Proving that the integral closure of $R$ in $S$ is a subring of $S.$

Question

Let $R \subset S$ be commutative rings. Prove that the integral closure of $R$ in $S$ is a subring of $S.$

Note that the integral closure of $R$ in $S$ consists of all $x \in S$ that is integral over $R$. And note that an element $x$ is integral over $R$ if there exists a monic polynomial $f(x) \in  R[x]$ such that $f(x) = 0$.

I was thinking about using the ordinary subring test for doing that but I was given a hint to use the following Lemma: $x$ is integral over $R$ iff $R[x]$ is a finitely generated $R$-module.

My question is:

I do not see why I should use this lemma and how it facilitates the steps of the proof, could anyone explain this to me please?

Answer 1

You prove that it is a subring by ordinary "subring test" but you use the lemma while doing so. Although in our commutative algebra course we used also the following version of lemma (we actually prove this first to get your lemma but I am not sure whether it is necessary to do it in this order. )

Lemma 1 $R\subset S$, $x\in S $ then $x$ is integral over $R$ if $R[x]$ is finitely generated $R-$module.

Lemma 2 $x$ is integral over $R$ iff there exists finitely generated $R-$module $R'$ such that $R\subset R[x]\subset R'\subset S$

Take $a,b\in S$ integral over $R$. By your lemma $R[a]$ is finitely generated $R$-module. But we can view $R[a]$ simply as a ring so $b$ is integral also over $R[a]$. Hence $R[a][b]$ is finitely generated module over $R[a]$.

Now $R[a][b]=R[a,b]$ as subrings of $S$ (they consist of the same elements) so $R[a,b]$ is also finitely generated as $R-$module.

Using lemma 2: Now set $x:=a+b$ and $R':=R[a,b]$. $R'$ is finitely generated $R-$module and $R\subset R[x]\subset R'\subset S$ so by lemma 2 we conclude that $x$ is integral.

Commentary I think that the reason we need to use lemma 2 instead of ypur version (lemma 1) is that you can't conclude that $R[a+b]$ is finitely generated from the fact that it is $R-$submodule of finitely generated $R-$module $R[a,b]$. there are submodules of fin. gen. modules that are not finitely generated (take ring $R$ as a module over itself it is generated by $1$ but unless it is Noetherian it has ideals that are not finitely generated)

However, in this particular setting, lemma 2 gives you that $a+b$ is integral so by lemma 1 is $R[a+b]$ finitely generated $R-$module after all.

Answer 2

Let $x,y\in S$ be integral over $R$. We wish to show that also $x+y$, $xy$ are integral over $R$. By the lemma $R[x]$ is a finite $R$-module. Note that $y$ is surely also integral over $R[x]$, so $R[x][y]$ is a finite $R[x]$-module. Show that this implies that $R[x][y]$ is a finite $R$-module and use the lemma again (we might need a slightly stronger version of the lemma, namely that if $z$ is contained in a ring that is a finitely generated $R$-module then $z$ is integral over $R$. This is often proved together with the lemma you cite, if not: the proof is almost identical using the Cayley-Hamilton theorem)