# Integral closure of ring A is integral closed?

Let $$A,B$$ be ring and $$A \subseteq B$$. Let $$C$$ be integral closure of $$A$$ in $$B$$.
Then $$C$$ is integral closed, isn't it?
Let D be integral closure of C. $$D \supseteq C$$ is clear. Let $$x \in D$$, then $$x$$ is integral over $$C$$. I want to prove that $$x$$ is integral over $$A$$. How can I prove this.

• If you take $A=B$, then $A$ is trivially integrally closed in $B$ but $A$ itself need not be integrally closed. Commented Jul 11 at 6:53
• @user6 I know that "integrally closed" is used when $A$ is a UFD, but here this is not the case. I think the op wants to show that $C$ is integrally closed in $B$. In your example, $A$ is integrally closed in $A$, so your counterexample doesn't work I think. Commented Jul 11 at 7:03
• See Integral dependence over rings is transitive. This is a well-known result that you can find in most of algebra books. See for instance the 5th chapter of Atiyah-Macdonald's book "Introduction to commutative algebra". Commented Jul 11 at 7:08
• the standard convention is: a domain $D$ is called integrally closed if it is integrally closed in its field of fractions. en.wikipedia.org/wiki/Integrally_closed_domain Commented Jul 11 at 16:50
• I know the convention, I'm just saying that the op does not recquire $A$ to be a domain. Commented Jul 12 at 6:38

The trick is to use a slightly different description of "integral closedness". You wish to show that $$x$$ is integral over $$A$$, i.e. that there is a relation of the form: $$x^n = \sum_i a_i x^i$$ with the $$a_i$$'s in $$A$$. Notice the following: this relation exists if and only if the $$A$$-module $$A[x]$$ is of finite type (the proof is quite easy).
So far, we only know that the $$C$$-module $$C[x]$$ is finitely generated. So there is a relation $$x^n = \sum_i c_i x^i$$ with the $$c_i$$'s in $$C$$. Let $$A'=A[c_0,...,c_{n-1}]$$, we obviously have $$A'[x]$$ finitely generated, so $$x$$ is integral over $$A'$$. Moreover, $$A'$$ is a finitely generated $$A$$-module (by definition of $$C$$). This shows that $$A[x,c_1,...,c_{n-1}]$$ is a finitely generated $$A$$-module, but we cannot a priori conclude that so is $$A[x]$$. Luckily, there is an additional, more involved, equivalent characterization of integral closedness (see wikipedia, or really any commutative algebra textbook):
• $$x \in D$$ is integral over $$A$$ if and only if there exists a finitely generated $$A$$-module $$M \subseteq D$$ which contains $$A[x]$$.
Taking this for granted, we are done: $$M=A'[x]$$ works. Implication $$\implies$$ is easy: take $$M=A[x]$$ itself. For the converse, we need the Cayley-Hamilton Theorem for modules. I will let you look up the details for yourself, but here is the general idea. Consider the endomorphism $$u$$ of $$M$$ given by multiplication by $$x$$ ; because $$M$$ is finitely generated by $$m$$ elements, the Cayley-Hamilton Theorem gives $$a_i$$'s in $$A$$ such that the endomorphism $$u^m - \sum_i a_i u^i$$ is zero. Now evaluate it at $$1 \in A \subseteq M$$, you find the relation for $$x$$ you were looking for.
This result makes sense: if "multiplication by $$x$$" satisfies some relation over $$A$$ on a big module $$M$$, it still satisfies it on any smaller submodule stable under that multiplication (which is the case for $$A[x]$$).