10
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 8
    $\begingroup$ EGA IV$_2$ 4.3 (relies on the same C-C Seminaire as Borel!). Borel meant $F/k$ to algebraic. In (a), by direct limits can assume $k'$ is $k$-finite, so $k'=k[x]/(f)$ for separable $f$, so $F_{k'} = F[x]/(f)$ is reduced (as ${\rm{gcd}}(f,f')=1$ over $F$ since OK over $k$); multiple primes for $F=k'$ nontrivial and $k$-finite. For (b) can assume char($k)=p>0$ and $k'$ is $k$-finite, so ${k'}^{p^e}\subset k$ for big $e$; non-reduced for $F=k'=k[x]/(x^p-a)$ for $a \in k-k^p$. For (c), with $k'=k(x_i)=S^{-1}k[x_i]$ for $S=k[x_i]-\{0\}$, we have $F_{k'} = S^{-1}F[x_i]$ is a localization of $F[x_i]$. $\endgroup$
    – nfdc23
    Commented Jan 23, 2016 at 22:01
  • $\begingroup$ cf. mathoverflow.net/questions/82083/when-is-the-tensor-product- of-two-fields-a-field for the result of Grothendieck mentioned in the question. $\endgroup$
    – ACL
    Commented Jan 24, 2016 at 14:54

2 Answers 2

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
5
$\begingroup$

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).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .