Hi! Let A be a subring of a ring $\displaystyle B$ which is integral over $\displaystyle A$. Then the following assertions hold:

(i) $\displaystyle B[X]$ is integral over $\displaystyle A[X]$

(ii) Let $\displaystyle b$ be an ideal of $\displaystyle B$ and define $\displaystyle a:=A \cap b$, then $\displaystyle B/b$ is integral over $\displaystyle A/a$.

ad (i)

Let $\displaystyle f \in B[X]$ (polynomial with degree n). I want to show that $\displaystyle (B[X])[f]$ is a finitely generated A[X]-module. I guess that $\displaystyle 1,f,f^2,...,f^(n-1)$ generates $\displaystyle (B[X])[f]$. Is this the right way? If yes, is it the right execution?

ad (ii)

Here, i have the following idea: Let $\displaystyle \overline{x} \in B/b$ and be $\displaystyle x$ any representative of this equivalence class. As$\displaystyle B$ is integer over $\displaystyle A$, there exists $\displaystyle f \in A[X]$, so that $\displaystyle f(x)=0$. Let $\displaystyle \pi$ be the canonical projection $\displaystyle B \rightarrow B/b$, then it follows that $\displaystyle \pi(f(x))=\overline{0}$, therefore $\displaystyle \pi(f(x))=\sum_{i=1}^n \overline{a_i} \cdot \overline{x}^i=\overline{0}$.

But is this polynomial now in $\displaystyle A/a[X]$?

Has somebody got answers or ideas?

Banach