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). □