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 (10.8), 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}]. □