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MA 252 Notes: Commutative Algebra (Distilled from [Atiyah-Macdonald])

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MA 252 Notes: Commutative Algebra (Distilled from [Atiyah-Macdonald]) Integral dependence MA 252 notes: Commutative algebra (Distilled from [Atiyah-MacDonald]) Dan Abramovich Brown University February 19, 2017 Abramovich MA 252 notes: Commutative algebra 1 / 24 Integral dependence Integral emlements Definition Let A ⊂ B. An element x 2 B is integral over A if there is a monic n n−1 polynomial f (t) = t + a1t + ··· + an 2 A[t] such that f (x) = 0. Examples p x = 2 2 C is integral over Z. x 2 Q is integral over Z if and only if x 2 Z (write x = m=n in rediced form, clear denominators in f (m=n) and reduce moduli a prime dividing m). p 1+ 5 2 x = 2 2 C is integral over Z, since α − α − 1 = 0, so appearance can be misleading. Abramovich MA 252 notes: Commutative algebra 2 / 24 Integral dependence Characterization Proposition The following are equivlent: (i) x 2 B integral over A. (ii) The subring A[x] ⊂ B is finite over A. (iii) A[x] is contained in a subring C ⊂ B finite over A. (iv) There is a faithful sub-A[x]-module M ⊂ B which is finite over A. (i))(ii) follows since A[x] = A1 + Ax + ··· + Axn−1 as n n−1 x = −(a1x + ··· + an) etc. (ii))(iii))(iv) are evident. (iv))(i) take the characteristic polynomial f (t) of x on generators of M, and we get f (x) = 0 2 End(M) by Cayley-Hamilton. Abramovich MA 252 notes: Commutative algebra 3 / 24 Integral dependence The set of integral elements Corollary If x1;:::; xn 2 B are integral over A then any A[fxi g] ⊂ B is finite. Indeed at each stage the extension is finite, and we have seen the result is finite. Corollary The set C ⊂ B of integral element over A is a subring. Indeed x1 ± x2; x1x2; x1=x2 2 A[x1; x2]. Definition The ring C ⊂ B of integral element over A is the integral closure of A in B. Remark For arbitrary f : A ! B, we consider f (A) ⊂ B. Abramovich MA 252 notes: Commutative algebra 4 / 24 Integral dependence Integral closure Definition Let C ⊂ B be the integral closure of A in B. A ⊂ B is integrally closed if C = A. We say B is integral over A if C = B. Corollary If A ⊂ B ⊂ C with B=A; C=B integral then C=A integral. n n−1 If x 2 C satisfies x + b1x + ··· + bn = 0, then A[b1;:::; bn][x] is finite over A[b1;:::; bn] which is finite over A. Corollary Let C be the integral closure of A in B. Then C is integrally closed in B. Indeed if x 2 B is integral over C then it is integral over A. Abramovich MA 252 notes: Commutative algebra 5 / 24 Integral dependence Integrality and fractions Proposition Say B=A integral. (i) b ⊂ B and a = A \ b then B=b is integral over A=a. (ii) If S ⊂ A multiplicative then B[S−1]=A[S−1] integral. (i) liftx ¯ 2 B=b to x 2 B and use the same equation, modulo a. (ii) For x=s 2 B[S−1] divide the equation of x by sn. Remark The opposite is false. Abramovich MA 252 notes: Commutative algebra 6 / 24 Integral dependence Integrality, fields, and primes Proposition If A ⊂ B are domains, B=A integral. Then B is a field if and only if A a field. n If A a field and y 6= 0 2 B then the equation y + ::: + an = 0 of −1 n−1 smallest degree has an 6= 0, so y = −(y + ··· + an−1)=an. If B a field, x 6= 0 2 A then y = 1=x 2 B has equation m n−1 n−1 y + a1y + ··· + an = 0 so y = −(a1 + ··· + anx ) 2 A. Corollary if q 2 Spec B, B=A integral, p = A \ q. Then q maximal if and only if p maximal. Abramovich MA 252 notes: Commutative algebra 7 / 24 Integral dependence Primes in integral extensions and surjectivity of Spec Corollary B=A integral, q; q0 2 Spec B, q \ A = q0 \ A. Then q = q0. We have Bp=Ap integral. Let m = pAp, a maximal ideal. Let 0 0 0 0 n = qBp, n = q Bp, so n ⊂ n and n \ Ap = n \ Ap = m. By the corollary n; n0 are maximal so n = n0. By the correspondence, q = q0. Theorem If A ⊂ B and B=A integral then Spec B ! Spec A surjective. Consider p 2 Spec A and the diagram A / B . For a maximal α β Ap / Bp ideal n ⊂ Bp, we have n \ Ap is maximal, so n \ Ap = pAp, in −1 −1 −1 particular q = β (n) restricts to α (n \ Ap) = α (pAp) = p: Abramovich MA 252 notes: Commutative algebra 8 / 24 Integral dependence Going up Theorem (Going up) Let B=A integral, p1 ⊂ · · · ⊂ pn ⊂ A primes, q1 ⊂ · · · ⊂ qm ⊂ B primes such that qi \ A = pi ; i = 1;:::; m. Then there is an extended sequence q1 ⊂ · · · ⊂ qn such that qi \ A = pi ; i = 1;:::; n. Suffices to prove: p1 ⊂ p2, p1 = q1 \ A then there is q1 ⊂ q2 such that q2 \ A = p2. Reducing to A=p1 ⊂ B=p2 we have by surjectivity ¯q2 2 Spec B=p2 with ¯q2 \ A=p1 = p¯2. This gives the required q2. Abramovich MA 252 notes: Commutative algebra 9 / 24 Integral dependence Integral closure and fractions Proposition If C is the integral closure of A in B and S ⊂ A multiplicative then C[S−1] is the integral closure of A[S−1] in B[S−1]. We have seen that C[S−1] is integral over A[S−1]. Let the equation of an integral b=s be n n−1 Q x + (a1=s1)x + ··· + (an=sn) = 0. Take t = si and multiply by (st)n, and get an integral equation for bt, so bt 2 C and b=s = (bt)=(st) 2 C[S−1]. Abramovich MA 252 notes: Commutative algebra 10 / 24 Integral dependence Integrally closed domains A domain A is integrally closed if it is integrally closed in its field of fractions K. E.g. every UFD (by the argument of Z). Proposition (Integral closedness is local) Fix a domain A. The following are equivalent: A integrally closed. Ap is integrally closed for all primes p. Am is integrally closed for all maximals m. Let C be the integral closure of A in K. Then Cp is the integral closure of Ap. A (resp. Ap) is integrally closed if and only if A ! C (resp. Ap ! Cp) surjective. Recall that surjectivity (just like injectivity) is a local property. (If we proved just injectivity, complete the proof!) Abramovich MA 252 notes: Commutative algebra 11 / 24 Integral dependence Integral closure of ideals Definition if a ⊂ A ⊂ B then b 2 B is integral over a if it satisfies f (b) = 0 n n−1 with f (t) = t + a1t + ··· an with ai 2 a. The integral closure of a in B is the set of all such. Lemma Let C be the integral closure of A ⊂ B and c = aC. Then the integral closure of a in B is the radical r(c). If x is integral over a then it is integral over A so x 2 C. The equation gives xn 2 aC = c so x 2 r(c). n P If x 2 r(c) then x = ai xi with ai 2 a and xi 2 C. The module n M = A[x1;:::; xn] is finite, and x M ⊂ aM, so taking characteristic polynomial we get that xn is integral over a, so x is integral over a. Abramovich MA 252 notes: Commutative algebra 12 / 24 Integral dependence Integrality and fields Proposition Let A ⊂ B be domains, assume A ⊂ K := A(0) integrally closed, and x 2 B integral over a ⊂ A. Then x is algebraic over K, and its n n−1 minimal polynomial f (t) = t + a1t + ··· + a1 has ai 2 r(a). Since x is integral over A it is algebraic over K. Let L=K be a normal extension containing x, so the conjugates xi 2 L. All xi are also integral over a, since f (t) divides the integrality equation. The coefficients of f (t) are polynimials in xi so integral over a (by the previous result). Since A ⊂ K integrally closed, ai 2 A. By the previous result they are in r(a). Abramovich MA 252 notes: Commutative algebra 13 / 24 Integral dependence Going down Theorem A ⊂ B integral, A; B domains, A ⊂ K integrally closed. A ⊃ p1 ⊃ · · · ⊃ pn and B ⊃ q1 ⊃ · · · ⊃ qm primes, such that qi \ A = pi . Then there is an extended chain B ⊃ q1 ⊃ · · · ⊃ qm · · · ⊃ qn of primes, such that qi \ A = pi . Again it suffices to take n = 2; m = 1. (Localizing at p1 we may assume it is maximal.) Abramovich MA 252 notes: Commutative algebra 14 / 24 Integral dependence Going down - proof There is q2 ⊂ q1 such that q2 \ A = p2. Consider the ring Bq1 . By that controversial proposition it suffices to prove that Bq1 p2 \ A = p2. Write y=s 2 Bq1 p2 with y 2 Bp2 and s 2 B r q1. Now y is r r−1 integral over p2. So its minimal equation y + u1y + ··· over K has coefficients ui 2 r(p2) = p2. If x = y=s 2 Bq1 p2 \ A then s = y=x 2 K so its minimal equation i i over K has coefficients vi = ui =x , hence x vi 2 p2.
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