integral closure is ring

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

Proof.  Let x be an arbitrary element of the integral closure A of A in B.  Then there are the elements a0,a1,,an-1 of A such that

a0+a1x++an-1xn-1+xn= 0

where  n>0.  If  f(X)=c0+c1X++cmXm  is a polynomialMathworldPlanetmathPlanetmathPlanetmath in A[X] with degree  m>n,  we have

f(x) =c0+c1x++cm-1xm-1+cmxm-n(-a0-a1x--an-1xn-1)

where  the elements ci 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,,xn.  Accordingly,


is a finitely generatedMathworldPlanetmathPlanetmathPlanetmath A-module.

Now we have evidently  AA.  Let y be another element of A.  Then


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.  Consequently, A is a subring of B (see the condition).


  • 1 M. Larsen & P. McCarthy: Multiplicative theory of ideals.  Academic Press, New York (1971).
Title integral closure is ring
Canonical name IntegralClosureIsRing
Date of creation 2013-03-22 19:15:40
Last modified on 2013-03-22 19:15:40
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 6
Author pahio (2872)
Entry type Theorem
Classification msc 13B22
Related topic PolynomialRing
Related topic RingAdjunction
Related topic IntegralClosuresInSeparableExtensionsAreFinitelyGenerated