Let $K/F$ be a field extension. I am interested in the situation where there exists a field extension $L/F$ such that the ring $L \otimes_FK$ is not a field.
If there exists $z\in K \setminus F$ such that $z$ is algebraic over $F$, then I think I can show that $K \otimes_F K$ is not a field. So my question is: If no such element exists in $K$, does it follow that for any field extension $L$ of $F$, $L \otimes_F K$ is a field?
On
As a graduate student I remember being disappointed that it was hard to find much information concerning tensor products of fields. Later, as with many things, I realized that it depends a good bit on knowing where to look: it turns out that the more standard topic (found in most "serious" treatments of field theory) of linear disjointness is closely related.
You can find some material on when $L \otimes_F K$ is a field (and also when it is a domain) in $\S 12$ of my field theory notes.
If $K/F$ is a non-trivial extension, then $K\otimes_FK$ is not a field. We have a natural $K$-algebra map $K\otimes_FK\rightarrow K$ with $x\otimes y\mapsto xy$. Take $x\in K\setminus F$. Then $x\otimes 1\neq 1\otimes x$ because there is a $K$-basis for $K\otimes_FK$ containing $1\otimes x$ and $1\otimes 1$. But $x\otimes 1-1\otimes x$ is in the kernel of the map in question, so $K\otimes_FK$ has a non-trivial, proper ideal, so it isn't a field.