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 ~~ generalized ten years later by Sharp who suppressed Grothendieck's hypothesis that $K\otimes_k L$ should be noetherian):

**Theorem (Grothendieck-Sharp )** 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:

At least one of $K,L$ is algebraic over $k$.

At least one of $K,L$ is primary over $k$

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.

**Edit (April 24th, 2016):Apologies to Sharp**

Due to EGA's abstruse cross-reference system I had missed that Grothendieck's formula is proved by him only under the supplementary hypothesis that $K\otimes_k L$ is *noetherian*.

It is indeed Sharp who first proved that formula in complete generality, without any noetherian hypothesis:

Rodney Y. Sharp,

The Dimension of the Tensor Product of Two Field Extensions, Bulletin of the London Mathematical Society9Issue 1 (1977) pp 42–48, doi:10.1112/blms/9.1.42

isa field as soon as the two extensions have relatively prime dimensions. (The simplest case is $\mathbb F_4 \otimes_{\mathbb F_2} \mathbb F_8=\mathbb F_{64}$.) $\endgroup$18more comments