# 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 $a_{0},\,a_{1},\,\ldots,\,a_{n-1}$ of $A$ such that

 $a_{0}\!+\!a_{1}x\!+\ldots+\!a_{n-1}x^{n-1}\!+\!x^{n}\;=\;0$

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

 $\displaystyle f(x)$ $\displaystyle\;=\;c_{0}\!+\!c_{1}x\!+\ldots+\!c_{m-1}x^{m-1}\!+\!c_{m}x^{m-n}(% -a_{0}\!-\!a_{1}x\!-\ldots-\!a_{n-1}x^{n-1})$ $\displaystyle\;=\;c_{0}^{\prime}\!+\!c_{1}^{\prime}x\!+\ldots+\!c_{m-1}^{% \prime}x^{m-1}$

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

 $A[x]\;=\;A+Ax+\ldots+Ax^{n}$

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).

## References

• 1 M. Larsen & P. McCarthy: Multiplicative theory of ideals.  Academic Press, New York (1971).
Title integral closure is ring IntegralClosureIsRing 2013-03-22 19:15:40 2013-03-22 19:15:40 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 13B22 PolynomialRing RingAdjunction IntegralClosuresInSeparableExtensionsAreFinitelyGenerated