ALGEBRA III DECEMBER 2, 2004

GABRIEL C. DRUMMOND-COLE

Okay, so, well, let’s see, I have various plans. How many classes do we still have? Three? That’s plenty. Today I will finish with integral dependence and then talk about Dedekind rings, which are interesting from every point of view, but you need integral extensions.

So let me try to go on with this. I want to write two things. This was the magical proposition that let us say that the integral extension of a ring forms a ring.

Proposition 1. The following are equivalent:

(1) x ∈ S is integral over R

(2) R[x] is a finitely generated R-module.

(3) There exists a ring C which is a finitely generated R[x]-module such that R[x] ⊂ C ⊂ S.

(4) There exists a faithful R[x]-module M which is finitely generated over R.

Corollary 1. The set of elements in S integral over R form a subring.

If x, y ∈ S are integral over R then R[x, y] was finitely generated, recall. You use part three. So let me pick from the proof the useful statement.

Corollary 2. If x_{1}, ⋯, x_{n} ∈ S are integral over R then R[x_{1}, ⋯, x_{n}] is a finitely generated R-module.

Let me phrase this as the following:

Corollary 3. Say R ⊂ S as rings. We call this extension integral if every element of S is integral over R. If S is a finitely-generated R-algebra, then S is a finitely generated R-module.

This is just a rephrasing. You’ll hear this as “integral plus finitely generated algebra equals finite (finitely generated module)”

You’ll wonder what is the geometric meaning. Integral extensions account for branched covers. The finiteness you see it everywhere, but this is more general. You need to know a little more about schemes.

What does it mean that I have an inclusion of rings? You have a map X → Y, affine algebraic varieties. Then you have A(Y) → A(X). So a regular function on Y will not vanish completely on X. What, topologically is that? This means that the image of f is not contained in any closed algebraic set, so the image is dense. A(Y) → A(X) has dense image if and only if X → Y is injective (the image is not contained in any closed propert algebraic subset of Y). When you have a dominant morphism you have an inclusion of rings. The fact that it is integral means it is finite and proper (preimage of compact is compact).

We’ll show in a moment finiteness. It’s part of what’s called going up. If I have a maximal ideal in A(Y) there are finitely many in A(X) that contract to it. They also have the same Krull dimension, so it is a cover with a finite number of leaves.

Okay, now, how do you show this corollary. The second one follows from the first corollary by induction. It’s transitivity of finiteness.

Let’s try to take care of the situation to show that these have the same Krull dimension. It’s the maximum length of a chain of primes.

So I clarified the meaning of injective, now what does it mean, integral.

Oh, I forgot.

Corollary 4. If B is integral over A and C is integral over B then C is integral over A.

Say x ∈ C. Then x satisfies a monic polynomial over B. Each of the coefficients of this polynomial is integral over A so A[b_{1}, ⋯, b_{n}] = B′ is a finitely generated A-module. Then x is a finitely generated B′-module since x is integral over B′. So B′[x] is finitely generated as an A-module. Then by my proposition, A[x] ⊂ B′[x] ⊂ S so x is integral over A.

The concrete stuff can have an advantage because you want to know the actual equation it satisfies.

Let me make two more definitions.

Definition 1. R ⊂ S is integrally closed if x ∈ S is integral over R implies x ∈ R.

If R is a domain, R is said to be integrally closed if R ⊂ Q(R) is integrally closed.

So ℤ is integrally closed. In particular, any UFD or PID is integrally closed.

Now, why is this notion meaningful? Because of the following corollary:

Computing an integral closure is to get rid of singularities.

Corollary 5. If R ⊂ S I can compute R ⊂ R̄ ⊂ S, the integral closure of R in S. It is all the integral elements of R over S, but it is integrally closed in S.

So typically, you are interested in computing integrally closed things. So I want, in an integral extension R and R̄ have the same Krull dimension and that this is finite, but the reverse I am not going to do; it takes a little more.

I’ll need three or four propositions. They will be easy, but bear with me. I’ll fix R ⊂ S integral, and relate prime ideals.

Proposition 2. Say A ⊂ B is integral.

(1) If I ⊂ B is an ideal and J = I^{c} = I ∩ A then I can look to A/J ↪ B/I. This is integral. You satisfy the same equation reduced modulo I.

(2) if S ⊂ A is a multiplicative system then S^{−1}A ↪ S^{−1}B is integral.

Pick x/s ∈ S^{−1}B. Since x ∈ B it satisfies an equation with coefficients a_{1}, ⋯, a_{n}. I want to write an equation for x/s. Divide by s^{n}. Then I get (x/s)^{n} + a_{1}/s(x/s)^{n−1} + ⋯ = 0 so we’re done.

Now let’s try to see that prime ideals are related in an integral extension. I’ll deal first with the zero ideal. When is it prime? When I’m talking about domains.

Corollary 6. Say A ⊂ B is integral, and these are domains.

Then B is a field if and only if A is a field.

This is relating maximality of the zero ideals in both. I claim that they have the same height.

Let’s prove this. Suppose A is a field. Let y ∈ B which is not 0. So y satisfies some equation y^{n} + ay^{n−1} + ⋯ = 0. Choose one of minimum degree. Now, can a_{n} be zero? No, because B is a domain so that this is not of minimal degree.

How do I invert? y(y^{n−1}+a_{1}y^{n−2}+⋯) = −a_{n}, so since a_{n} ∈ A and A is a field, y(−a_{n})^{−1}(y^{n−1}+ a_{1}y^{n−2} + ⋯) = 1. So B is a field.

Conversely, suppose B is a field. I want to show that A is a field. Pick x ∈ A. you want to show that x^{−1} ∈ B is also in A. So this satisfies an equation (x^{−1})^{n} + (x^{−1})^{n−1}a_{1} + ⋯ + a_{n} = 0. So multiply with x^{n−1} and you get x^{−1} = −(a_{1} + a_{2}x + ⋯).

So this relates the zero ideals, as I said. It has an immediate consequence.

Corollary 7. Let A ⊂ B be integral and let q ⊂ B be prime, p = A ∩ q. Then p is maximal if and only if q is maximal.

A/p ↪ B/q is integral, then apply the previous lemma.

Corollary 8. “Incomparability.” These are part of the Noether normalization package.

A ⊂ B integral and q ⊂ q′ in B prime ideals such that q ∩ A = q′ ∩ A = p. Then q = q′.

So any two primes, and there will be some collection, above p, are incomparable.

So we want to localize, so let S be the complement of p. Then A_{p} ↪ B_{p} (by abuse of notation). This is integral by the above. Now in A_{p} I have the ideal pA_{p} which is maximal. So n = qB_{p} and n′ = q′B_{p} are two ideals, and these are both prime. Why? Because q restricts to p so neither meet the multiplicative system. This was one of the facts of localization. So n ⊂ n′. But n ∩ A_{p} is pA_{p}, and the same for n′. So we’re done because we have an inclusion of a maximal ideal into another, which means a coincidence of them.

But their contractions back in B are q and q′ so they are also equal.

As I said, integral extensions correspond to finite maps. How do you prove surjectivity?

Theorem 1. Let A ⊂ B be an integral extension and p ⊂ A prime. Then there exists prime q ⊂ B with q ∩ A = p.

The proof goes like this: Localize at p and get A_{p} ↪ B_{p} is integral. Now I have the square

[figure: A_{p}; B_{p}; A; B]

So I have pA_{p} ⊂ A_{p} a maximal ideal. Then any maximal ideal m in B_{p} contracts to a maximal ideal, the unique one, pA_{p}. Then take the preimage of m, which has to contract to the preimage of pA_{p}, which is p.

That’s it, very easy.

The going up theorem says the following. I’m not going to prove it.

Theorem 2. If A ↪ B and I have a chain of primes p_{1} ⊂ ⋯ ⊂ p_{s} ⊂ ⋯ ⊂ p_{m} in A and q_{1} ⊂ ⋯ ⊂ q_{s} with q_{i} ∩ A = p_{i} then I can continue. You go by induction so you only need one square.

\xymatrix{p_{1} \ar[r]^{⊂} \ar[d]_{⊂} & p_{2} \ar@{.>}[d]^{⊂} \\ q_{1} \ar@{.>}[r]_{⊂} & q_{2}}

This roughly says that they have the same Krull dimension.