TITLE: If $A\otimes_k l$ is a normal integral domain then $K(A)\otimes_k l$ is a field.
QUESTION [5 upvotes]: I am trying to solve Ex. 5.4.M in Vakil's notes. Quoting the text:

Suppose $A$ is a $k$-algebra, and $l/k$ is a finite extension of fields. (Most likely your proof will not use finiteness; this hypothesis is included to avoid distraction by infinite-dimensional vector spaces.) Show that if $A\otimes_k l$ is a normal integral domain, then $A$ is a normal integral domain as well.

At a certain point the hint suggests to show that $K(A)\otimes_k l$ is a field ($K(A)$ is the fraction field of $A$).
I showed that $K(A)\otimes_k l$ is an integral domain and that a $K(A)$-basis of it is given by $\{1\otimes b_i\}_{i \in I}$, where $\{b_i\}_{i \in I}$ is a $k$-basis of $l$. In particular, if $l/k$ is a finite extension then $K(A)\otimes_k l$ is a field.
How to proceed in the infinite case?

REPLY [3 votes]: Let $A=K$ be a field extension of $k$, and $l=k(x)$ a purely transcendental extension of $k$. Then $A\otimes_kl\simeq S^{-1}K[x]$, where $S=k[x]\setminus\{0\}$. This is clearly a normal integral domain which is not necessarily a field.
However, if the extension $k\subset l$ is algebraic, and $K\otimes_kl$ is an integral domain, then it is a field since the ring extension $K\subset K\otimes_kl$ is generated by algebraic elements ($1\otimes a$, $a\in l$), so it is algebraic.