IV. Commutative rings If noncommutative rings are at the heart of the theory of lin- ear representations of finite groups, then commutative rings under- pin that of algebraic varieties and schemes. In this final stretch of the course, we will cover just enough commutative algebra to reach the Hilbert Nullstellensatz, which will give a bijective correspon- dence between (i) varieties (solution sets of polynomial equations in n variables) over an algebraically closed field k and (ii) radical ideals (which contain f if they contain a power of f ) in the polyno- mial ring k[x1,..., xn]. This is, in some sense, the foundational result of algebraic geometry. It is helpful to have a running example in mind while reading through the definitions and proofs, starting with localization. For some purposes R = Z will do, but R = C[x] (or C[x, y]) gives a more geometric feel to the material, as it comprises the regular (think “al- gebraic holomorphic”) functions on the “complex line” C. So when localization away from an element r 2 R tells us to consider the ring 1 of fractions R[ r ], one can take r = x − a 2 C[x] and think of the ( ) Q x Q functions of the form (x−a)k ( polynomial), which are the regular functions on Cnfag (hence the “away from”). Similarly, when localization at a prime ideal P of R defines RP to r consist of fractions s (r 2 R, s 2 RnP), one can take P = (x − a) and Q(x) − ( ) think of rational functions R(x) with x a - R x . These are regular functions on “C minus everything but a” (by which one means the complement of some finite point set not including a, or an “inverse limit” of such complements). The upshot is that this leaves one with only one maximal ideal, namely (x − a), in C[x](x−a), and quotient- ing by it evaluates these rational functions at a (hence the “at”). 193 194 IV. COMMUTATIVE RINGS IV.A. Localization Let R be a commutative ring. The aim of this section is to con- struct rings of fractions which are “intermediate” between R and its fraction field, in the sense that there are restrictions on the denomi- nators which are allowed, and the result is (in general) not a field. IV.A.1. DEFINITION. A subset S ⊂ R is said to be multiplicative if S contains 1, does not contain 0, and a, b 2 S =) ab 2 S. IV.A.2. EXAMPLE. Given a prime ideal P ⊂ R, RnP is multiplica- k tive (why?). Given an element r 2 R, so is fr gk≥0. These are the most important examples. Given a multiplicative subset S, define an equivalence relation “∼” on R × S by 0 0 0 0 (IV.A.3) (r, s) ∼ (r , s ) () 9 s1 2 S such that s1(s r − sr ) = 0. The RHS is equivalent to rs0 = r0s provided R has no zero-divisors.1 Taking the quotient by (IV.A.3), put R × S (IV.A.4) S−1R := . ∼ By “0” and “1” in S−1R we’ll mean the equivalence classes of (0, 1) and (1, 1), respectively. These are distinct, since s1(1 · 1 − 0 · 1) = 0 is impossible. More generally, denote the equivalence-class of (r, s) by r −1 r r0 0 0 s 2 S R, and write s = s0 to mean that (r, s) ∼ (r , s ). We notice tr r right away that ts = s for any t 2 S, since 1(trs − rts) = 0. Clearly, r it’s consistent with all of this to write r = 1 . − IV.A.5. PROPOSITION. S 1R is a commutative ring, which is a do- main if R is. (In the special case where R is a domain and S = Rnf0g, it is the fraction field of R.) 1Clearly, it would have been a terrible idea to allow 0 2 S, because then everything would be equivalent (and the quotient (IV.A.4) would be zero)! IV.A. LOCALIZATION 195 SKETCH. There are a lot of little checks here, beginning with the r r0 rr0 r r0 well-definedness of multiplication s · s0 := ss0 and addition s + s0 := 0 0 rs +r s r r0 0 . For instance, suppose = ; then for some s 2 S, we have ss s s0 1 0 2 0 2 0 0 0 0 s1(rs0 − sr0) = 0 =) s1frs0(s ) − sr0(s ) + ss0s r − ss0r s g = 0 0 0 0 0 0 0 r0s +r s0 rs +r s r0 r r r =) 0 = 0 =) + 0 = + 0 . So addition is well- s0s ss s0 s s s defined. I leave the distributivity and associativity to you. r r0 rr0 0 Clearly, a product s · s0 = ss0 is zero iff s1rr = 0 for some s1(6= 0). If R is a domain, this only happens if r = 0 or r0 = 0, in which r r0 case s or s0 is zero. For the parenthetical on the fraction field, see [Algebra I, III.F.10]. The natural ring homomorphism −1 fS : R !S R (IV.A.6) r r 7! r (:= 1 ) r 0 is evidently injective iff 1 = 1 =) r = 0, which is equivalent to S containing no zero divisors. If S ⊂ R∗ then (IV.A.6) is also surjective r (hence an isomorphism) because every s already exists on the LHS. Indeed, the whole point of these fraction rings is to turn the elements −1 ∗ 1 of S into units: we have fS (S) ⊂ (S R) since s · s = 1. (This does no good if they already were units!) In fact, (IV.A.6) has the universal property of factoring any map sending S to units: IV.A.7. THEOREM. Let f : R ! T be a homomorphism satisfying f (S) ⊂ T∗. Then there exists a unique homomorphism f¯: S−1R ! T such that f¯ ◦ fS = f . ¯ r −1 ∗ PROOF. Set f ( s ) := f (s) f (r), which we can do since f (s) 2 T . r r0 0 0 This is well-defined, since s = s0 =) s1(rs − sr ) = 0 =) 0 0 0 0 ∗ f (s1) f (rs − sr ) = 0 =) f (rs − sr ) = 0 (since f (s1) 2 T ) =) −1 0 −1 0 f (s) f (r) = f (s ) f (r ). Clearly f¯(fS (r)) = f (r), and checking fS a homomorphism is easy. −1 For the uniqueness, suppose g : S R ! T has g ◦ fS = f . Then 1 1 r 1 −1 1 = g(s · s ) = g(s) · g( s ) =) g( s ) = g( s )g(r) = g(s) g(r) = − f (s) 1 f (r) =) g = f¯. 196 IV. COMMUTATIVE RINGS Next we want to consider ideals in the fraction rings. Given an ideal I ⊂ R, we can define one in S−1R by S−1 I := (I × S)/ ∼, and this is compatible with sums, products, and intersections (of ideals). But it is only interesting if I avoids the multiplicative subset: − − IV.A.8. PROPOSITION. (i) S 1 I = S 1R () S \ I 6= Æ. (ii) Every ideal in S−1R is of the form S−1 I. s −1 PROOF. (i) ( (= ): s 2 S \ I =) 1S−1R = s 2 S I. −1 −1 −1 −1 ı ( =) ): fS (S I) = fS (S R) = R =) 1 = fS (1R) = s for some ı 2 I =) s1(ı − s) = 0 for some s1 2 S =) s1s = s1ı 2 S \ I. −1 −1 (ii) Let J ⊂ S R be an ideal. Then fS (J) =: I is an ideal in −1 −1 −1 r R, and S I = S R · fS (I) ⊂ S R · J ⊂ J. But if s 2 J, then r s r −1 r −1 fS (r) = 1 = 1 · s 2 J =) r 2 fS (J) = I =) s 2 S I. Given the central role of prime ideals in the structure of commu- tative rings, the compatibility of this construction with primality is of particular importance: IV.A.9. LEMMA. If P ⊂ R is a prime ideal with P \S = Æ, then S−1P ⊂ S−1R is prime, and −1 −1 (IV.A.10) fS (S P) = P. P −1 PROOF. First we check (IV.A.10): clearly fS (P) (= 1 ) ⊂ S P. −1 −1 −1 r p Conversely, r 2 fS (S P) =) fS (r) 2 S P =) 1 = s (p 2 P, s 2 S) =) s1(sr − p) = 0 (for some s1 2 S) =) (s1s)r = s1 p 2 P. But s1s 2 S hence2 / P, so we must have r 2 P since P is prime. −1 r r0 −1 rr0 p Now for primality of S P: suppose · 0 2 S P. Then 0 = , s s ss s0 0 0 whence s1s0rr = s1ss p 2 P. Since P is prime (and s1s0 2/ P), we 0 0 r r0 must have rr 2 P hence r 2 P or r 2 P, and then s or s0 belongs to − S 1P. IV.A.11. THEOREM. There is a bijection prime ideals in R ! prime ideals in S−1R disjoint from S IV.A. LOCALIZATION 197 induced by P 7−! S−1P. − PROOF. The assignment P 7! S 1P is injective by (IV.A.10). For −1 −1 the surjectivity, let J ⊂ S R be prime; then P := fS (J) is an ideal with S−1P = J, by the proof of IV.A.8(ii). It remains to show that P is prime: so let ab 2 P. Then fS (ab) = fS (a)fS (b) 2 J =) fS (a) −1 or fS (b) 2 J =) a or b 2 fS (J) = P, done.