Polynomial ring with integral coefficients is integral
Solution 1:
The integral closure of $A[X]$ over $B[X]$ will obviously contain :
- $B$, because $B$ is integral over $A$, and then over $A[X]$
- 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 $A[X]$.