Integral dependence

MA 252 notes: (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 ∈ B is integral over A if there is a monic n n−1 polynomial f (t) = t + a1t + ··· + an ∈ A[t] such that f (x) = 0.

Examples √ x = 2 ∈ C is integral over Z. x ∈ Q is integral over Z if and only if x ∈ Z (write x = m/n in rediced form, clear denominators in f (m/n) and reduce moduli a prime dividing m). √ 1+ 5 2 x = 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 ∈ B integral over A. (ii) The 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 ∈ End(M) by Cayley-Hamilton.

Abramovich MA 252 notes: Commutative algebra 3 / 24 Integral dependence The set of integral elements

Corollary

If x1,..., xn ∈ B are integral over A then any A[{xi }] ⊂ 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 ∈ A[x1, x2]. Definition The 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 ∈ 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 ∈ 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 ¯ ∈ B/b to x ∈ B and use the same equation, modulo a. (ii) For x/s ∈ 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 ∈ 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 ∈ A then y = 1/x ∈ B has equation m n−1 n−1 y + a1y + ··· + an = 0 so y = −(a1 + ··· + anx ) ∈ A. Corollary if q ∈ 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 ∈ Spec B, q ∩ A = q0 ∩ A. Then q = q0.

We have Bp/Ap integral. Let m = pAp, a maximal . 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 ∈ 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 ∈ 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 ∈ C and b/s = (bt)/(st) ∈ 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 ∈ B is integral over a if it satisfies f (b) = 0 n n−1 with f (t) = t + a1t + ··· an with ai ∈ 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 ∈ C. The equation gives xn ∈ aC = c so x ∈ r(c). n P If x ∈ r(c) then x = ai xi with ai ∈ a and xi ∈ 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 ∈ 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 ∈ r(a).

Since x is integral over A it is algebraic over K. Let L/K be a containing x, so the conjugates xi ∈ 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 ∈ 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 ∈ Bq1 p2 with y ∈ Bp2 and s ∈ B r q1. Now y is r r−1 integral over p2. So its minimal equation y + u1y + ··· over K has coefficients ui ∈ r(p2) = p2.

If x = y/s ∈ Bq1 p2 ∩ A then s = y/x ∈ K so its minimal equation i i over K has coefficients vi = ui /x , hence x vi ∈ p2. Since s is integral over A we have vi ∈ A. But if x 6∈ p2 then vi ∈ p2 would imply s ∈ r(q1) = q1, a contradiction, so x ∈ p − 2 as required.

Abramovich MA 252 notes: Commutative algebra 15 / 24 Integral dependence Integral closure in a finite extension field

Proposition Let A be integrally closed in K, L/K a finite separable extension, B the integral closure of A in L. Then there is a basis vi of L/K P such that B ⊂ Avi .

P r−i Suppose v ∈ L. Then it satisfies ai v = 0 for ai ∈ A, a0 6= 0. It follows that a0v is integral over A. So we can rescale elements of a basis to get a basis {ui } ⊂ B. Since L/K separable the bilinear form Tr(xy) is nondegenerate. Let {vi } ⊂ L be the dual basis: Tr(vi uj ) = δij . P For x ∈ B write x = xi vi . Now xui ∈ B so Tr(xui ) ∈ A as the coefficient of the polynimial of xui . P P Now Tr(xui ) = Tr( j xj vj ui ) = j xj Tr(vj ui ) = xi ∈ A so P B ⊂ Avj . Challenge: what about purely inseparable?

Abramovich MA 252 notes: Commutative algebra 16 / 24 Integral dependence Valuation rings

Definition

Let B be a domain, K = B(0). We say B is a if every x ∈ K has either x ∈ B or x−1 ∈ B.

The condition says that two elements can be compared: x 4 y if y/x ∈ B, with x ≈ y if x/y ∈ B×. Fields are valuation rings. Z(p) and any Dm for PID D and m maximal is a valuation ring. Zp is a valuation ring.

Abramovich MA 252 notes: Commutative algebra 17 / 24 Integral dependence Basic properties

Proposition Assume B ⊂ K a valuation ring. Then (1) B is local, (2) B ⊂ B0 ⊂ K ⇒ B0 a valuation ring, and (3) B is integrally closed.

× Writ m = B r B . We claim m an ideal (and then it is maximal and B local). Indeed if a ∈ B, x ∈ m ⇒ ax 6∈ B× ⇒ ax ∈ m. Ifx, y ∈ m nonzero then either xy −1 or yx−1 ∈ B, so x + y = (xy −1 + 1)y = x(1 + yx−1) ∈ m, giving (1). (2) is immediate. n n−1 If x ∈ K is integral then x + b1x + ··· + bn = 0 so if we had −1 −1 −(n−1) x ∈ B then x = −(b1 + b2x + ··· + bnx ∈ B anyway.

Abramovich MA 252 notes: Commutative algebra 18 / 24 Integral dependence Construction of many valuation rings

Consider K and an algebraically closed field Ω. Consider the set P = {(A, f )|A ⊂ K, f : A → Ω}, partially ordered by containment/restrictions. Zorn’s Lemma applies, so maximal elements dominating any given one (A, f ) exist. If Ω ⊃ K then B = K itself is a maximal element. If B a valuation ring and Ω ⊃ B/m the algebraic closure then it is already maximal. What about the others? Theorem maximal elements (B, g) are valuation rings.

Abramovich MA 252 notes: Commutative algebra 19 / 24 Integral dependence Lemmas

Lemma B is local and m = Kerg the .

g(B) ⊂ Ω a domain, so m is prime. Extend Bm → Ω by g(b/s) = g(b)/g(s), well defined as s 6∈ m. By maimality B = Bm local and the kernel is m. Lemma For 0 6= x ∈ K consider the subring B[x] ⊂ K and ideal m[x]. Then either m[x] 6= B[x] or [x−1] 6= B[x−1].

P −j P i If vj x = ui x are minimal degeree relations with ui , vj ∈ m, then multiply one to cancel out and get e.g. n n−1 (1 − v0)x = v1x + ··· + vn, but 1 − v0 is a unit so we get n Pn i x = i=1 wi x which can be substitute to shorten the given one, contradicting assumption.

Abramovich MA 252 notes: Commutative algebra 20 / 24 Integral dependence Proof of theorem

If (B, g) maximal then B a valuation ring.

May assume m[x] 6= B[x] ⊂ K. So m[x] ⊂ n ⊂ B[x] maximal. m ⊂ n ∩ B ( B so frn ∩ B = m. get inclusion B/m ⊂ B[x]/n = B/m[¯x]. Whether or notx ¯ algebraic it has a place in Ω, so by maximality B[x] = B and x ∈ B.

Abramovich MA 252 notes: Commutative algebra 21 / 24 Integral dependence Chracterizing integral closure

Corollary

If A˜ the integral closure of A ⊂ K then A˜ = ∩A⊂B valuation of K B.

First, any B ⊃ A˜ since B integrally closed. Note that if x ∈ A[x−1] then x ∈ A˜. Assume x 6∈ A˜ then x−1 6∈ A[x−1]×. So there is a maximal ideal x−1 ∈ m ⊂ A[x−1]×. The morphism A → A[x−1]/m extends to B → Ω for some valuation ring B containing A, x−1, and since x−1 ∈ m we have x 6∈ B.

Abramovich MA 252 notes: Commutative algebra 22 / 24 Integral dependence Extensions of homomorphisms

Proposition A ⊂ B fin. gen., domain, 0 6= v ∈ B. There is 0 6= u ∈ A s.t. any f : A → Ω with f (u) 6= 0 extends to g : B → Ω, g(v) 6= 0.

By induction assume B = A[x]. If x transcendental, write v = h(x) for some polynomial h ∈ A[t], and take u = a0 say. Write h¯[t] ∈ Ω[t] for the image under f , nonzero since f (a0) 6= 0. we have some ξ ∈ Ω such that h¯(ξ) 6= 0, and define g : A[x] → Ω by g(x) = ξ. −1 P i If x is algebraic then v is algebraic. Set ai x = 0 and P −i bi v = 0. Let u be the product of leading terms. Then f extends to A[u−1] since Ω a field, and to a morphism g 0 : C → Ω on a valuation ring C containing A[u−1] by the big extension theorem. Note x, v −1 are integral over A[u−1]. So x, v −1 ∈ C, in particular B ⊂ C and v ∈ C ×, so 0 0 g (v) 6= 0. The restriction g = gB works. Abramovich MA 252 notes: Commutative algebra 23 / 24 Integral dependence Weak Nullstellensatz, first installement

Proposition A ⊂ B fin. gen., domain, 0 6= v ∈ B. There is 0 6= u ∈ A s.t. any f : A → Ω with f (u) 6= 0 extends to g : B → Ω, g(v) 6= 0.

Theorem A finitely generated algebra B over a field k is a field if and only if it is a finite extension of k. Indeed, one can take v = 1 ∈ B and consider f : A → Ω = k¯.

Abramovich MA 252 notes: Commutative algebra 24 / 24