Advanced Algebra II
Localization
Before we going to the study of dimension theory, we need to recall some basic notion of localization.
Definition 0.1. A subset S ⊂ R is said to be a multiplicative set if (1) 1 ∈ S,
(2) if a, b ∈ S, then ab ∈ S.
Given a multiplicative set, then one can construct a localized ring S−1R which I suppose the readers have known this. In order to be self-contained, I recall the construction:
In R × S, we define an equivalent relation that (r, s) ∼ (r0, s0) if (rs0 − r0s)t = 0 for some t ∈ S. Let rs denote the equivalent class of (r, s). One can define addition and multiplication naturally. The set of all equivalent classes, denoted S−1R, is thus a ring. There is a natural ring homomorphism ı : R → S−1R by ı(r) = r1.
Remark 0.2.
(1) If 0 ∈ S, then S−1R = 0. We thus assume that 0 6∈ S.
(2) If R is a domain, then ı is injective. And in fact, S−1R ,→ F naturally, where F is the quotient field of R.
(3) Let J C S−1R. We will use J ∩ R to denote the ideal ı−1(J).
(If R is a domain, then J ∩ R = ı−1(J) by identifying R as a subring of S−1R).
I would like to recall the most important example and explain their geometrical meaning, which, I think, justify the notion of localization.
Example 0.3. Let f 6= 0 ∈ k[x1, ..., xn] and let S = {1, f, f2...}. The localization S−1k[x1, ..., xn] is usually denoted k[x1, ..., xn]f. This ring can be regarded as ”regular functions” on the open set Uf := Ank−V(f ).
One notices that Uf is of course the maximal open subset that the ring k[x1, ..., xn]f gives well-defined functions.
Example 0.4. Let x = (a1, ..., an) ∈ Ank and mx = (x1− a1, ..., xn− ak) be its maximal ideal. Take S = k[x1, ..., xn] − mx, then the localization is denoted k[x1, ..., xn]mx. It is the ring of regular functions ”near x”.
Recall that for a R-module M, one can also define S−1M which is naturally an S−1R-module. And we have:
Proposition 0.5.
(1) If I C R, then S−1I C S−1R. Moreover, every ideal J C S−1R is of the form S−1I for some I C R.
(2) For J C S−1R, then S−1(J ∩ R) = J.
(3) S−1I = S−1R if and only if I ∩ S 6= ∅.
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(4) There is a one-to-one correspondence between {p ∈ Spec(R)|p ∩ S = ∅} and {q ∈ Spec(S−1R)}.
(5) In particular, the prime ideals of the local ring Rp are in one- to-one correspondence with the prime ideals of R contained in p.
Proof.
(1) If xs,yt ∈ S−1I, that is, x, y ∈ I, then xs + yt = xt+ysst ∈ S−1I.
And if rs ∈ S−1R and xt ∈ S−1I, then rsxt = rxst ∈ S−1I. Hence S−1I is an ideal.
Moreover, let J C S−1R. Let I := ı−1(J) C R. We claim that J = S−1I. To see this, for x ∈ I, x1 ∈ J. Hence xs = x11s ∈ J for all s ∈ S. It follows that S−1I ⊂ J. Conversely, if xs ∈ J, then
x
1 = xss1 ∈ J and x1 = ı(x). So xs ∈ S−1I.
(2) If xs ∈ J ∩ R, then xs = y1 for some y ∈ R , i.e. (x − sy)t0 = 0, for some t0 ∈ S. Then look at yt ∈ S−1(J ∩ R). It’s clear that
y
t = xs1t ∈ J.
Conversely, if xs ∈ J, then x1 ∈ J ∩ R. And hence xs ∈ S−1(J ∩ S).
(3) If x ∈ I ∩ S, then 11 = xx ∈ S−1I. Conversely, if 11 = xs ∈ S−1I, then (x − s)t = 0. Therefore, xt = st ∈ S ∩ I.
(4) For q ∈ Spec(S−1R). It’s clear that q ∩ R = ı−1q ∈ Spec(R).
(q ∩ R) ∩ S = ∅, otherwise q = S−1(q ∩ R) = S−1R which is impossible.
Conversely, let p ∈ Spec(R) and p ∩ S = ∅. We would like to show that S−1p is a prime ideal. First of all, if 11 ∈ S−1p, then
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1 = xs for some x ∈ p. It follows that (x − s)t = 0 for some t ∈ S. Thus st = xt ∈ p which is a contradiction. Therefore, S−1p 6= S−1R.
Moreover, if xsyt ∈ S−1p. Then xyst = xs00 for some x0 ∈ p and s0 ∈ S. Then (xys0 − stx0)t0 = 0 for some t0 ∈ S. Hence xys0t0 ∈ p. It follows that xy ∈ p since s0t0 6∈ p. Thus either x or y in p, and either xs or yt in S−1p.
It remains to show that the correspondence is a one-to-one correspondence. We have seen that
ı−1 : Spec(S−1R) → {p ∈ Spec(R)|p ∩ S = ∅}
is surjective. By (2), it follows that this is injective.
(5) Let S = R − p, then S ∩ q = ∅ if and only q ⊂ p.
¤ But for I C R, then S−1I ∩ R ⊃ I only. Indeed, if x ∈ I C R. Then x = xsx ∈ S−1I ∩ R. Conversely, for x ∈ S−1I ∩ R, then x1 = yt for some y ∈ I. Thus (y − xt)s = 0 for some s, t ∈ S. We can not get y ∈ I in general. However, this is the case if I is prime and S ∩ I = ∅. Thus we have
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Proposition 0.6. If p C R is a prime ideal and S ∩ p = ∅. Then S−1p ∩ R = p.
Corollary 0.7. If R is Noetherian (resp. Artinian), then so is S−1R.
Proof. This is an immediate consequence of (1) of the Proposition. ¤ Example 0.8. Let p ∈ Spec(R), then Rp is a local ring with the unique maximal ideal pRp. To see that, if there is a maximal ideal m. By the correspondence, m = qRp for some q ⊂ p. Thus m ⊂ pRp and thus must be equal.
A ring with a unique maximal ideal is called a local ring. Thus Rp is a local ring.
Proposition 0.9. The operation S−1 commutes with formation of fi- nite sums, product, intersection and radicals.
Proof. These can be checked directly. ¤
Another important feature is that
Proposition 0.10. The operation S−1 is exact. That is, if M0 −→ Mf −→ Mg 00
is an exact sequence of R-module, then
S−1M0 S−→ S−1f −1M S−→ S−1g −1M00 is exact as S−1R-module.
Proof. By the construction, it follows directly that S−1g ◦ S−1f = 0.
It suffices to check that Ker(S−1g) ⊂ Im(S−1f ). If xs ∈ Ker(S−1g), then g(x)s = 0 ∈ S−1M00. That is tg(x) = 0 for some t ∈ S. We have tg(x) = g(tx) = 0. And then tx = f (y) for some y ∈ M0. Therefore,
x s = xt
st = f (y)
st = S−1f (y st).
¤ Corollary 0.11. The operation S−1 commutes with passing to quo- tients by ideals. That is, let I C R be an ideal and ¯S the image of S in R := R/I. Then S¯ −1R/S−1I ∼= ¯S−1R.¯
Proof. By considering 0 → I → R → ¯R → 0 as an exact sequence of R-modules. We have S−1R/S−1I ∼= S−1R as S¯ −1R-modules. We claim that there is a natural bijection from S−1R to ¯¯ S−1R by¯ xs¯ 7→ xs¯¯ which is compatible with all structures.
One can also try to prove this directly. Basically, construct a surjec- tive ring homomorphism S−1R → ¯S−1R and shows that the kernel is¯
S−1I. We leave it as an exercise. ¤
