on the rank of M: For n = 1, M = R, and R is Noetherian. For rank(M) = n > 1, M = R ⊕ L, L a free R-module of rank n − 1. L is Noetherian by induction hypothesis, M/L = R is Noetherian, and therefore M is Noetherian by the previous Lemma. □

7.7 Theorem (Hilbert’s basis theorem). Let R be a Noetherian ring. Then R[x] is Noetherian.

Proof. Let J ≠ (0) be an ideal of R[x]. For n ∈ ℕ, let I_{n} be the set of leading coefficients of polynomials of degree n or less in J and set I = ⋃_{n∈ℕ} I_{n}. The I_{n} are ideals of R with I_{0} ⊆ I_{1} ⊆ I_{2} ⊆ … and therfore I is an ideal of R too. The fact that both I and all I_{n} are finitely generated now implies that J is finitely generated: let g_{1}, …, g_{m} ∈ J be polynomials whose leading coefficients generate I and let N = max_{1≤k≤m} deg g_{k}. To g_{1}, …, g_{m} we add for each n < N a finite set of elements of J of degree n or less whose leading coefficients generate I_{n}. That the resulting finite set of polynomials generates J is now easily seen by induction on the degree of an element of J. □

7.8 Corollary. Let R be a Noetherian ring. Then R[x_{1}, …, x_{n}] and therefore every finitely generated R-algebra (as a homomorphic image of a polynomial ring R[x_{1}, …, x_{n}]) is Noetherian.

7.9 Lemma. Let R be a commutative ring and P an ideal of R that is not finitely generated and which is maximal with respect to this property. Then P is prime.

Proof. Suppose ab ∈ P and a ∉ P, b ∉ P. By maximality of P, P + (a) is finitely generated, by elements p_{1} + r_{1}a, …, p_{k} + r_{k}a (with p_{i} ∈ P, r_{i} ∈ R), and similarly, P + (b) is finitely generated, by elements q_{1} + s_{1}b, …, q_{m} + s_{m}b (with q_{i} ∈ P, s_{i} ∈ R).

Also, the ideal (P : a) = {r ∈ R | ra ∈ P } properly containes P (because P ⊂ P + (b) ⊆ (P : a)), and is therefore finitely generated, by j_{1}, …, j_{h}, say.

If c∈P, then c∈P +(a) and there exist t_{1}, …, t_{k} ∈R with c=t_{1}p_{1}+…+t_{k}p_{k}+ (t_{1}r_{1} + … + t_{k}r_{k})a. Let d = t_{1}r_{1} + … + t_{k}r_{k}, then da = c − (t_{1}p_{1} + … + t_{k}p_{k}) ∈ P and therefore d ∈ (P : a). Let t_{1}′, …, t_{h}′ ∈ R such that d = t_{1}′j_{1} + … + t_{h}′j_{h}.

Then c = t_{1}p_{1} + … + t_{k}p_{k} + t_{1}′j_{1}a + … + t_{h}′j_{h}a, and p_{1}, …, p_{k} as well as j_{1}a, …, j_{h}a are in P. We have shown that p_{1}, …, p_{k}, j_{1}a, …, j_{h}a generate P, a contradiction. □

7.10 Theorem (Cohen). Let R be a commutative ring. If every prime ideal of R is finitley generated, then R is Noetherian.

Proof. The set of non-finitely-generated ideals of a commutative ring clearly satisfies the hypothesis of Zorn’s lemma. Therefore, if this set is non-empty, it has a maximal element by Zorn’s lemma. Such a maximal element is a prime ideal by the previous lemma. Therefore, every non-Noetherian ring has a prime ideal that is not finitely generated. □

8. Associated ideals

∗ ∗ ∗ Zero-divisors and associated primes ∗ ∗ ∗

The following theorem holds for (not necessarily Noetherian) commutative rings:

8.1 Theorem (Prime Avoidance). Let R be a commutative ring and P_{1}, …, P_{n} ideals of R such that at most two of the P_{i} are not prime. Let I be a ring (possibly without a unit element) such that I ⊆ ⋃_{1≤i≤n} P_{i}. Then there exists an i with S ⊆ P_{i}.

Proof. Induction on n. Let n = 2, I ⊆ P_{1}∪P_{2}, and suppose there exists a_{1} ∈ S ⊆ P_{1} and a_{2} ∈ S ⊆ P_{2}. Then a_{1} + a_{2} is in I, but neither in P_{1} nor in P_{2}, a contradition. Now let n > 2, I ⊆ ⋃_{1≤i≤n} P_{i}, but for every 1 ≤ k ≤ n, I is not contained in the union of the P_{i} with i ≠ k (otherwise we are done by induction hypothesis). For every k pick a_{k} in I and not in the union of the P_{i} with i ≠ k. Then a_{k} ∈ P_{k}. As n > 2, at least one of the P_{i} is prime, say P_{1} is prime, and we set b = a_{1} +a_{2} … a_{n}. Then b is in I, but not in any P_{i}, a contradiction □

8.2 Definition. Let M be an R-module.

(i) For m ∈ M, the annihilator of m is Ann_{R}(m) := {r ∈ R | rm = 0}.

(ii) r ∈ R is a zero-divisor of M if rm = 0 for some non-zero m ∈ M.

(iii) The set of zero-divisors of M is denoted by Z(M), in other words,

Z(M) = ⋃_{{m∈M \atop m≠0}} Ann_{R}(m).

(iv) A prime ideal P of R is called an associated prime ideal of M if P = Ann_{R}(m) for some non-zero m ∈ M.

8.3 Lemma. Let M be a Noetherian R-module and N an R-submodule generated by the set S. Then N is generated by a finite subset of S.

Proof. Consider the (clearly non-empty) set of submodules of N generated by finite subsets of S. This set of submodules has a maximal element N′, generated by s_{i_{1}}, …, s_{i_{n}} ∈ S. For any s ∈ S, the submodule of N generated by s, s_{i_{1}}, …, s_{i_{n}} contains N′ and is therefore equal to N′ by maximailty of N′. We see that every s ∈ S is already in N′ and therefore N = N′. □

8.4 Remark: For any commutative ring R and R-module M, the set R \ Z(M) of non-zerodivisors of M is a saturated multiplicative set. Therfore, by 10.3, its complement, Z(M), is a union of prime ideals. In the Noetherian case, we will see that only finitely many prime ideals are needed.

8.5 Proposition (Prime and maximal annihilators). Let R be a commutative ring and M ≠ 0 an R-module.

(i) Every maximal annihilator of a non-zero element is prime.

(ii) If R is Noetherian then every annihilator of a non-zero element is contained

in a maximal one. (iii) If M is a Noetherian module then there are only finitely many maximal

annihilators of non-zero elements of M. (iv) If R is a Noetherian ring and M a Noetherian module, then Z(M) is a

finite union of prime ideals, each of which is an annihilator of a non-zero element of M:

Z(M) = Ann_{R}(m_{1}) ∪ … ∪ Ann_{R}(m_{n}), Ann_{R}(m_{i}) prime, for 1 ≤ i ≤ n

Proof.

(i) Let I = Ann_{R}(m) be maximal among annihilators of non-zero elements. I ≠ R since M is unitary and non-zero. Suppose ab ∈ I and b ∉ I. Then bm ≠ 0 and a ∈ Ann_{R}(bm). Now Ann_{R}(m) ⊆ Ann_{R}(bm) implies, by maximality of I, that Ann_{R}(m) = Ann_{R}(bm) and therefore a ∈ I. (ii) For I = Ann_{R}(m), consider the set S of annihilators of non-zero elements containing I (which is non-empty since I ∈ S). Since R is Noetherian S has a maximal element which is clearly maximal among annihilators of non-zero elements and contains I. (iii) Suppose P_{λ} = Ann_{R}(m_{λ}), λ ∈ Λ are all the maximal annihilators of non-zero elements of M. Consider the submodule N of M generated by the m_{λ}, λ ∈ Λ. There exists a finite subset m_{λ_{1}}, …, m_{λ_{n}} of these generators that already generate N. For any m_{μ} with μ ∈ Λ there exist r_{1}, …, r_{n} ∈ R, such that m_{μ} = r_{1}m_{λ_{1}}, + … + r_{n}m_{λ_{n}}, which implies P_{λ_{1}} … P_{λ_{n}} ⊆ Ann_{R}(m_{μ}). Since Ann_{R}(m_{μ}) = P_{μ} is prime we have P_{λ_{i}} ⊆ P_{μ} for some 1 ≤ i ≤ n. By maximality of P_{λ_{i}} it follows that P_{μ} = P_{λ_{i}}. Therfore every Ann_{R}(m_{μ}), μ ∈ Λ is one of P_{λ_{1}}, …, P_{λ_{n}}.

(iv) follows from (i)–(iii). □

8.6 Theorem. Let R be a Noetherian ring and M a Noetherian R-module. If I is a subring (possibly without a unit) of R contained in Z(M) then there exists a non-zero element m ∈ M with Im = {0}.

Proof. By the previous lemma, Z(M) is a union of finitely many prime ideals of the form Ann_{R}(m). By prime avoidance 8.1, I is contained in a single Ann_{R}(m). □

8.7 Corollary (Universal zero divisor). Let R be a Noetherian ring, and I an ideal (or non-unitary subring) consisting entirely of zero-divisiors of R. Then there exists r ∈ R such that ir = 0 for all i ∈ I.

∗ ∗ ∗ Associated primes ∗ ∗ ∗

8.8 Definition. If M is an R-module, an associated prime of M is a prime ideal P of R that is the annihilator of an element of M, P = Ann(m). The set of all associated primes of M is denoted by Ass_{R}(M), or Ass(M), when R is understood.

8.9 Definition. The associated primes of an ideal I of an integral domain R are the associated primes of the R-module R/I.

The apparent ambiguity of the definition of the associated primes of an ideal I is not a problem, as the only associated prime of the R-module I is (0) (in an intergal domain, Ann(r) = 0 for every non-zero r ∈ R), so this is a concept we can disregard.

Let R be an integral domain with quotient field K and k ∈ K. Consider the “conductor ideal” (R:_{R} k) = {r ∈ R | rk ∈ R}. Note that (R:_{R} k) = R iff k ∈ R. Also, if k is written as a fraction of elements of R, k = \frac{a}{b} then (R:_{R} k) = (Rb:_{R} a), which is the annihilator of (the residue class of) a in R/Rb. Therefore conductor ideals (of elements of the quotient field) that are prime are exactly the associated primes of non-zero principal ideals of R.

9. Motivation from algebraic geometry

Let K be a field and n ∈ ℕ. We are interested in subsets of the n-dimensional K-vectorspace that are characterized by a system of polynomial equations in n variables with coefficients on K.

To every subset S of the polynomial ring K[x_{1}, …, x_{n}] we associate the set of its zeros

Z(S) = {(a_{1}, …, a_{n}) ∈ K^{n} | ∀f ∈ S f(a_{1}, …, a_{n}) = 0}

and to every subset A of the n-dimensional K-space the set of polynomials that are zero on A

I(A) = {f ∈ K[x_{1}, …, x_{n}] | ∀(a_{1}, …, a_{n}) ∈ A f(a_{1}, …, a_{n}) = 0}.

We can consider Z as a function whose arguments are subsets of K[x_{1}, …, x_{n}] and whose values are subsets of K^{n} (and vice versa for I), or, without losing any information, we can restrict the arguments of Z and the values of I to ideals of K[x_{1}, …, x_{n}]. This is so because I(A) is an ideal of K[x_{1}, …, x_{n}] for every A ⊆ K^{n}, and Z(S) = Z((S)), where (S) is the ideal of K[x_{1}, …, x_{n}] generated by S, for every set of polynomials S ⊆ K[x_{1}, …, x_{n}]. (Exercise: check this.)

Subsets of K^{n} of the form A = Z(J) for some ideal J ⊴ K[x_{1}, …, x_{n}] are called algebraic sets.

9.1 Remark: The more obvious properties of I and Z: For all A, B ⊆K[x_{1}, …, x_{n}], and all C, D ⊆ K^{n}

(i) A ⊆ B ⟹ Z(A) ⊇ Z(B)  and  C ⊆ D ⟹ I(C) ⊇ I(D)
(ii) A ⊆ I(Z(A))  and  C ⊆ Z(I(C))
(iii) Z(A) = Z(I(Z(A)))  and  I(C) = I(Z(I(C)))

(iv) There is a bijective correspondence between ideals of K[x_{1}, …, x_{n}] of the form J = I(S) (for some S ⊆ K^{n}) and subsets of K^{n} of the form A = Z(L) (for some ideal L of K[x_{1}, …, x_{n}]) given by the restrictions of I and Z to arguments of this form.

(i) and (ii) are easy to see. By (i) and (ii), I and Z constitute a Galois correspondence between subsets of K^{n} and ideals of K[x_{1}, …, x_{n}] (each ordered by set-theoretic inclusion). (iii) and (iv) then follow from the Galois correspondence.

9.2 Definition. A Galois Correspondence between two partially ordered sets (X , ≤) and (Y, ≤) consists of functions f: X → Y and g: Y → X satisfying the following conditions for all A, B ∈ X and all C, D ∈ Y:

(i) A ≤ B ⟹ f(A) ≥ f(B)  and  C ≤ D ⟹ g(C) ≥ g(D)
(ii) A ≤ f(g(A))  and  C ≤ g(f(C))

9.3 Remark: The conditions defining a Galois correspondence are symmetric in f and g. To reflect this symmetry and to allow us to express two statements in one, we can write both f and g in prime notation: for A ∈ X, A′ := f(A) and for A ∈ Y, A′ := g(A). Conditions (i) and (ii) become

(i) A ≤ B ⟹ A′ ≥ B′

(ii) A ≤ A′′

Given a Galois correspondence, we call an element A of X or Y closed with respect to the Galois correspondence, if it satisfies A′′ = A.

It is easy to see that conditions (i) and (ii) above imply (iii) A′ = A′′′,

which in turn implies (iv) The subset of elements A in X (or Y) satisfying A = A′′ is exactly the

subset of elements of the form B′ for some B ∈ Y (or X, respectively).

(v) There is a bijective correspondence, given by A ↦ A′ (in either direction) between the set of closed elements in X and the set of closed elements in Y.

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

Like in other situations in mathematics, the nontrivial facts connected with a Galois correspondence between two sets consist of characterizations of the closed elements in either set from a different point of view, independent of the Galois structure.

The following Lemma shows that there is a topology on K^{n} whose closed sets are exactly the subsets of K^{n} of the form Z(J) (for some ideal J of K[x_{1}, …, x_{n}]).

9.4 Lemma.

1) If A, B ⊆ K^{n} are algebraic sets then so is A ∪ B. Indeed, for all sets

S, T ⊆ K[x_{1}, …, x_{n}],

Z(S) ∪ Z(T) = Z(ST),

where ST = {st | s ∈ S, t ∈ T }.

2) If A_{λ} is an algebraic set for all λ ∈ Λ (an arbitrary index set) then ⋂_{λ∈Λ} A_{λ} is an algebraic set. Indeed,

⋂_{λ∈Λ} Z(S_{λ}) = Z( ⋃_{λ∈Λ} S_{λ}).

3) ∅ and K^{n} are algebraic sets: ∅ = Z(1) = Z(K[x_{1}, …, x_{n}]) and K^{n} = Z(0).

Proof. Ad 1) Clearly, Z(S)∪Z(T) ⊆ Z(ST). For the reverse inclusion, we consider a point a ∈ Z(ST) that is not in Z(S) and show that it is in Z(T). Fix s_{0} ∈ S with s_{0}(a) ≠ 0. For all t ∈ T, 0 = (s_{0} · t)(a) = s_{0}(a)t(a), therefore for all t ∈ T t(a) = 0, i.e., a ∈ Z(T).

Ad 2) ⋂_{λ∈Λ} Z(S_{λ}) = {a ∈ K^{n} | ∀λ ∈ Λ ∀f ∈ S_{λ}f(a) = 0} = Z(⋃_{λ∈Λ} S_{λ}). □

9.5 Remark:

1) If I and J are ideals of a ring R then IJ is defined as

IJ = {i_{1}j_{1} + … + i_{k}j_{m} | m ∈ ℕ, i_{k} ∈ I, j_{k} ∈ J},

which is the ideal generated by {ij | i ∈ I, j ∈ J}.

With this definition of IJ, we have, for ideals I, J of K[x_{1}, …, x_{n}]:

Z(I ∩ J) = Z(IJ) = Z(I) ∪ Z(J).

Proof: {ij | i ∈ I, j ∈ J} ⊆ IJ ⊆ I ∩J implies Z(I ∩J) ⊆ Z(IJ) ⊆ Z({ij | i ∈ I, j ∈ J}); Z({ij | i ∈ I, j ∈ J}) = Z(I)∪Z(J) holds by lemma 9.4, and Z(I)∪Z(J) ⊆ Z(I ∩J) for purely logical reasons.

2) If S_{λ} is a subset of K^{n} for all λ ∈ Λ then

I( ⋃_{λ∈Λ} S_{λ}) = ⋂_{λ∈Λ} I(S_{λ}).

In particular, for any set A ⊆ K^{n},

I(A) = ⋂_{a∈A} I(a).

9.6 Definition. Let R be a commutative ring and I an ideal of R. The radical of I is defined by

\sqrt{I} = \{r ∈ R | ∃n ∈ ℕ r^{n} ∈ I\}.

Note that I ⊆ \sqrt{I} for every ideal I and that \sqrt{I} = R if and only if I = R. It is easy to see that \sqrt{Q} = Q whenever Q is a prime ideal or, more generally, an intersection of prime ideals. (We shall see in chapter 4 that \sqrt{I} is the intersection of all prime ideals containing I.)

9.7 Theorem (Hilbert’s Nullstellensatz). Let K be an algebraically closed field and f_{1}, …, f_{m}, g ∈ K[x_{1}, …, x_{n}] such that g(a) = 0 for all a ∈ Z(f_{1}, …, f_{m}) then

g ∈ \sqrt{(f_{1}, …, f_{m})} .

Proof. Will be proved in chapter 3. □

The importance of Hilbert’s Nullstellensatz lies in the following corollary.

9.8 Corollary. Let J be an ideal of K[x_{1}, …, x_{n}], where K is an algebraically closed field. Then

I(Z(J)) = \sqrt{J} .

Proof. The Nullstellensatz says I(Z(J)) ⊆ \sqrt{J} for finitely generated J, and from Hilbert’s basis theorem we know that every ideal of K[x_{1}, …, x_{n}] is finitely generated. The reverse inclusion is easy; it holds even if K is not algebraically closed: if g∈ \sqrt{J} then for some m∈ℕ g^{m} ∈J and therefore g^{m}(a)=g(a) … g(a)=0 for every a ∈ Z(J). Since K is an integral domain, g(a) = 0 follows. □

9.9 Lemma (maximal ideals). Let K be a field, a = (a_{1}, …, a_{n}) ∈ K^{n}, and define Mā to be the ideal of K[x_{1}, …, x_{n}] generated by x_{1} − a_{1}, …, x_{n} − a_{n}. Then Mā = I({a}) and Mā is a maximal ideal of K[x_{1}, …, x_{n}].

Proof. The substitution homomorphism φ: K[x_{1}, …, x_{n}] → K[x_{1}, …, x_{n}] with φ(x_{i}) = x_{i} − a_{i} and φ(k) = k for k ∈ K is an automorphism of K[x_{1}, …, x_{n}]. Therefore the residue class ring of Mā is isomorphic to the residue class ring of (x_{1}, …, x_{n}), which is K. As K is a field, Mā is a maximal ideal. Therefore Mā ⊆ I({a}) ⊂ K[x_{1}, …, x_{n}] implies Mā = I({a}). □

With this in mind, \sqrt{J} = I(Z(J)) becomes

\sqrt{J} = ⋂_{a∈Z(J)} I(a) = ⋂_{J⊆I(a)} I(a) = ⋂_{J⊆Mā} Mā,

the statement that the radical of every ideal of K[x_{1}, …, x_{n}] (K algebraically closed) is an intersection of ideals of the form Mā. (Since this also holds for maximal ideals, every maximal ideal is of the form Mā.) Therefore, one way to see the Nullstellensatz is:

9.10 Theorem (Variant of Hilbert’s Nullstellensatz). Let K be an algebraically closed field.

(1) (“K[x_{1}, …, x_{n}] is a Hilbert ring.”) In K[x_{1}, …, x_{n}], the radical of every ideal is an intersection of maximal ideals.

(2) (“Weak Nullstellensatz”) Every maximal ideal of K[x_{1}, …, x_{n}] is of the form Mā, a ∈ K^{n}.

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

9.11 Definition. A subset Y of a topological space X is called irreducible if Y ≠ ∅ and, whenever Y ⊆ B ∪ C, where B, C are closed sets, it follows that A ⊆ B or A ⊆ C.

For a closed set A, this is equivalent to: A ≠ ∅; and, whenever A = B ∪ C with B, C closed, it follows that A = B or A = C.

9.12 Remark: For a subset S of K^{n} with Zariski topology, S is irreducible if S ≠ ∅ and, whenever S ⊆ Z(J) ∪ Z(L) for some J, L ⊴ K[x_{1}, …, x_{n}], it follows that S ⊆ Z(J) or S ⊆ Z(L).

For an algebraic subset A = Z(I) of K^{n} this means: A ≠ ∅ and, whenever A = Z(J) ∪ Z(L), it follows that A = Z(J) or A = Z(L).

9.13 Proposition. Let A ⊆ K^{n}. Then A is irreducible iff I(A) is a prime ideal.

Proof. (⇒) Suppose A is irreducible and JL ⊆ I(A) for some J, L ⊴ K[x_{1}, …, x_{n}]; to show J ⊆ I(A) or L ⊆ I(A).

Z(J) ∪ Z(L) = Z(JL) ⊇ Z(I(A)) ⊇ A

By irreducibility of A we get A ⊆ Z(J) or A ⊆ Z(L), say the former, which implies I(A) ⊇ I(Z(J)) ⊇ J.

(⇐) Suppose I(A) is prime and A ⊆ Z(J) ∪ Z(L); to show A ⊆ Z(J) or A ⊆ Z(L).

I(A) ⊇ I(Z(J) ∪ Z(L)) = I(Z(J)) ∩ I(Z(L)) ⊇ I(Z(J))I(Z(L))

I(A) being prime we have I(Z(J)) ⊆ I(A) or I(Z(L)) ⊆ I(A), say the former. Then A ⊆ Z(I(A)) ⊆ Z(I(Z(J))) = Z(J). □

9.14 Corollary. For I ⊴ K[x_{1}, …, x_{n}], Z(I) is irreducible iff \sqrt{I} is prime.

The weak Nullstellensatz establishes a bijection between the points of K^{n} and the maximal ideals of K[x_{1}, …, x_{n}] (when K is algebraically closed). Thus the topology on K^{n} (whose closed sets are the algebraic sets) translates to a topology on MaxSpec(K[x_{1}, …, x_{n}]) (the set of maximal ideals of K[x_{1}, …, x_{n}]). A set S of maximal ideals is closed if and only if there exists an ideal I of K[x_{1}, …, x_{n}] such that S is exactly the set of maximal ideals containing I.

This topology can be generalized to the spectrum of an arbitrary commutative ring R.

9.15 Definition. Let R be a commutative ring and Spec(R) the set of prime ideals of R. For every ideal I of R we define a set of prime ideals V (I) = {P ∈ Spec(R) | I ⊆ P }.

Zariski topology on Spec(R) is defined by the convention: a set S of prime ideals of R is closed if and only if there exists an ideal I of R such that S = V (I).

To show that this is a valid definition of a topology, we have to check:

9.16 Lemma.

1) Finite unions of closed sets are closed. Indeed, V (I) ∪ V (J) = V (IJ).

2) Arbitrary intersections of closed sets are closed. Indeed,

⋂_{λ∈Λ} V (I_{λ}) = V (( ⋃_{λ∈Λ} I_{λ})).

3) ∅ and Spec(R) are closed: ∅ = V (R) and Spec(R) = V ((0)).

We have a Galois connection between ideals of R on one hand and sets of prime ideals of R on the other hand given by I ↦ V (I) for I ⊴ R and S ↦ ⋂_{P ∈S} P for S ⊆ Spec(R), which gives a bijection between radical ideals of R and Zariski-closed sets of prime ideals.

It is clear that the Zariski-closure of a set S of prime ideals consists of all prime ideals containing ⋂_{P ∈S} P.

9.17 Definition. A topological space X is called Noetherian, if it satisfies the descending chain condition on closed sets, i.e., whenever A_{1} ⊇ A_{2} ⊇ … A_{m} ⊇ A_{m+1} ⊇ … with A_{i} closed for all i, then there exists N ∈ ℕ such that for all i > N A_{i} = A_{N}.

9.18 Remark: It is easy to see that X is Noetherian if and only if every set A of closed sets has a minimal element with respect to inclusion.

If R is a Noetherian ring then Spec(R) with Zariski topology is easily seen to be a Noetherian topological space.

9.19 Theorem. Let X be a Noetherian topological space. Then every closed set A is a finite union of irreducible closed sets A = A_{1} ∪ … ∪ A_{m} with A_{i} ⊆/A_{j} for i ≠ j and these sets A_{i} are unique.

Proof. Existence: Suppose the set A of all closed sets that are not a finite union of closed irreducible sets is nonempty. Then it contains a mimimal A. This A is not irreducible, so there exist B, C closed with A = B ∪ C and A ≠ B, A ≠ C. By minimality of A ∈ A, B and C are not in A and are each representable as a finite union of irreducible closed sets. But then, so is A, a contradiction. Once a closed set is represented as finite union of irreducible closed A_{i} we can delete those A_{i} that are contained in some A_{j} for j ≠ i.

Uniqueness: Suppose A_{1} ∪ … ∪ A_{m} = A′_{1}cup … ∪ A′_{n}. For 1 ≤ i ≤ m A_{i} ⊆ A′_{1}cup … ∪ A′_{n}. As A_{i} is irreducible, there exists j with A_{i} ⊆ A′_{j}. By the same token, there exists k, such that A′_{j} ⊆ A_{k}. Then A_{i} ⊆ A_{k}. Therefore i = k and A_{i} = A′_{j}. By setting φ(i) = j with A_{i} = A′_{j} we get an injective map φ: {1, … m} → {1, … n}. φ is also surjective: take k ∈ {1, … n}. Then, as above, there exists i with A′_{k} = A_{i}, and also A_{i} = A′_{φ(i)}, so A′_{k} = A′_{φ(i)}, which implies k = φ(i). □

10. Primary ideals and the radical.

∗ ∗ ∗ Multiplicatively closed setes. ∗ ∗ ∗

10.1 Definition. A non-empty subset S of a commutative ring R is called multiplicatively closed, or multiplicative, if s, t ∈ S implies st ∈ S.

A multiplicatively closed set S is called saturated if it contains every divisor of each of its elements, i.e., if s = ab ∈ S implies a ∈ S.

The set S̄ = {t ∈ R | ∃s ∈ S t | s} of divisors of elements of a multiplicatve set S – clearly the smallest saturated multiplcative set containing S – is called the saturation of S.

Examples of multiplicatively closed sets (1–3 are saturated):

1) R \ P, where P is a prime ideal of R,

2) R \ ⋃_{α∈A} P_{α}, where all P_{α} are prime ideals of R

3) the set of all non-zerodivisors in R

4) the set of powers of a fixed element {r^{n} | n ∈ ℕ_{0}}.

10.2 Lemma. Let S be a multiplicatively closed subset of R and I an ideal of R with I ∩ S = ∅. Then there exists a prime ideal P of R with I ⊆ P and P ∩ S = ∅.

Proof. Apply Zorn’s Lemma to J = {J ⊴ R | I ⊆ J and S ∩ J = ∅}. Since I ∈ J, J ≠ ∅. Every chain of ideals in J has an upper bound in J: its union is in J. Let P be a maximal element of J. We will show that P is prime. S ≠ ∅ implies P ≠ R. Suppose ab ∈ P, a ∉ P and b ∉ P. By maximailty of P there exists s ∈ S ∩ (P + Ra) and t ∈ S ∩ (P + Rb). But then st ∈ S ∩ P, a contradiction. □

10.3 Corollary.

(i) Let S be a saturated multiplicatively closed subset of R. Then S =R\⋃_{P ∈P} P, where P is the set of all prime ideals P of R with S ∩ P = ∅.

(ii) The saturated multiplicative subsets of R are precisely the complements of unions of prime ideals. (The trivial case S = R corresponds to the empty union of prime ideals).

Proof. Ad (i). Clearly, S ⊆ R \ ⋃_{P ∈P} P. For the reverse inclusion, we show that for every u ∉ S there exists some prime ideal P with S ∩ P = ∅ and u ∈ P. S being saturated, u ∉ S implies Ru ∩ S = ∅. By Lemma 10.2 there exists a prime ideal P with Ru ⊆ P and S ∩ P = ∅. Since the complement of a union of prime ideals is a saturated multiplicateve set, (ii) follows. □

10.4 Corollary. The set Z(R) of zero-divisors of a commutative ring R is a union of prime ideals of R. More generally, the set Z(M) of zero-divisors of any R-module M is a union of prime ideals of R.

Proof. Verify that R \ Z(R), and more generally, R \ Z(M), is a saturated multplicative set. □

10.5 Lemma. Let I be an ideal of R and P a prime ideal containing I. Then there exists a minimal prime P′ of I with I ⊆ P′ ⊆ P.

Proof. Let S = R \ P and consider the set S of multiplicatively closed sets T with S ⊆ T and T ∩I = ∅. Then S is non-empty because S ∈ S. Every chain of elements of S has an upper bound in S (its union). Let S_{0} be a maximal element of S (which exists by Zorn’s Lemma), and set U = R \ S_{0}. By Lemma 10.2, there exists a prime ideal P′ containing I and disjoint from S_{0}, i.e., I ⊆ P′ ⊆ U. It follows that P′ = U and P′ is a minimal prime of I, since for any prime ideal Q with I ⊆ Q ⫋ U, its complement R \ Q would be an element of S stricltly containing S_{0}. □

∗ ∗ ∗ The radical. ∗ ∗ ∗

10.6 Definition. Let I be an ideal of R. The radical of I is defined by

\sqrt{I} = \{r ∈ R | ∃n ∈ ℕ r^{n} ∈ I\}.

Convince yourself of the following easy facts:

10.7 Remark:

(1) I ⊆ \sqrt{I}

(2) \sqrt{\sqrt{I}} = \sqrt{I}

(3) \sqrt{I} = R ⟺ I = R

(4) If Q is an intersection of prime ideals then \sqrt{Q} = Q.

Recall that an element r ∈ R is called nilpotent if there exists a natural number n with r^{n} = 0.

10.8 Definition. The radical of (0), consisting of all nilpotent elements of R, is called the nilradical of R:

Nil(R) = {r ∈ R | ∃n ∈ ℕ r^{n} = 0}.

10.9 Proposition.

\sqrt{I} = ⋂ ; {P prime \atop I ⊆ P} P.

Proof. Clearly, \sqrt{I} is contained in all prime ideals containing I. Conversely, if r ∉ \sqrt{I} , then S = \{r^{n} | n ∈ ℕ\} is a multiplicatively closed set with S ∩ I = ∅ and by Lemma 10.2, there exists a prime ideal P with I ⊆ P and r ∉ P. □

10.10 Corollary. The nilradical of R is the intersection of all prime ideals of R.

10.11 Lemma. Some rules of arithmetic for radicals:

(1) \sqrt{IJ} = \sqrt{I ∩ J} = \sqrt{I} ∩ \sqrt{J}

(2) \sqrt{I + J} = \sqrt{\sqrt{I} + \sqrt{J}}

(3) \sqrt{I} + \sqrt{J} = R ⟺ I + J = R

Proof. Exercise. □

10.12 Lemma. Let P be a prime ideal of R and I an ideal of R.

(i) P^{n} ⊆ I ⊆ P for some n ∈ ℕ implies \sqrt{I} = P. (ii) If R is Noetherian, then \sqrt{I} = P implies for some n P^{n} ⊆ I ⊆ P.

Proof. (i) is easy. Ad (ii), P = (p_{1}, …, p_{m}) = \sqrt{I} implies for some k ∈ ℕ, p_{1}^{k}, …, p_{m}^{k} ∈ I. Every element q of P is an R-linear combination of the p_{i}. By the multinomial theorem, (r_{1}p_{1} + … + r_{m}p_{m})^{m}k is a sum of terms each of which is divisible by p^{k} for some i, therefore q^{n} ∈ I for every q ∈ P, n = km. □

∗ ∗ ∗ Primary ideals ∗ ∗ ∗

10.13 Definition. An ideal Q of R is called primary iff Q ≠ R and for all a, b ∈ R with ab ∈ Q either a ∈ Q or ∃n ∈ ℕ b^{n} ∈ Q.

The definition of primary tends to be confusing at first. If Q is primary, then ab ∈ Q implies a ∈ Q or b ∈ \sqrt{Q} , as well as, by symmetry, b ∈ Q or a ∈ \sqrt{Q} . Therefore, for an ideal Q ≠ R, being primary is equivalent to

ab ∈ Q ⇒ a ∈ Q ∨ b ∈ Q ∨ a, b ∈ \sqrt{Q}

10.14 Easy exercise. If Q is a primary ideal then \sqrt{Q} is a prime ideal.

10.15 Definition. A primary ideal Q with \sqrt{Q} = P is called P-primary.

Not every ideal whose radical is prime is primary, however. For instance, \sqrt{P^{n}} = P for every prime ideal P and n ∈ ℕ, but not every power of a prime ideal is primary. In a more positive vein, we have the following facts:

10.16 Lemma. Let M be a maximal ideal of R and Q an ideal with \sqrt{Q} = M. Then Q is M -primary.

Proof. Suppose ab ∈ Q and a ∉ M; we must show b ∈ Q. We have M + aR = R, i.e., \sqrt{Q} + aR = R. This implies Q + aR = R, so 1 = q + ar for some q ∈ Q and r ∈ R, and finally b = qb + abr ∈ Q. □

10.17 Corollary. Let M be a maximal ideal of R and I ≠ R an ideal with M^{n} ⊆ I for some n ∈ ℕ. Then I is M -primary.

Proof. I is not contained in any prime ideal P other than M: if I ⊆ P then for every m ∈ M, m^{n} ∈ I ⊆ P implies m ∈ P, hence M = P. Since I ≠ R, this shows \sqrt{I} = M and we are done. □

Just like prime ideals, primary ideals behave nicely with respect to ring homomorphisms:

10.18 Lemma. Let f: R → T be a ring homomorphism (with f(1) = 1, as always) and I, Q ideals of T. Then

(1) f^{−1}(\sqrt{I} ) = \sqrt{f^{−1}(I)} .

(2) If Q is P -primary then f^{−1}(Q) is f^{−1}(P)-primary.

Again, like prime ideals, primary ideals can be characterized by a property of their quotient ring.

10.19 Proposition. An ideal Q is primary if and only if R/Q is a non-zero ring in which every zero-divisor is nilpotent.

Proof. Easy. □

Together with the previous lemma this implies that a canonical projection induces a bijection between primary ideals containing the kernel and primary ideals in the quotient ring. (Note that everything we show for canonical projections holds mutatis mutandis for any surjective ring homomorphism f: R → T, since f is just the canonical projection π: R → R/Kerf followed by the isomorphism f̄: R/Kerf → T.)

10.20 Proposition. Let I be an ideal of R and π: R → R/I the canonical projection. Then a bijection between all primary ideals Q of R with I ⊆ Q and all primary ideals of R/I is given by Q ↦ π(Q). Its inverse is Q̃ ↦ π^{−1}(Q̃) (for Q̃ ⊴ R/I).

Primary ideals also behave just like prime ideals with respect to the canonical homomorphism into a ring of fractions f_{S}: R → R_{S} and with respect to the canonical projection onto a residue class ring:

10.21 Proposition. Let R_{S} be a ring of fractions of R and f_{S}: R → R_{S} the canonical homomorphism.

(1) If Q ⊴ R is primary and \frac{r}{s} ∈ Q_{S} then r ∈ Q.

(2) A bijection between the primary ideals Q of R with S ∩ Q = ∅ and all primary ideals of R_{S} is given by Q ↦ Q_{S} (for Q a primary ideal of R). Its inverse is Q̃ ↦ f_{S}^{−1}Q̃ (for Q̃ a primary ideal of R_{S}).

This is shown just like the analogous statement for prime ideals. (Note that

Q ∩ S = ∅ implies \sqrt{Q} ∩ S = ∅.)

11. Hilbert rings and the Nullstellensatz

∗ ∗ ∗ Goldman-Krull proof of Hilbert’s Nullstellensatz ∗ ∗ ∗

If P is an ideal of a commutative ring R, we know that

P is prime ⟺ R/P is an integral domain
P is maximal ⟺ R/P is a field.

We are now going to see a similar characterization for a concept that falls in between prime and maximal.

P is a prime ideal that is not an intersection of strictly larger prime ideals. ⟺ R/P is an integral domain whose quotient field is generated (as a ring over R/P) by a single element.

11.1 Definition. An ideal P is a G-ideal (or Goldman-ideal) if P is a prime ideal which is not an intersection of prime ideals strictly containing P.

Note that P is a G-ideal iff R/P is an integral domain in which the intersection of all non-zero prime ideals is not (0).

11.2 Definition. A domain D with quotient field K is a G-domain if K is generated by a single element as a ring over D, i.e., ∃z ∈ K such that K = D[z].

Some equivalent characterizations of G-domains:

11.3 Lemma. Let D be a domain and K its quotient field. The following are equivalent:

(1) There exists u ∈ D \ {0} such that K = D[u^{−1}].

(2) K is generated by a single element as a ring over D.

(3) K is finitely generated as a ring over D.

Proof. (3 ⇒ 1) If K = D[\frac{a_{1}}{b_{1}} , …, \frac{a_{n}}{b_{n}}] and u = b_{1} · . . . · b_{n} then K = D[u^{−1}]. □

11.4 Lemma. Let D be a domain, K its quotient field, and 0 ≠ u ∈ D. Then K = D[u^{−1}] iff u is in the intersection of all prime ideals P ≠ (0) of D.

Proof. Suppose K = D[u^{−1}] and let P be a non-zero prime ideal. Pick b ∈ P \{0}. Then b^{−1} = cu^{−n} for some c ∈ D and n ∈ ℕ. Now bc = u^{n}, therefore u^{n} ∈ P and finally u ∈ P.

Conversely, suppose u ∈ P for every prime ideal P ≠ (0). Let 0 ≠ b ∈ D. Then u ∈ \sqrt{Rb} , meaning there exists n ∈ ℕ with u^{n} = cb for some c ∈ D. Rewritten as b^{−1} = cu^{−n} this shows b^{−1} ∈ D[u^{−1}]. As b was arbitrary, K = D[u^{−1}]. □

As a consequence of the previous two lemmata:

11.5 Theorem. Let D be a domain. The intersection of all non-zero prime ideals of D is not (0) iff the quotient field of D is generated by a single element as a ring over D.

Applying this to residue class rings we get:

11.6 Corollary. Let P ⊴ R. Then P is a G-ideal iff R/P is a G-domain.

Note that a commutative ring R is a G-domain if and only if (0) is a G-ideal of R.

11.7 Lemma. Let I be an ideal in a commutative ring R. Then

\sqrt{I} = ⋂ ; {P G-ideal \atop I ⊆ P} P.

Proof. Clearly, \sqrt{I} is contained in every prime ideal containing I. Conversely, if u ∉ \sqrt{I} then there exists a G-ideal P containing I with u ∉ P: As in the proof of \sqrt{I} = ⋂_{I⊆P prime} P (10.9), let P be maximal among the ideals containing I and disjoint with S = {u^{n} | n ∈ ℕ}. Then P is prime. Also, P is a G-ideal, because every prime ideal strictly containing P intersects S and therefore contains u, but u ∉ P. □

11.8 Lemma and Definition. Let R be a commutative ring. Then the following are equivalent

1) Every G-ideal of R is maximal.

2) For every ideal I of R, \sqrt{I} is an intersection of maximal ideals.

A commutative ring satisfying one (and hence both) of the above conditions is called Hilbert ring (or Jacobson ring).

Proof. (2 ⇒ 1) Let P be a G-ideal. Then \sqrt{P} = P, because P is prime. By 2), P is an intersection of maximal ideals, but, as a G-ideal, P is no intersection of prime ideals strictly containing P, therefore P must be maximal. □

11.9 Lemma. Let R be a commutative ring. Then R[x] is not a G-domain.

Proof. If R is not a domain, then R[x] is not a domain, let alone a G-domain. Now let R be a domain with quotient field K. We must show that the intersection of all non-zero prime ideals of R[x] is (0).

Since P ∩ R[x] is a non-zero prime ideal of R[x] for every non-zero prime ideal P of K[x], it suffices to show that the intersection of all non-zero prime ideals of K[x] is (0).

This is the case because there are infinitely many monic irreducible polynomials in K[x]: If there were only finitely many, p_{1} = x, p_{2}, …, p_{n}, then p_{1} · . . . · p_{n} + 1 (which is not a unit because its degree is not zero), would not be divisible by any monic irreducible polynomial, a contradiction to K[x] being a unique factorization domain. □

11.10 Corollary. Let K be a field. Then K[x] is a Hilbert ring.

Proof. K[x] is a principal ideal domain, so the only non-maximal prime ideal is (0), which is not a G-ideal by 11.9. Therefore every G-ideal of K[x] is maximal. □

11.11 Lemma. Let S ⊆ T be domains, T = S[t] with t algebraic over S. Then T is a G-domain if and only if S is a G-domain.

Proof. Let K and L be the quotient fields of S and T respectively. Then L = K[t] is algebraic over K (and hence over S). Suppose T is a G-domain. Then L = T [c^{−1}] = S[t, c^{−1}] for some c ∈ T. Both c^{−1} and t are algebraic over S. Let a and b be the leading coefficients of non-zero polynomials f and g in S[x] with f(c^{−1}) = 0 and g(t) = 0 and let S̃ = S[a^{−1}, b^{−1}]. Then c^{−1} and t are integral over S̃. Hence L is integral over S̃, which makes S̃ a field (4.17). Clearly, S̃ = K. Therefore K is finitely generated as a ring over S, and S is a G-domain.

The converse is easy. If S is a G-domain then K = S[s^{−1}] for some s ∈ S and the quotient field of T = S[t] (with t algebraic over S) is K(t) = K[t] = S[s^{−1}, t] = T [s^{−1}]. □

11.12 Lemma. Let R be a commutative ring. If Q is a G-ideal of R[x] then P = Q ∩ R is a G-ideal of R.

Proof. Let Q be a G-ideal of R[x] and P = Q ∩ R. Then Q contains P [x] and this containment is strict, because R[x]/Q is a G-domain by hypothesis and R[x]/P [x] ≃ (R/P)[x] is not a G-domain by Lemma 11.9.

Let S = R/P and T = R[x]/Q. We know that T is a G-domain and want to show that S is a G-domain.

S is contained in T via r + P ↦ r + Q, which is injective, because P = Q ∩ R. Also, T = S[t] for t = x + Q.

We show that t is algebraic over S: Clearly, t = x+Q ∈ T is a root of f for any f ∈ Q, and there exists some f ∈ Q which is not the zero polynomial in (R/P)[x], because Q ⫌ P R[x]. We can now apply Lemma 11.11. □

11.13 Lemma. Let R be a commutative ring and P a G-ideal of R. Then there exists a maximal ideal M of R[x] with M ∩ R = P.

Proof. Let S = R/P and K the quotient field of S. As S is a G-domain, K = S[b] for some b ∈ K. Let π: S[x] → S[b] be the substitution homomorphism with π(x) = b, π|_{S} = id_{S}. Since the image of π, S[b], is a field, the Ker(π) = M is a maximal ideal of S[x], and clearly Ker(π) ∩ S = (0).

Now S[x] = (R/P)[x] ≃ R[x]/P [x], and every maximal ideal M of R[x]/P [x] corresponds to a maximal ideal Q of R[x] containing P [x] (the inverse image of M under the canonical projection of R[x] onto (R/P)[x], and M ∩ S = (0) translates to Q ∩ R = P. □

Lemma 11.13 and Lemma 11.12 combined yield:

11.14 Theorem. Let R be a commutative ring and P an ideal of R. The following are equivalent:

(1) P is a G-ideal of R.

(2) There exists a maximal ideal M of R[x] with M ∩ R = P.

(3) There exists a G-ideal Q of R[x] with Q ∩ R = P.

11.15 Proposition. Let R be a commutative ring.

(1) R[x] is a Hilbert ring if and only if R is a Hilbert ring.

(2) If R is a Hilbert ring, then every maximal ideal Q of R[x] is generated by P = Q ∩ R together with some f ∈ R[x], which represents an irreducible polynomial in (R/P)[x].

Proof. Suppose R is a Hilbert ring, Q a G-ideal of R[x], and P = Q ∩ R.

P = Q ∩ R is a G-ideal of R by 11.12, and hence maximal by hypothesis. Q is the inverse image of some G-ideal Q̄ of R[x]/P R[x] ≃ (R/P)[x] under the canonical projection π : R[x] → R[x]/P R[x]. As (R/P)[x] is the polynomial ring over a field, Q̄ is maximal by 11.10, and therefore Q is a maximal ideal of R[x].

Furthermore, Q̄ is generated by a single irreducible polynomial in f̄∈(R/P)[x], so Q is generated by P and f, a representative in R[x] of f̄.

We have shown (2), and the “if” direction of (1). The “only if” direction of (1) is easy: R is a homomorphic image of R[x] and the Hilbert ring property clearly carries over to homomorphic images. □

11.16 Theorem. Let K be an algebraically closed field and n ∈ ℕ. Then

(1) K[x_{1}, …, x_{n}] is a Hilbert ring.

(2) Every maximal ideal of K[x_{1}, …, x_{n}] is of the form (x_{1} − a_{1}, …, x_{n} − a_{n}), for some a_{1}, …, a_{n} ∈ K.

Proof. (1) follows from 11.15 by induction. (Note that a field is trivially a Hilbert ring: every prime ideal is maximal.)

Similarly, we show (2) by induction, using 11.15: n = 1 : if K is algebraically closed, every maximal ideal of K[x] is generated by a monic linear polynomial.

Now let M be a maximal ideal of K[x_{1}, …, x_{n}] and P = M ∩ K[x_{1}, …, x_{n−1}]. P is a G-ideal by 11.12, and hence maximal by (1). P = (x_{1}−a_{1}, …, x_{n−1}−a_{n−1}), by induction hypothesis, therefore K[x_{1}, …, x_{n−1}]/P ≃ K, and every irreducible polynomial of (K[x_{1}, …, x_{n−1}]/P)[x_{n}] has the form x_{n}−(a_{n}+P), for some a_{n} ∈K.

Applying part (2) of 11.15 to R = K[x_{1}, …, x_{n−1}], M is generated by x_{1} − a_{1}, …, x_{n−1} − a_{n−1} and x_{n} − a_{n}. □

11.17 Corollary (Version of Hilbert’s Nullstellensatz). Let K be an algebraically closed field and I ⊴ K[x_{1}, …, x_{n}]. Then \sqrt{I} is the intersection of all ideals of the form M_{a} = {f ∈ K[x_{1}, …, x_{n}] | f(a) = 0} (with a ∈ K^{n}) containing I.

11.18 Corollary (Hilbert’s Nullstellensatz). Let K be an algebraically closed field and n ∈ ℕ. If f_{1}, …, f_{m}, g are polynomials in K[x_{1}, …, x_{n}] such that g(a) = 0 for all a ∈ K^{n} for which f_{i}(a) = 0 for 1 ≤ i ≤ m then g ∈ \sqrt{(f_{1}, …, f_{m})} .

∗ ∗ ∗ classical proof of Hilbert’s Nullstellensatz ∗ ∗ ∗

11.19 Lemma. Let K ⊆ F be fields. If F = K[z_{1}, …, z_{n}] (F is generated by z_{1}, …, z_{n} as a ring over K) then z_{1}, …, z_{n} are algebraic over K.

Proof. Induction on n. If x is transcendental over K then the polynomial ring K[x] is not a field (every polynomial of positive degree is a non-unit).

Now let F = K[z_{1}, …, z_{n}], n > 1. F = K(z_{1})[z_{2}, …, z_{n}], so by induction hypothesis z_{2}, …, z_{n} are algebraic over K(z_{1}). There exists an element r ∈ K[z_{1}] (r = g(z_{1}) for some g ∈ K[x]) such that rz_{2}, …, rz_{n} are integral over K[z_{1}]. Therefore, for every f ∈ K[z_{1}, …, z_{n}] there exists m ∈ ℕ such that r^{m}f is integral over K[z_{1}]. In particular, this holds for every f ∈ K(z_{1}) ⊆ K[z_{1}, …, z_{n}].

Suppose z_{1} is transcendental over K. Then K[z_{1}] is a unique factorization domain with infinitely many irreducibles. K[z_{1}] is integrally closed in its quotient field K(z_{1}). By the previous paragraph, every element of K(z_{1}) has a representation as a fraction whose denominator is a power of a fixed polynomial g(z_{1}). This is false. Therefore, z_{1} is algebraic over K. □

11.20 Theorem (weak Nullstellensatz). Let K be an algebraically closed field. Then every maximal ideal of K[x_{1}, …, x_{n}] is of the form (x_{1} − a_{1}, …, x_{n} − a_{n}) for some a_{1}, …, a_{n} ∈ K.

Proof. Let M be a maximal ideal of K[x_{1}, …, x_{n}], and F = K[x_{1}, …, x_{n}]/M the residue field. The canonical projection π: K[x_{1}, …, x_{n}] ⟶ K[x_{1}, …, x_{n}]/M, restricted to K is injective, so the elements k+M (with k ∈K) form an isomorphic copy of K in F, and F is generated as a ring over K by the images of the x_{i} under π, F = K[y_{1}, …, y_{n}], with y_{i} = x_{i} + M.

By 11.19, y_{1}, …, y_{n} are algebraic over K. As K is algebraically closed, the only monic irreducible polynomials in K[x] are linear, so each y_{i} is a zero of x−a_{i} for some a_{i} ∈ K, i.e., (x_{i} + M) − (a_{i} + M) = 0 + M. This implies x_{i} − a_{i} ∈ M for 1 ≤ i ≤ n. Since the ideal generated by x_{1} − a_{1}, …, x_{n} − a_{n} is maximal, it must be M. □

A really neat argument by Rabinowitsch shows that the Nullstellensatz for n variables follows from the weak Nullstellensatz for n + 1 variables:

11.21 Theorem. Let K be an algebraically closed field, and f_{1}, …, f_{m}, f polynomials in K[x_{1}, …, x_{n}] such that f(a) = 0 for every common zero a ∈ K^{n} of

f_{1}, …, f_{m}. Then f ∈ \sqrt{(f_{1}, …, f_{m})} .

Proof. Let f_{1}, …, f_{m}, f ∈ K[x_{1}, …, x_{n}] as in the theorem. The polynomials f_{1}, …, f_{m} and g = 1 − x_{n+1}f in K[x_{1}, …, x_{n+1}] have no common zero in K^{n+1}. By the weak Nullstellensatz they are not contained in any maximal ideal of K[x_{1}, …, x_{n+1}], so there exist h_{1}, …, h_{m}, h ∈ K[x_{1}, …, x_{n+1}] such that

f_{1}h_{1} + … + f_{m}h_{m} + (1 − x_{n+1}f)h = 1.

We now substitute 1/f for x_{n+1} and multiply both sides by f^{r}, where r is the maximal degree in x_{n+1} of any h_{i}. This gives f_{1}g_{1} + … + f_{m}g_{m} = f^{r}, where g_{1}, …, g_{m} ∈R[x_{1}, …, x_{n}], so f is in the radical of the ideal generated by f_{1}, …, f_{m} in K[x_{1}, …, x_{n}]. □

12. A little point-set topology.

∗ ∗ ∗ Open and closed sets ∗ ∗ ∗

12.1 Definition. A subset τ of the power set P(X) of a set X is called a topology on X if the following axioms hold: (O1) ∅ ∈ τ and X ∈ τ (O2) S, T ∈ τ ⟹ S ∩ T ∈ τ (O3) S_{i} ∈ τ for all i ∈ I (I an arbitrary index set) ⟹ ⋃_{i∈I} S_{i} ∈ τ (X, τ) is then called a topological space. The members of τ are called open sets.

If (X, τ) is a topological space, then the complements of open sets, A = X \ S with S ∈ τ, are called closed sets. The collection of closed sets C = {X \O | O ∈ τ} satisfies (C1) ∅ ∈ C and X ∈ C. (C2) A, B ∈ C ⟹ A ∪ B ∈ C. (C3) A_{i} ∈ C for all i ∈ I (I an arbitrary index set) ⟹ ⋂_{i∈I} A_{i} ∈ C

Conversely, if C ⊆ P(X) satisfys C1–C3, then the complements of elements of C form a topology on X, whose closed sets are precisely the elements of C.

12.2 Example: Zariski topology on the spectrum of a ring. Let R be a commutative ring. A topology on the spectrum of R (Spec(R) = {P | P a prime ideal of R}) is defined by specifying its closed sets as sets of prime ideals of the form

V (I) = {P ∈ Spec(R) | P ⊇ I},

for some ideal I of R.

There’s nothing to prevent sets from being open and closed at the same time. Sets both open and closed are often called clopen.

12.3 Definition. Let (X, τ) be a topological space.

A collection B of open sets such that every open set is a union of elements of B is called a basis of the topology τ.

A collection S of open sets such that every open set is a union of finite intersections of elements of S is called a subbasis of the topology τ.

Topologies can be defined by specifying a basis or subbasis: If B ⊆ P(X) is closed with respect to finite intersections, then the unions of (arbitrarily many) members of B form a topology on X, of which B is a basis.

If S ⊆ P(X) is any collection of subsets of X, then arbitrary unions of finite intersections of members of S form a topology on X, of which S is a subbasis.

12.4 Example: Order Topology. If (X, ≤) is a totally ordered set, then order topology on X is defined by specifying “open rays”, i.e., sets of the form (a, ∞) = {x ∈ X | a < x} and (−∞, b) = {x ∈ X | x < b}, for a, b ∈ X as a subbasis.

If (X, ≤) has neither a maximal nor a minimal element, then “open intervals” (a, b) = {x ∈ X | a < x < b} form a basis of order topology. (If X does have a maximal or minimal element, then open rays of the form (a, ∞), or (−∞, b), respectively, have to be added to the open intervals to get a basis.)

12.5 Definition. If (X, τ) is a topological space and Y a subset of X then Y inherits a topological structure from X (called subspace topology) through the convention: a subset U of Y is open (in Y ) iff there exists an open subset O of X with U = O ∩ Y .

If Y is an open subset of X then U ⊆ Y is open in Y if and only it is open in X; if Y is a closed subset of X then A ⊆ Y is closed in Y if and only it is closed in X.

12.6 Remark: If (X, ≤) is a totally ordered set and Y ⊆ X, then Y inherits a topology from the order topology of X, and at the same time Y inherits an order relation from X which makes (Y, ≤) a totally ordered set, for which order topology may be defined. These two topologies on Y in general do not agree. (Examples can be found among subsets of the real numbers.)

∗ ∗ ∗ Neighborhoods and neighborhood bases ∗ ∗ ∗

Perhaps a more intuitive approach to topology is through neighborhoods of a point, which (as sets containing an open ball around the point) are already familiar from the study of metric spaces.

12.7 Definition. (∗) Let (X, τ) be a topological space and p ∈ X. A neighborhood of p is a set U such that there exists an open set O with p ∈ O ⊆ U. The set of all neighborhoods of a point p is called the neighborhood filter of p. We will denote it by U(p).

12.8 Remark: Let (X, τ) be a topological space and p ∈ X. The neighborhood filter of p has the properties (U1) ∀U ∈ U(p) p ∈ U. (U2) U, V ∈ U(p) ⟹ U ∩ V ∈ U(p)

(U3) U ∈ U(p) and V ⊇ U ⟹ V ∈ U(p) (U4) ∀U ∈ U(p) ∃V ∈ U(p) ∀v ∈ V U ∈ U(v). Also,

(U)  O ∈ τ ⟺ ∀p ∈ O ∃U ∈ U(p) U ⊆ O

Conversely, given a set X and for each p ∈ X a set U(p) ⊆ P(X) such that U1–U4 hold, we can define a topolgy τ on X by (U), and, what is more, the neighborhood filter of each point in the resulting topology τ is exactly the U(p) we started out with.

As with metric spaces, it suffices to know a system of “basic” neighborhoods of a point – with the property that every neighborhood contains one of them – to know all neighborhoods.

12.9 Definition. Let (X, τ) be a topological space. A collection B(p) ⊆ U(p) of neighborhoods of p is called a neighborhood basis of p if for every U ∈ U(p) there exists B ∈ B(p) with B ⊆ U.

If (X, τ) is a topological space, and for each p ∈ X, B(p) is a neighborhood basis, then, for every p ∈ X (B1) ∀B ∈ B(p) p ∈ B. (B2) U, V ∈ B(p) ⟹ ∃B ∈ B(p) B ⊆ U ∩ V (B3) ∀U ∈ B(p) ∃V ∈ B(p) ∀v ∈ V ∃B ∈ B(v) B ⊆ U).

Also,

(B)  O ∈ τ ⟺ ∀p ∈ O ∃B ∈ B(p) B ⊆ O

Conversely, if we are given for every p ∈ X a collection B(p) of subsets of X satisfying B1–B3, we can define a topology on X by (B), and in this topology B(p) will be a neighborhood basis of p.

12.10 Definition. A topological space in which every point has a countable neighborhood basis is said to satisfy the first countability axiom.

12.11 Example: Let (X, d) be a metric space, and B_{ε}(p) = {x ∈ X | d(p, x) < ε} (for ε > 0, p ∈ X) the open ball of radius ε around p. If we define B(p) as the set of all B_{ε}(p) with ε > 0, then B1–B3 hold. In other words, every metric induces a topology in which the open ε-balls with center p form a neighborhood basis of p. Actually, countably many balls B_{\frac{1}{n}} (p), n ∈ ℕ, already form a neighborhood basis of p in this topology. We see that every metric space satisfies the first countability axiom.

12.12 Example: I-adic topology Let R be a commutative ring and I an ideal of R. I-adic topology on R is defined by specifying a neighborhood base of r ∈ R:

B(r) = {r + I^{n} | n ∈ ℕ}.

Note that these basic neighborhoods are both open and closed.

∗ ∗ ∗ Closure and interior ∗ ∗ ∗

12.13 Definition. Let A be a subset of a topological space X. The closure of A, denoted Ā, is defined to be the intersection of all closed sets containing A.

Clearly, Ā is a closed set containing A and every closed set B that contains A also contains Ā.

The closure operator A ↦ Ā on P(X) has the properties: (A1) ¯∅ = ∅. (A2) A ⊆ Ā. (A3) ¯ Ā = Ā. (A4) Ā ∪ B̄ = \overline{A ∪ B}.

Conversely, any operator A ↦ Ā on the power set of a set X with properties A1–A4 may be used to define a topology on X by declaring the closed sets to be precisely the sets of the form Ā for some A ∈ P(X). (To see that these sets satisfy C3, first note that A4 implies A ⊆ C ⟹ Ā ⊆ C̄. From this we get that \overline{⋂_{i∈I} Ā_{i}} ⊆ Ā_{i} for all i ∈ I, i.e., \overline{⋂_{i∈I} Ā_{i}} ⊆ ⋂_{i∈I} Ā_{i} and the reverse inclusion is just A2.)

12.14 Definition. Let (X, τ) be a topological space and A ∈ P(X) then the open interior (or interior for short) of A, denoted A^{∘}, is the union of all open sets contained in A.

Clearly, A^{∘} is open and A^{∘} ⊆ A. Also, (X \ A)^{∘} = X \ Ā. (Note that the interior of a non-empty set may be empty.)

∗ ∗ ∗ Boundry points, accumulation points and such ∗ ∗ ∗

The following properties of the closure of a set (which we defined as the intersection of all closed sets containing it) and the interior of a set (which we defined as the union of all open sets contained in it) are often used as definitions:

12.15 Lemma.

• Ā consists of precisely those points p ∈ X such that U ∩ A ≠ ∅ for every neighborhood U ∈ U(p) and

• A^{∘} consists of precisely those points p∈X such that there exists a neighborhood U ∈ U(p) with U ⊆ A.

12.16 Definition. Let (X, τ) be a topological space, p ∈ X and A ⊆ X then

• p is called a boundary point of A if for every neighborhood U ∈ U(p) both A ∩ U ≠ ∅ and (X \ A) ∩ U ≠ ∅. The set of all boundary points of A is the boundary of A, denoted by δA.

• p is called an interior point of A if p ∈ A^{∘}, i.e., if there exists a neighborhood U ∈ U(p) with U ⊆ A.

• p is called an accumulation point of A if every neighborhood of p contains an element of A other than p.

• p is called an isolated point of A if there exists a neighborhood U of p with U ∩ A = {p}. Note that interior points of A and isolated points of A are necessarily in A,

while boundary points and accumulation points may or may not belong to A.

By purely logical arguments we see that every set A induces a partition of X into three disjoint parts in two different ways ( ˙∪ denotes disjoint union):

(i) X = A^{∘} ˙∪ δA ˙∪ (X \ A)^{∘} and (ii) X = {isolated points of A} ˙∪{accumulation points of A} ˙∪(X \ A)^{∘}.

Also, (iii) Ā = A^{∘} ˙∪ δA and (iv) Ā = {isolated points of A} ˙∪ {accumulation points of A} The last two partitions of Ā follow from the characterization of Ā as the set

of those p such that A ∩ U ≠ ∅ for every U ∈ U(p). They are incomparable in the sense that all four combinations of belonging to one set of one partition and one set of the other are possible, an accumulation point of A can be either an interior point or a boundary point of A, etc. Also, we have seen that

(v) Ā = A ∪ δA and (vi) Ā = A ∪ {accumulation points of A} Unlike (iii) and (iv), the unions (v) and (vi) are in general not disjoint.

∗ ∗ ∗ Continuous functions ∗ ∗ ∗

12.17 Definition. Let (X, τ) and (Y, τ′) be topological spaces. A function f: X →Y is called continuous if f^{−1}(O) is open for every open set O ⊆ Y .

12.18 Remark: Inverse image commutes with arbitrary unions and intersections

f^{−1}(⋃_{i∈I} S_{i}) = ⋃_{i∈I} f^{−1}(S_{i}) and f^{−1}(⋂_{i∈I} S_{i}) = ⋂_{i∈I} f^{−1}(S_{i}).

Therefore, for f: X → Y to be continuous, it suffices that f^{−1}(O) be open for all O in some fixed subbasis of Y . Also,

f^{−1}(Y \ S) = X \ f^{−1}(S).

Therefore, f: X → Y is continuous if and only if f^{−1}(A) is closed for every closed set A ⊆ Y .

In terms of neighborhoods, a function f: X → Y is continuous, if and only if for every x ∈ X, for every U ∈ U(f(x)) there exists a V ∈ U(x) with f(V) ⊆ U. The familiar ε-δ-definition of continuous functions is easily seen to be the specialization to metric spaces of this topological characterization.

12.19 Definition. Let (X, τ) and (Y, τ′) be topological spaces. A function f: X →Y is called open if f(O) is open for every open set O ⊆ X.

A bijective function both open and continuous is called a homeomorphism.

A topology τ_{1} on X is called stronger (or finer) than another topology τ_{2} on the same set X if τ_{1} ⊇ τ_{2} (every τ_{2}-open set is τ_{1}-open); τ_{2} is then called weaker or coarser than τ_{1}. Two trivial topologies exist on every set X: discrete topology τ = P(X) (the finest topology on X) and chaotic topology τ = {∅, X} (the coarsest topology on X).

If τ_{1} and τ_{2} are two topologies on a set X then τ_{1} is stronger than τ_{2} iff id_{X} : (X, τ_{1}) → (X, τ_{2}) is continuous; τ_{1} is weaker than τ_{2} iff id_{X} : (X, τ_{1}) → (X, τ_{2}) is open.

∗ ∗ ∗ Connectedness ∗ ∗ ∗

12.20 Definition. A topological space X is connected, if, whenenver O_{1} and O_{2} are open sets with O_{1} ∪ O_{2} = X and O_{1} ∩ O_{2} = ∅, it follows that O_{1} = ∅ or O_{2} = ∅. A subset Y of X is connected if it is conneced in subspace topology.

12.21 Exercise. If X is connected and f: X → Y continuous, then f(X) is connected.

12.22 Lemma. If X_{i} is a connected subset of X for every i ∈ I (I an arbitrary index set) and ⋂_{i∈I} X_{i} ≠ ∅ then ⋃_{i∈I} X_{i} is connected.

Proof. Easy exercise. □

12.23 Lemma and Definition. The following relation ∼ is an equivalence relation on X: x ∼ y if and only if there exists a connected subset C of X with x, y ∈ C.

The equivalence classes with respect to ∼ are called the connected components of X.

From the above definition it is clear that the connected components of X form a partition of X. Also, the component of x ∈ X is the union of all connected subsets of X containing x, and it is therefore the unique largest connected subset of X containing x.

12.24 Lemma. If Y is a connected subset of X then every set C with Y ⊆ C ⊆ \overline{Y} is connected.

Proof. Exercise. □

12.25 Definition. A topological space X is locally connected if every x ∈ X has a neighborhood basis consisting of connected neighborhoods.

12.26 Lemma. The connected components of a topological space X are closed sets. If every x ∈ X has a connected neighborhood (in particular, if X is locally connected) then they are also open.

Proof. By the lemma above, the closure of a connected component is again connected and therefore contained in the component. If a point x possesses a connected open neighborhood U_{x} then the compoment of x (being the union of all connected sets containing x) contains U_{x}. □

12.27 Definition. A topological space X is totally disconnected if it doesn’t contain any connected set with more than one element; equivalently, if its connected components are singletons.

∗ ∗ ∗ Filters ∗ ∗ ∗

12.28 Definition. Let X be a set. A filter on X is a set F ⊆ P(X) with the properties

(1) ∅ ∉ F

(2) If A, B ∈ F then A ∩ B ∈ F.

(3) If A ∈ F and X ⊇ C ⊇ A then C ∈ F.

12.29 Definition. Let F and G be filters on X. We say that F is finer than G (or, equivalently, G is coarser than F) if F ⊇ G.

12.30 Definition. An ultrafilter on X is a filter F with the property:

∀S ⊆ X : S ∈ F ∨ (X \ S) ∈ F.

12.31 Definition. If F is a filter on X then a subset F′ of F is called a filter base of F if ∀F ∈ F ∃F′ ∈ F′ with F′ ⊆ F.

12.32 Remark: If F is a filter then F′ ⊆ F is a filter base of F if and only if F consists precisely of all supersets of elements of F′. We can easily give a criterion for a collection of sets F′ ⊆ P(X) to be the base of a filter on X. The set of supersets of elements of F′ is a filter if and only if F′ has the finite intersection property, that is, the intersection of any finite number of sets in F′ is non-empty.

12.33 Lemma. If F is a filter on X containing neither S nor X \ S, then there exists a filter F_{1} ⊇ F ∪ {S} and a filter F_{2} ⊇ F ∪ {(X \ S)}.

Proof. No F ∈ F is contained in S (otherwise S would be in F), and likewise, no F ∈ F is contained in X \S. Therefore F ∩(X \S) ≠ ∅ for all F ∈ F and F ∩S ≠ ∅ for all F ∈ F. Now take F ∪ {S} as a filter base for F_{1} and F ∪ {(X \ S)} as a filter base for F_{2}. □

12.34 Lemma. A filter is maximal with respect to refinement (i.e., inclusion) if and only if it is an ultrafilter.

Proof. It is clear that no sets can be added to an ultrafilter without violating proprty (1) or (2) in the definition of a filter. Conversely, 12.33 shows that a filter that is not an ultrafilter has a proper refinement. □

12.35 Lemma. For every filter F on X there exists an ultrafilter on X finer than F. F is the intersection of all ultrafilters on X finer than F.

Proof. Consider the set S of all filters on X containing F, ordered by inclusion. Since the union of a chain of filters is again a filter, every chain in S has an upper bound in S. By Zorn’s Lemma, there exists a maximal element in S, which is an ultrafilter containing F, by 12.34. By 12.33, the intersection of all ultrafilters containing F contains no other sets than the elements of F. □

∗ ∗ ∗ Nets ∗ ∗ ∗

12.36 Definition. A directed set is a set I With a binary relation ≤ such that ∀i, j, k ∈ I

(1) i ≤ i.

(2) if i ≤ j and j ≤ k then i ≤ k.

(3) ∃n ∈ I with n ≥ i and n ≥ j.

Note that we do not require anti-symmery.

12.37 Definition. A net in X is a function from a directed set to X, ψ: I → X, usually written (like a sequence) as a list of values indexed by arguments, (x_{i})_{i∈I}.

12.38 Definition. Let ψ: I → X, written as (x_{i})_{i∈I}, be a net in X, J a directed set and and φ: J → I an increasing function that is cofinal in I, that is,

∀i, i′ ∈ I (i ≤ i′ ⟹ φ(i) ≤ φ(i′))
∀i ∈ I ∃k ∈ K : φ(k) ≥ i.

Then the composition of maps ψ ∘ φ: J → X is called a subnet of (x_{i})_{i∈I}, and is written (x_{i_{j}})_{j∈J}.

12.39 Definition. Let (x_{i})_{i∈I} be a net in X and S ⊆ X. We say that (x_{i}) is eventually in S if there exists n ∈ I such that x_{i} ∈ S for all i ≥ n. We say that (x_{i}) is frequently in S, if for all n ∈ I there exists i ∈ I with i ≥ n and x_{i} ∈ S.

12.40 Remark: A set of the form {x_{i} | i ≥ n} for some n ∈ I is called a tail of the net (x_{i})_{i∈I}. A net is eventually in a set S if and only if some tail is contained in S; it is frequently in S if and only if all its tails intersect S nontrivially.

12.41 Definition. An ultranet in X is a net such that for every subset S of X, the net is eventually in S or eventually in X \ S.

12.42 Exercise. Let f: X → Y be any function. If (x_{i})_{i∈I} is an ultranet in X then (f(x_{i}))_{i∈I} is an ultranet in Y . If F is an ultrafilter on X then {f(F) | F ∈ F} is an ultrafilter on f(X) and {S ⊆ Y | ∃F ∈ F : f(F) ⊆ S} is an ultrafilter on Y .

∗ ∗ ∗ Convergence ∗ ∗ ∗

12.43 Definition. Let F be a filter on X, x ∈ X, and U(x) the neighborhood filter of x.

F converges to x if and only if U(x) ⊆ F. In this case, x is called a limit point of F. x is a cluster point of F if and only if F ∩ U ≠ ∅ for all F ∈ F and all U ∈ U(x) (or equivalently, if x ∈ F̄ for all F ∈ F).

12.44 Proposition. Let G ⊆ F be filters on X and x ∈ X.

(1) If G converges to x then the finer filter F converges to x.

(2) If x is a cluster point of F then x is a cluster point of the coarser filter G.

Proof. Follows immediately from the definition of filter convergence and cluster points. □

12.45 Definition. Let x ∈ X and U(x) the neighborhood filter of x.

A net in X converges to x if and only if for every U ∈ U(x), the net is eventually in U. In this case, x is called a limit point of the net. x is a cluster point of a net on X if and only if for every U ∈ U(x), the net is frequently in U.

12.46 Proposition. Let (x_{n_{k}}) be a subnet of the net (x_{n}) on X and x ∈ X. If (x_{n}) converges to x then the subnet (x_{n_{k}}) converges to x. If x is a cluster point of the subnet (x_{n_{k}}) then x is a cluster point of (x_{n}).

Proof. Follows immediately from the definition of subnet. □

12.47 Proposition. An ultrafilter converges against each of its cluster points. Similarly, an ultranet converges against each of its cluster points.

Proof. Suppose F is a filter and x ∈ X such that for all F ∈ F and all U ∈ U(x), U ∩ F ≠ ∅. Then for all U ∈ U(x), (X \ U) ∉ F. If F is an ultrafilter, U ∈ F for all U ∈ U(x) follows. The case of nets is similar. □

The following constructions of a net from a filter and a filter from a net often allow to translate statements about filters to statements about nets and vice versa:

12.48 Lemma and Definition. Let (x_{n})_{n∈N} be a net on X and x ∈ X. The filter constructed from (x_{n})_{n∈N} is defined by taking the set of ends {x_{n} | n ≥ n_{0}} for n_{0} ∈ N as a filter basis.

Then the filter constructed from (x_{n}) converges to x if and only if (x_{n}) converges to x. Also, x is a cluster point of the filter constructed from (x_{n}) if and only if x is a cluster point of (x_{n}).

Proof. Easy exercise. □

12.49 Lemma and Definition. Let F be a filter on X and x ∈ X. The net constructed from F is indexed by the set I = {(F, y) | F ∈ F, y ∈ F } with (F, y) ≥ (F′, y′) :⟺ F ⊆ F′; and x_{(F,y)} = y.

Then the net constructed from F converges to x if and only if F converges to x. Also, x is a cluster point of the net constructed from F if and only if x is a cluster point of F.

Proof. Easy exercise. □

12.50 Exercise. Let F be a filter on X and x ∈ X. For each F ∈ F choose x_{F} ∈ F. Does the net (x_{F})_{F ∈F} (indexed by F directed by F′ ≥ F :⟺ F′ ⊆ F), also satisfy the equivalences of 12.49?

12.51 Remark: For a filter F to converge to x ∈ X, it suffices that F contains, for a fixed subbasis S of the topology, every Y ∈ S with x ∈ Y . Similarly, for a net to converge to x, it suffices that it is eventually in Y for every Y ∈ S with x ∈ Y .

Proof. Easy exercise. □

∗ ∗ ∗ Compactness ∗ ∗ ∗

12.52 Definition. Let X be a topological space and Y ⊆ X. An open cover of Y is a set C of open sets such that Y ⊆ ⋃_{O∈C} O. Y is compact if every open cover of Y admits a finite subcover, that is, there exist O_{1}, …, O_{n} ∈ C with Y ⊆ O_{1} ∪ … ∪ O_{n}.

Be aware that many authors require compact sets to be Hausdorff, and call our notion of compact “quasi-compact”.

12.53 Exercise. If X is compact and f: X → Y continuous, then f(X) is compact.

12.54 Theorem. Let X be a topoological space, and S a subbasis of the topology. The following are equivalent:

(1) X is compact, i.e., every open cover of X has a finite subcover.

(2) Every cover of X consisting of elements of S has a finite subcover.

(3) Every ultrafilter on X converges to some x ∈ X.

(4) Every filter on X has a cluster point x ∈ X.

Proof. (1 ⇒ 2) a fortiori.

(2 ⇒ 3) Suppose the ultrafilter U doesn’t converge. For every x ∈ X choose U_{x} ∈ S with x ∈ U_{x} and U_{x} ∉ U (possible by 12.51) and let A_{x} = X \ U_{x}. Then A_{x} ∈ U. Also, {U_{x} | x ∈ X} covers X so there exists a finite set Y ⊆ X with ⋃_{x∈Y} U_{x} = X. Therefore ∅ = ⋂_{x∈Y} A_{x} ∈ U, a contradiction.

(3 ⇒ 4) By 12.35, every filter F on X can be refined to an ultrafilter. This ultrafilter converges to some x ∈ X and then x is a cluster point of F by 12.44.

(4 ⇒ 1) Suppose C is an open cover of X that has no finite subcover. Then we may use {X \ O | O ∈ C} as base for a filter F on X. Let x ∈ X be a cluster point of F and O_{x} ∈ C with x ∈ O_{x}. Then F ∩ O_{x} ≠ ∅ for all F ∈ F. But (X \ O_{x}) ∈ F, a contradiction. □

12.55 Corollary. Let X be a topologial space. Let B ⊆ P(X) be a set of closed sets such that every closed subset of X is representable as an arbitrary intersection of finite unions of elements of B (or, equivalently, such that S = {(X \ A) | A ∈ B} is a subbasis of the topology on X). Then the following are equivalent.

(1) X is compact. (1’) For every set A ⊆ P(X) of closed subsets of X it is true that: if ⋂_{A∈A} A = ∅

then there exists a finite subset {A_{1}, …, A_{n}} ⊆ A with ⋂_{1≤i≤n} A_{i} = ∅. (2’) Like (1’), but restricted to sets of closed sets A ⊆ B. (3’) Every ultranet in X converges to some x ∈ X. (4’) Every net in X has a cluster point x ∈ X.

Proof. (1 ⇔ 1′) and (2 ⇔ 2′) by de Morgan. (3 ⇔ 3′) and (4 ⇔ 4′) by 12.48 and 12.49. □

12.56 Exercise. If X is Hausdorff, then we can separate disjoint compact sets C_{1}, C_{2} by open sets, i.e., there are open sets O_{1}, O_{2} such that C_{1} ⊆ O_{1}, C_{2} ⊆ O_{2} and O_{1} ∩ O_{2} = ∅. (First show that we can separate a compact set C from a point x ∈ X \ C by open sets.)

12.57 Exercise.

(i) Every closed subset of a compact space is compact. (ii) If X is Hausdorff, then every compact subset of X is closed.

∗ ∗ ∗ Product topology ∗ ∗ ∗

12.58 Definition. For i ∈ I (an arbitrary index set) let X_{i} be a topological space. Product topology on the cartesian product ∏_{i∈I} X_{i} is defined by a subbasis consisting of all sets of the form

S(j, O_{j}) = {(x_{i})_{i∈I} ∈ ∏_{i∈I} X_{i} | x_{j} ∈ O_{j}},

for some j ∈ I, and some O_{j} open ⊆ X_{j}. (Equivalently, the sets O_{j} could be restriced to members of a given basis or subbasis of the topology of X_{j}.)

Remark: the finer topology on ∏_{i∈I} X_{i} given by the basis

B = {∏_{i∈I} O_{i} | ∀i O_{i} open ⊆ X_{i}}

is called box topology.

Note that the projections p_{j}: (∏_{i∈I} X_{i}) → X_{j}, p_{j}((x_{i})_{i∈I}) = x_{j}, are continuous, both for product topology and for box topology.

12.59 Proposition. A net (x_{λ})_{λ∈Λ} in X = ∏_{i∈I} X_{i} converges to y = (y_{i})_{i∈I} in product topology, if and only if for every i ∈ I, its projection to X_{i}, (p_{i}(x_{λ}))_{λ∈Λ} converges to y_{i} in X_{i}.

Proof. Easy exercise. □

12.60 Theorem (Tychonoff). X = ∏_{i∈I} X_{i} (with product topology) is compact if and only if each X_{i} is compact.

Proof. Easy direction: if ∏_{i∈I} X_{i} is compact, then for each i, X_{i} is compact as the image of ∏_{i∈I} X_{i} under the projection onto the i-th coordinate, which is continuous. Conversely, to show compactness of X = ∏_{i∈I} X_{i}, consider an ultranet on X. The projection to the i-the coordinate is an ultranet on X_{i}, which converges, since X_{i} is compact. As all coordinates of the ultranet converge, the ultranet itself converges. □

We give another proof of Tychonoff’s theorem using a different criterion for compactness. Proof. Easy direction: if ∏_{i∈I} X_{i} is compact and C is an open cover of X_{i}, then {S(i, O) | O ∈ C} is an open cover of X, which has a finite subcover S(i, O_{1}),…, S(i, O_{n}). Clearly, O_{1},…, O_{n} cover X_{i} and constitute a finite subcover of C.

Now assuming compactness of each X_{i}, to show compactness of X = ∏_{i∈I} X_{i}, consider a cover C by subbasis elements S(i, O). There must be some coordinate j ∈ I such that the open sets O occurring in sets S(j, O) ∈ C cover X_{j}. (Otherwise, by the axiom of choice, there would be a point (x_{i})_{i∈I} such that for all i, x_{i} is in none of the sets O with S(i, O) ∈ C, and therefore (x_{i})_{i∈I} is not covered by C.) As X_{j} is compact, there is a finite cover of X_{j} by open sets O_{1}, … O_{n} with S(j, O_{k}) ∈ C. Clearly then S(j, O_{1}), …, S(j, O_{n}) cover X. □