I'm thinking about the following problem:
Assume that $K$ is field with fixed algebraic closure $L$. Assume we have two finite field extensions $E$ and $F$ of $K$ such that $E$ is contained in $F$. Then we have an inclusion map $\iota: E \to F$. This induces an inclusion map $\iota \otimes id: E \otimes_K L \to F \otimes_K L$. Obviously $\iota$ is an integral map but is $\iota \otimes id$ also integral?
$\textbf{Attempt on proof}$
I think I found the proof.
It suffices to show that any element of the form $f \otimes l \in F \otimes L$ is integral over $E \otimes L$. We can also assume that $l$ is not zero because otherwise it's trivial. Let $p \in E[X]$ be a polynomial such that $p(f) = 0$. We can write $$ p = e_0 + ... + e_n X^n$$ for some $e_i \in E$. Set $a_i = e_i \otimes 1/{l^i} \in E \otimes L$. Then $$a_i (f \otimes l)^i = (e_i \otimes 1/{l^i})(f \otimes l)^i = (e_i \otimes 1/{l^i})(f^i \otimes l^i) = e_i f^i \otimes 1.$$ Thus we have $$a_0 + ... + a_n (f \otimes l)^n = (e_0 \otimes 1) + ... _ (e_n f^n \otimes 1) = (e_0 + ... + e_n f^n) \otimes 1 = p(f) \otimes 1 = 0.$$