integral extension, going up and going down theorems In this section, we are going to explore more about the notion of integral extension. The goal is to show the going up and going down theorems.
Let A, B be rings. We say B is an extension over A if A ⊂ B.
An element x ∈ B is said to be integral over A if it satisfies a monic polynomial in A[x]. B is integral over A if every element of B is integral over A.
The following Properties are more or less parallel to the theory of algebraic extensions. Proofs are similar. The reader should find no difficulty working them out.
Proposition 0.1. Let A ⊂ B be an extension. The followings are equivalent:
(1) x ∈ B is integral over A.
(2) A[x] is a finitely generated A-module.
(3) A[x] is contained is a subring C ⊂ B such that C is a finitely generated A-module.
Corollary 0.2. Let A ⊂ B be an extension. If xi ∈ B is integral over A for i = 1, .., n. Then A[x1, ..., xn] is a finitely generated A-module.
Corollary 0.3. Let A ⊂ B ⊂ C be extensions. If C is integral over B and B is integral over C, then C is integral over A.
Corollary 0.4. Let A ⊂ B be an extension. The integral closure of A in B, which is the set of elements in B integral over A, is a ring (subring of B).
Let A ⊂ B be an extension. A is said to be integrally closed in B if the integral closure of A is A itself.
Corollary 0.5. Let A ⊂ B be an extension. And let C be the integral closure of A in B. Then C is integrally closed in B.
Proposition 0.6. Let A ⊂ B be an integral extension.
(1) If b C B and a := b ∩ A, then B/b is integral over A/a.
(2) Let S be a multiplicative set of A (hence of B), then S−1B is integral over S−1A.
Proof. (1) For ¯b ∈ B/b, one notices that b ∈ B and bn+ an−1bn−1+ ...+a0 = 0 for some ai ∈ A. It’s clear that ¯bn+an−1¯bn−1+... ¯a0 = 0 ∈ ¯B := B/b. And hence ¯b is integral over ¯A := A/a.
(2) For bs ∈ S−1B. One notice that bn+ an−1bn−1+ ... + a0 = 0 for some ai ∈ A. Hence
(b
s)n+an−1 s (b
s)n−1+ ... + a0 sn = 0.
1
And we are done.
¤ Lemma 0.7. Let A ⊂ B be an integral extension and B is an domain.
Then A is a field if and only if B is a field.
Proof. Suppose that A is a field. For any b 6= 0 ∈ B, bn+ an−1bn−1+ ... + a0 = 0 for some ai ∈ A. We may assume that a0 6= 0 ∈ A because B is a domain. Then
b(bn−1+ an−1bn−2+ ... + a1) = −a0. Therefore, b is invertible in B and so B is a field.
Conversely, let B be a field. For a 6= 0 ∈ A, a−1 ∈ B. Thus (a−1)n+ an−1(a−1)n−1+ ... + a0 = 0. In particular, a−1 = −(an−1+ ... +
a0an−1) ∈ A. ¤
Let A ⊂ B be an integral extension, and q ∈ SpecB, p ∈ SpecA. We say that q is lying over p if q ∩ A = p.
We are going to study the relation between prime ideals of integral extension.
Proposition 0.8. Keep the notation as above with q is lying over p.
Then q is maximal if and only if p is maximal.
Proof. B/q is again integral over A/p. Since B/q is a field if and only
if A/p is a field, we are done. ¤
An consequence is the following corollary which assert the ”unique- ness in a chain of prime ideal”:
Corollary 0.9. Keep the notation as above. If q1 ⊂ q2 are prime ideals lying over p. Then q1 = q2.
Proof. Let S = A−p. (Note that qi∩S = ∅ for i = 1, 2.) Then we have Bp integral over Ap. Moreover, q1Bp ⊂ q2Bp. Note that qiBp∩ Ap = (qi ∩ A)Ap = pAp is maximal for i = 1, 2. Hence both q1Bp ⊂ q2Bp are maximal ideal lying over pAp. We then have q1Bp = q2Bp. By the correspondence of prime ideals, we have q1 ⊂ q2. ¤ It’s also desirable to have existence of prime ideal lying over a specific one.
Proposition 0.10. Let A ⊂ B be an integral extension. Let p ∈ Spec(A). Then there exist a q ∈ Spec(B) lying over A.
Proof. Let S = A − p. We consider Ap ⊂ Bp. (This is injective since S−1 is exact.) Take a maximal ideal m of Bp. We claim that q := m∩B is a prime ideal lying over p. To see this,
q ∩ A = (m ∩ B) ∩ A = (m ∩ Ap) ∩ A = pAp∩ A = p.
This is because m lying over a maximal ideal of Ap and the only max-
imal ideal of Ap is pAp. ¤
Theorem 0.11 (Going-up theorem). Let B be an integral extension over A. Let p1 ⊂ p2 ∈ Spec(A) and q1 ∈ Spec(B) lying over p1. Then there is q2 ∈ Spec(B) containing q1 lying over p2.
Proof. Let ¯A := A/p1and ¯B := B/q1. Then ¯B is integral over ¯A. There is a prime ideal ¯q2 lying over ¯p2. Lift to B, then we are done. ¤ As we have seen, let B be an integral extension over A. Then every chain of distinct prime ideals of B restricts to a chain of distinct prime ideals of A and conversely, every chain of distinct prime ideals of A extends to a chain of distinct prime ideals of B. It follows that
dimA = dimB.
Before we going to dimension theory. We would like to investigate integral extension a little bit more which will be useful in Dedekind domain and DVRs.
Integral extension is very similar to algebraic extension.
Definition 0.12. A domain A is said to be integrally closed if it is integrally closed in its quotient field.
For the rest of this section, we are going to assume that B is a domain integral over A and A is integrally closed (i.e. in its quotient field K). We remark that this setting is closely related to algebraic field extensions.
Lemma 0.13. Let A ⊂ B be an extension and C is the integral closure of A in B. Let a C A be an ideal. We say that b ∈ B is integral over a is b satisfies a polynomial with coefficient (except the leading term) in a. Then the integral closure of a in B is √
aC.
Proof. If b ∈ B is integral over a, then
bn+ an−1bn−1+ ... + a0 = 0.
Since b ∈ C, we have
bn = −an−1bn−1− ... − a0 ∈ aC.
Thus b√ aC.
Conversely, let b ∈ √
aC. Then bn =Pr
i=1aixi, with ai ∈ a, xi ∈ C.
Then A[x1, ..., xr] is a finite module over A. Since bnA[x1, ..., xr] ⊂ aA[x1, ..., xr]. It follows that the multiplication by bn behave like a matrix on A[x1, ..., xr] with entries in a. Hence bn satisfies the char- acteristic polynomial with coefficient in a. It follows that b is integral
over a. ¤
Remark 0.14. Keep the notation as above, let f (x) be an integral polynomial of b ∈ B and p(x) be its minimal polynomial over K. We remark that there is NO notion of ”minimal integral polynomial” in general because the division algorithm doesn’t holds in A.
However, if A is UFD, then by Gauss lemma, we have p(x)|f (x) not only in K[x] but also in A[x]. Hence the minimal polynomial is integral.
Lemma 0.15. Keep the notation as above, the minimal polynomial of b ∈ B is in A[x]. If b ∈ B is integral over a C A, i.e. the integral polynomial has coefficients in a, then the minimal polynomial of b ∈ B is in √
a[x].
Proof. Assume now b is integral over a with minimal polynomial p(x).
Take a splitting field of p(x), say L/K. And let b = b1, ..., bn be con- jugates of b. Then they satisfy f (x) as well. Thus bi is integral over a It’s not difficult to see that p(x) = Qn
i=1(x − bi)mi for some mi ≥ 0.
The coefficient are combination of bi hence integral over a. Apply the above Lemma to the extension A ⊂ K, we have integral closure of a is
√a. Hence minimal polynomial is in √
a[x]. ¤
Let A be an integrally closed domain with quotient field K. Given an extension L/K, one can consider B to be the integral closure of A in L. (We may assume that L is algebraic over K, or even to be the splitting field of B.) Especially, in number theory, we usually consider a number field which is a finite extension over Q. And let O be the domain of algebraic integers. The extension Z ⊂ O justify our setup.
Proposition 0.16. Let A be an integrally closed domain with quotient field K. Given a normal extension L/K, one can consider B to be the integral closure of A in L. Let p ∈ SpecA, then prime ideals in B lying over p are conjugate. That is, for q1, q2 ∈ SpecB lying over p, there is σ ∈ GalKL such that σ(q1) = q2.
Proof. We will only prove this under the assumption that B is finitely generated over A. (Then we may assume that [L : K] is finite).
If q2 6= σj(q1) for all j then q2 6⊂ σj(q1) for all j.
claim. there is x ∈ q2 such that x 6∈ σj(q1) for all j. We leave this claim as an exercise.
Let y = Q
σj∈GalKLσj(x). Then y is invariant under GalKL. We may assume that ypl is separable over K for some l ≥ 0. (If char(K)=0, then l = 0.) Let S be the separable closure of K in L. Then S is Galois over K with GalKS = GalKL (cf. Thm. 12 of lecture 1). Hence ypl ∈ K.
One notice that B ∩ K = A. It follows that ypl ∈ A. And then y ∈ K is integral over A. Therefore, y ∈ A. We Clearly, y = xyx ∈ q2. Hence y ∈ A ∩ q2 = p ⊂ q1. Hence σj(x) ∈ q1 for some j. This leads to a
contradiction. ¤
Corollary 0.17. Keep the notation as above. Assume furthermore that B is finitely generated over A. Then for p ∈ SpecA, there are only finitely many prime ideal in B lying over p. And any two of them are
”conjugate” to each other.
Proof. Let L be the splitting field of B over K. One sees that L/K is finite. Let C be the integral closure of A in L. Clearly, A ⊂ B ⊂ C.
We know that prime ideal in C lying over p is conjugate to each other.
Since the Galois group is finite. There are only finitely many prime ideals. Restrict to B, then there are only finitely many prime ideals in
B lying over p. ¤
Theorem 0.18 (Going down theorem). Keep the notation as above, i.e. B is an domain integral over an integrally closed domain A. If there are p2 ⊂ p1 ∈ SpecA and q1 in B lying over p1. Then there is q2 ⊂ q1 ∈ SpecB lying over p2
Proof. Let L be the normal closure of B over K and C be the integral closure of A in L. It’s clear that C is integral over B. There is a prime ideal r1 in C lying over q1. And there is a prime ideal r2 in C lying over p2. By Going up theorem, there is a prime ideal r01 ⊃ r2 in C lying over p1. Therefore, there is σ ∈ GalKL such that σ(r1) = r01. Then σ−1(r2) ⊂ r1. Let q2 := σ−1(r2) ∩ B. Then we are done. ¤
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