TITLE: What are the basic possibilities for a tensor product of two fields?
QUESTION [10 upvotes]: Let $k$ be a field, with $F,k'$ field extensions of $k$.  The ring $k' \otimes_k F$ is denoted by $F_{k'}$.  In Borel's Linear Algebraic Groups, it is claimed (I believe erroneously) that "each of [$F_{k'}$'s] prime ideals is minimal."  Indeed, by a result due to Grothendieck, the Krull dimension of $F_{k'}$ is the minimum of the transcendence degrees of $F/k, k'/k$.  So there may be a tacit assumption that $F/k$ is algebraic, I am not sure.  In any case, Borel claims results for three possibilities:

(a) $k'$ is separable algebraic over $k$.
Then $F_{k'}$ is reduced, but it may have more than one prime ideal.
(b) $k'$ is algebraic and purely inseparable over $k$.
Then $F_{k'}$ has a unique prime ideal (consisting of the nilpotent elements), but $F_{k'}$ need not be reduced.
(c) $k'$ is purely transcendental over $k$.
Then $F_{k'}$ is clearly an integral domain.

I was wondering if anyone might be able to provide references or proofs for any of the statements (a), (b), (c) as well as examples, e.g. an example where $F_{k'}$ has more than one prime ideal.  Unfortunately, the reference Borel gives is an old algebraic geometry conference from the 1950s which I have had a lot of difficulty understanding.

REPLY [8 votes]: Looks like nfdc23 has explanations for (a),(b), and (c). 
But: indeed, primes of $F_{k'}$ are not in general minimal if $F/k$ is not algebraic. Let $k'=k(x)$ and $F=k(y)$ be transcendental extensions of $k$. 
Then $F_{k'}$ identifies with a subalgebra of the field $k(x,y)$, hence is an integral domain [assertion (c)] - i.e. $\{0\}$ is prime.
There is a homomorphism $\phi:F_{k'} = k' \otimes_k F \to k(t)$ - where $t$ is again transcendental - which on pure tensors is given by $f(x) \otimes g(y) \mapsto f(t) \cdot g(t)$. And then $x\otimes 1 - 1\otimes y$ is a non-zero element of $P = \ker \phi$, hence $P$ is a prime ideal which is not minimal.

REPLY [5 votes]: For a), if $L$ is a finite Galois extension of $k$ with Galois group $G$, then
$$L \otimes_k L \cong \prod_{g \in G} L$$
has $|G|$ prime ideals. For b), if $k = \mathbb{F}_p(a)$ and $L = k[x]/(x^p - a)$, then
$$L \otimes_k L \cong L[x]/(x^p - a) \cong L[x]/(x - \sqrt[p]{a})^p$$
is not reduced. I haven't thought about c).