2

Let $B$ be a ring and $A\subset B$ a subring. Assume that $B$ is integral over $A$. I have to prove that $B[X]$ is integral over $A[X]$.

I tried writing down an integral relation for $f(X)\in B[X]$ over $A[X]$ but I don't really think that's the right way. Any hint?

user26857
  • 53,190
Angelo Rendina
  • 1,923

1 Answers1

7

The integral closure of $A[X]$ over $B[X]$ will obviously contain :

  1. $B$, because $B$ is integral over $A$, and then over $A[X]$
  2. the element $X$, because $X$ is obviously an integral element over $A[X]$

The smallest subring of $B[X]$ that contains both $B$ and the element $X$ is $B[X]$ itself. As the integral closure of $A[X]$ over $B[X]$ is a subring of $B[X]$, it is $B[X]$.

So $B[X]$ is integral over $A[X]$.

Xoff
  • 10,493