Integral dependence and field extension

Question

Let $R$ be a domain (commutative with unity). $k$ is field algebraically dependent on $k_0$. $A$ is some ideal of $R \otimes_{k_0} k$ and $A_0$ = $A \cap R$.

How to prove that $(R \otimes_{k_0} k)/A$ is integrally dependent on $R/A_0$

Answer 1

First of all $R$ must be a $k_0$-algebra in order to define $R \otimes_{k_0} k$.

Then we can see a general frame of this problem which is easily proven:

If $R\subset S$ is an integral extension and $J\subset S$ an ideal, then $R/J\cap R\subset S/J$ is also an integral extension.

In your case take $S=R \otimes_{k_0} k$. But why then $R\subset S$ is integral? Since $k_0\subset k$ is algebraic. (It's enough to prove that the simple tensors, that is, elements of the form $r\otimes a$ with $r\in R$ and $a\in k$, are integral over $R$. Moreover, note that it's enough to prove this for $1\otimes a$.)