integral closure is ring

Theorem.  Let $A$ be a subring of a commutative ring $B$ having nonzero unity.  Then the integral closure of$A$ in $B$ is a subring of $B$ containing $A$.

Proof.  Let $x$ be an arbitrary element of the integral closure $A^{\prime }$ of $A$ in $B$.  Then there are the elements $a_0,a_1,\mathrm{\dots },a_{n-1}$ of $A$ such that

$$a_0+a_{1}x+\mathrm{\dots }+a_{n-1}{x}^{n-1}+x^n=\mathrm{\hspace{0.33em}0}$$

where  $n>0$.  If  $f(X)=c_0+c_{1}X+\mathrm{\dots }+c_m{X}^m$  is a polynomial in $A[X]$ with degree  $m>n$,  we have

$f(x)$$\mathrm{\hspace{0.33em}}=c_0+c_{1}x+\mathrm{\dots }+c_{m-1}{x}^{m-1}+c_m{x}^{m-n}(-a_0-a_{1}x-\mathrm{\dots }-a_{n-1}{x}^{n-1})$

$\mathrm{\hspace{0.33em}}=c_0^{\prime }+c_1^{\prime }x+\mathrm{\dots }+c_{m-1}^{\prime }{x}^{m-1}$

where  the elements $c_i^{\prime }$ belong to $A$.  This procedure may be repeated until we see that $f(x)$ is an element of the $A$-module generated by $1,x,\mathrm{\dots },x^n$.  Accordingly,

$$A[x]=A+Ax+\mathrm{\dots }+A{x}^n$$

is a finitely generated $A$-module.

Now we have evidently  $A\subseteq A^{\prime }$.  Let $y$ be another element of $A^{\prime }$.  Then

$$A[x,y]=A[x][y]$$

is a finitely generated $A[x]$-module, whence  $A[x,y]$  is a finitely generated $A$-module.  Because the elements$x-y$ and $xy$ belong to  $A[x,y]$,  they are integral over $A$ and thus belong to $A^{\prime }$.  Consequently, $A^{\prime }$ is a subring of $B$ (see the http://planetmath.org/node/2738subring condition).