Proof that a domain is normal (2)

If $t \in K[\alpha]$ is integral over $A[\alpha]$, then $A[\alpha,t]$ is a finitely generated $A[\alpha]$-module. But $A[\alpha]$ is a finitely generated $A$-module, so that $B=A[\alpha,t]$ is a finitely generated $A$-module.

This is sufficient to see that $t$ is integral over $A$. Indeed, there is a finite free $A$-module $B_1$ with an $A$-surjection to $B$ and a map $\tau: B_1 \rightarrow B_1$ that reduces (in $B \rightarrow B$) to the multiplication by $t$. As $B_1$ is finite free over $A$, Cayley-Hamilton states that $\tau$ is integral over $A$. Thus the multiplication by $t$ (as a map $B \rightarrow B$) is itself integral over $A$, which means that $t$ is integral over $A$.