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:



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


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.



What are the basic possibilities for a tensor product of two fields?
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.


