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Consider two extension fields $K/k, L/k$ of a field $k$.
A frequent question is whether the tensor product ring $K\otimes_k L$ is a field. The answer is "no" and this answer is often justified by some particular case of the following result:

Proposition Given a strict field extension $k \subsetneq K$ , the tensor product $K\otimes_kK$ is not a field.
Proof The multiplication $m:K\otimes_kK\to K:x\otimes y \mapsto xy$ cannot be injective for dimension reason, hence it has a kernel which is a non-zero ideal of the ring $K\otimes_kK$ and thus that ring cannot be a field.
Corollary If the extensions $K/k, L/k$ contain finite subextensions $k\subsetneq K'\subset K, k \subsetneq L'\subset L$ which are $k$-isomorphic ( $K' \stackrel {k}{\simeq} L'$), then $K\otimes_k L$ is not a field.

The most powerful and beautiful tool in this context is Grothendieck's underrated result ( often attributed to Sharp who redicovered it ten years after Grothendieck! cf. this answer in math.stackexchange ):
Theorem (Grothendieck) The Krull dimension of the tensor product of the field extensions $K/k, L/k$ is given by the formula
$$ \dim_{\mathrm{Krull}}(K\otimes_k L) = \min(\operatorname{trdeg}_k(K),\operatorname{trdeg}_k(L)) $$

This shows that we can only hope that $K\otimes_k L$ will be a field if at least one of the extensions $K,L$ is algebraic over $k$. An example where we do obtain a field is when the extension fields $K,L$ are finite dimensional over $k$ with relatively prime dimensions.
[To see this, embed $K$ and $L$ into an algebraic closure $\overline k$ of $k$ and notice that the canonical morphism $K\otimes_k L\to K\cdot L\subset \overline k$ is an isomorphism because it is surjective and because $K\cdot L$ has the same dimension as $K\otimes_k L$ by the relative primeness assertion]

A fairly general criterion for obtaining a field is the following.
A sufficient condition The tensor product $K\otimes_k L$ is a field if the three conditions below simultaneously hold:

  1. At least one of $K,L$ is algebraic over $k$.
  2. At least one of $K,L$ is primary over $k$
  3. At least one of $K,L$ is separable over $k$

    Proof
    The ring $K\otimes_k L$ is zero-dimensional by 1) and Grothendieck's formula.
    Once divided by its nilpotent radical it is a domain by 2).
    However, by 3), its nilpotent radical is zero.
    So $K\otimes_k L$ is a zero-dimensional domain, hence a field.

[Reminder: a field extension $E/k$ is primary if the algebraic closure of $k$ in $E$ is purely inseparable over $k$. In that case for any field extension $F/k$ the quotient $E\otimes_k F/Nil (E\otimes_k F)$ is a domain. In other words $Spec(E\otimes_k F) $ is irreducible.]

I feel that all these results are a little fragmentary and my not very precise question is , as you have guessed :
Question Is there a general procedure for deciding whether the tensor product $K \otimes_k L$ of two field extensions is a field?

Bibliography Grothendieck's result is to be found in EGA IV, Quatrième partie, page 349 , Remarque (4.2.1.4). This is in the Errata et Addenda to the volume!

Edit Since linearly disjointness keeps getting mentioned in the comments, let me insist that it makes no sense to say that $K$ and $L$ are linearly disjoint unless they are provided with embeddings into an extension $E$ of $k$.
For example take $K=L=k(x)$ ($x$ an indeterminate over $k$) and consider the extension $k \subset E=k(y,z)$, the function field in two indeterminates over $k$.
If you embed $K$ (resp. $L$) into $E$ by sending $x\mapsto y$ (resp.$x\mapsto z$), the images will be linearly disjoint.
However if you embed $K$ (resp. $L$) into $E$ by sending $x\mapsto y$ (resp. $x \mapsto y$), the images will be equal and certainly not linearly disjoint.
However the $k$-algebra $k(x)\otimes_k k(x) $ does not care about all these embeddings: Grothendieck has decreed that it is not a field, and that's it.
(Our friend Pete Clark has a section on these questions in his extremely well-written online notes, page 65. According to Pete, that section was inspired by an exchange he had concerning a question asked by our other friend Andrew Critch )

New edit: Is all this a real problem? Since we know so many conditions ensuring that $K\otimes_k L$ is a field and so many conditions ensuring that it isn't, I wonder if someone could come up with a tensor product of extensions $K\otimes_k L$ for which MO users couldn't (immediately) say whether it is a field or not.
I would be very happy to consider such a challenge as an answer, to upvote it and possibly to accept it.

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In Grothendieck's formula, the last K should be L, presumably? –  Vladimir Dotsenko Nov 28 '11 at 13:51
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Dear Ralph, a) your $KL$ only makes sense if $K,L$ are subfields of a big extension of $k$. In this case your condition on dimensions is equivalent to linear disjointness. b) Your remark that $K\otimes _k L$ is not a field if $K\leq L$ also follows from my Corollary. c) However your last sentence "In particular..." is not correct: for example the tensor product of two finite extensions of a finite field is a field as soon as the two extensions have relatively prime dimensions. (The simplest case is $\mathbb F_4 \otimes_{\mathbb F_2} \mathbb F_8=\mathbb F_{64}$.) –  Georges Elencwajg Nov 28 '11 at 16:52
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Dear @Ralph, concerning a): yes you can $k$-embed $K$ and $L$ into $\bar k$ . The problem is that this is non canonical and the $KL$ you obtain depends on the embeddings. For example, take $K=\mathbb Q, K=\mathbb Q(\sqrt [3] 2), L=\mathbb Q(\omega\sqrt [3] 2)$ where $\omega =e^{2i\pi/3}$. You can embed $K,L$ naturally into $\bar {\mathbb Q}\subset \mathbb C$ in which case you obtain $KL=\mathbb Q(\omega,\sqrt [3] 2)$. But you can also embed $L$ onto $\mathbb Q(\sqrt [3] 2)$, in which case you obtain $KL=\mathbb Q(\sqrt [3] 2)$. (Anyway $K\otimes_{\mathbb Q} L$ is not a field by my Corollary) –  Georges Elencwajg Nov 28 '11 at 18:26
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@Will : One can take a degree 4 irreducible polynomial $f$ with Galois group $S_4$ and no real roots, then $K=\mathbf{Q}[X]/f$ and $L=\mathbf{R}$ will work ($K$ has no non-trivial subfield and $K \otimes_{\mathbf{Q}} L$ is not a field). –  François Brunault Nov 29 '11 at 17:15
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This suggests that the Galois closure of $K$ sharing a subfield with $L$ is enough to cause problems. –  Will Sawin Nov 30 '11 at 6:50

2 Answers 2

A remark concerning Georges general question: if both $K$ and $L$ are separable algebraic extensions, and $k\subseteq K$ is finite, then $K\otimes_k L = L[x]/fL[x]$, where $f\in k[x]$ is the minimal polynomial of a primitive element of $K$. So to decide whether the tensor product is a field amounts to deciding whether $f\in L[x]$ is irreducible. This seems to indicate that an answer to Georges questions highly depends on the nature of the field $k$, more precisely on its Galois theory.

Note also that if $k\subseteq K$ is separable algebraic and $k\subseteq L$ is purely inseparable, then regardless of the embeddings into the algebraic closure of $k$, the resulting extensions are linearly disjoint over $k$. Hence the tensor product is a field in this case. I guess that one can exploit this fact to reduce the whole problem to considering two separable algebraic or two purely inseparable extensions.

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Dear Hagen, thank you for your thoughtful answer: your remark on the separable algebraic case is quite relevant. That the the tensor product of a separable algebraic extension K and a purely inseparable extension L is a field follows from A sufficient condition in the question since a purely inseparable extension is trivially primary. –  Georges Elencwajg Nov 29 '11 at 13:50

We can very nearly solve the separable portion. The transcendental portion seems simple enough, and I'm not sure about the inseparable. This consists of most of the interesting part of the problem for me. I'm not sure what you think.

The tensor product is a field if the Galois closures of $K$ and $L$ do not contain any subfields which are isomorphic. Proof:

We can reduce to the case with $K/k$ and $L/l$ Galois, since $K \otimes L$ is contained in the tensor product of the Galois closures, and an Artin ring inside a field is a field.

Let $K/k$ and $L/k$ Galois, then $K\otimes L$ contains $k(K,L)$. We must show that the degrees of this extension is large enough. Consider the Galois group $G$, which contains subgroups $H_K$ and $H_L$ that fix $K$ and $L$. $H_K$ and $H_L$ are normal, and they are not together contained in any subgroup. Therefore $H_KH_L=G$, so $|H_K||H_L|\geq |G|$, so $|G/H_K||G/H_L|\leq G$, so $[k(K,L):k]\geq[K:k][L:k]$, so the extension is a field.

Edit: If one of the fields is algebraic, adding transcendentals to the second one couldn't possibly make it not a field, so we only need to consider the algebraic parts of the second.

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(Will, you have actually stated the contrapositive of the Corollary in the original post (I nearly made the same mistake myself, awhile back)). –  B R Dec 15 '11 at 0:11
    
Not true. My argument proves that things are fields if they satisfy certain conditions. –  Will Sawin Dec 15 '11 at 1:28
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Apologies, I had forgotten why I didn't make my comment. I should have said "follows from linear-disjointness". See Pete's linked notes, results 107, 108, and 111. Though I agree that this covers most of the interesting stuff (and I'm not exactly clear what is left over). –  B R Dec 15 '11 at 2:55
    
(The results are on pages 66 and 67) –  B R Dec 15 '11 at 2:56
    
For the purely inseparable case, how about some criterion like: there does not exist $a \in (K - k)$ and $b \in (L - k)$ such that $a^p = b^p$? This is also implied by the statement: the fields $L$ and $K$ do not contain any nontrivial extensions of $k$ which are isomorphic. –  name Mar 20 '12 at 3:54

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