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:

  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 $k$

    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 ( 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 Society 9 Issue 1 (1977) pp 42–48, doi:10.1112/blms/9.1.42

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    $\begingroup$ 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 $\mathbb F_4 \otimes_{\mathbb F_2} \mathbb F_8=\mathbb F_{64}$.) $\endgroup$ Nov 28, 2011 at 16:52
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    $\begingroup$ 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) $\endgroup$ Nov 28, 2011 at 18:26
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    $\begingroup$ @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). $\endgroup$ Nov 29, 2011 at 17:15
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    $\begingroup$ Small question: why is the canonical morphism $K \otimes_k L \to KL$ surjective? For example, if $x\in K, y\in L$, then, what is the preimage of $(x+y)^{-1}$? As far as I understand, $K.L$ is the smallest field containing both $K$ and $L$. The problem is in expressing an inverse of the sum of two elements as a sum itself. $\endgroup$
    – FNH
    Jan 7, 2020 at 4:41
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    $\begingroup$ Dear @Fawzy: By $K \cdot L$ I mean the $k$-algebra generated by $K$ and $L$, not the $k$-field extension generated by those fields. I think this is the prevailibng convention: Bastida for example adopts it and writes $K\vee L$ for the field extension, so that $K\vee L=\operatorname {Frac} (K \cdot L)$ (the field of fractions of the domain $K \cdot L$). $\endgroup$ Jan 7, 2020 at 10:11

3 Answers 3


This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the separable algebraic and algebraic closures of $k$ in $K$ and $L$. The result is the following.

Theorem. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then

  1. $K \otimes_k L$ is irreducible if and only if $k \subseteq K^{\operatorname{sep}}$ and $k \subseteq L^{\operatorname{sep}}$ are linearly disjoint with respect to one (equivalently, every) choice of embedding into $\bar k$.
  2. $K \otimes_k L$ is reduced if and only if the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ inside $\Omega_k \otimes_k (K \otimes_k L)$ is generically trivial.
  3. $K \otimes_k L$ has dimension $0$ if and only if $k \subseteq K$ or $k \subseteq L$ is algebraic.

In particular, $K \otimes_k L$ is a field if and only if all three criteria hold.

Here we say a property holds generically if it holds for the localisation at all minimal primes. The second criterion is not very nice in general, but if $K \otimes_k L$ has dimension $0$ we can remove the word 'generically', so we get a rather clean criterion. This applies in particular in the case the OP is interested in, for $K \otimes_k L$ to be a field. On the other hand, in Example 3 we show that 'generically' in part 2 of the theorem cannot be removed, even in the case $K = L$.

In Proposition 3 below we give a more conceptual (but more technical) criterion equivalent to the one stated here.

Statements 1, 2, and 3 of the theorem generalise without difficulty to finite tensor products of fields. The statement is the 'obvious generalisation' of the version for binary tensor products. For example, in 1 we need that $K_1^{\operatorname{sep}} \otimes_k \ldots \otimes_k K_n^{\operatorname{sep}} \to \bar k$ is injective for one (equivalently, any) choice of embeddings into $\bar k$. In 2, we need that generically the sum of the subspaces is direct. In 3, we need that at most one of the extensions has a transcendental part.

Remark. We will use the algebro-geometric properties of field extensions: a field extension $k \subseteq \ell$ is

  1. algebraic if and only if it is integral (obvious);
  2. separable if and only if it has geometrically reduced fibres (Tag 030W);
  3. separable algebraic if and only if it is integral with geometrically reduced fibres (combine 1 and 2);
  4. purely inseparable (algebraic) if and only if it is radicial (by definition, see Tag 01S3);
  5. primary if and only if it has geometrically irreducible fibres (partially explained in Tag 037Q);
  6. regular if and only if it has geometrically integral fibres (combine 2 and 5).

This is useful because not all rings we encounter are fields, and the algebro-geometric properties are preserved by base change. In fact all these properties descend under faithfully flat morphisms (see [EGA IV$_2$, Prop 2.6.1(v)] for radicial, [EGA IV$_2$, Prop 2.7.1(xvii)] for integral, and obvious for all other properties), so $k \to K$ has one of the properties above if and only if $L \to K \otimes L$ does.

Below we will treat parts 1, 2, and 3 of the theorem separately; see Lemma 1, Proposition 3, and Lemma 4. Also have a look at examples 1, 2, 3, and 4, because they provide some insight into the types of behaviours you can expect.

1. Irreducibility

The precise version of part 1 of the theorem is as follows.

Lemma 1. Let $k \subseteq K$ and $k \subseteq L$ be field extensions, and let $\bar k$ be a separable algebraic closure of $k$. Then the following are equivalent:

  1. $K \otimes_k L$ is irreducible;
  2. $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is irreducible;
  3. $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field;
  4. for every pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K)$ and $j(L)$ of $\bar k$ are linearly disjoint;
  5. for one pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ of $\bar k$ are linearly disjoint.

Proof. Note that $K^{\operatorname{sep}} \to K$ and $L^{\operatorname{sep}} \to L$ are primary, hence flat with geometrically irreducible fibres. Then the same goes for $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to K \otimes_k L$, so (1) $\Leftrightarrow$ (2) follows from [EGA IV$_2$, Prop. 4.5.7] (or modify the proof of Tag 038F). Since $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is reduced of dimension $0$, it is a field if and only if it is irreducible, proving (2) $\Leftrightarrow$ (3).

If $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ are linearly disjoint for one choice of $i$ and $j$, then $$i \otimes j \colon K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to \bar k$$ is injective, so $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field. The converse is also clear, showing (3) $\Leftrightarrow$ (4) $\Leftrightarrow$ (5). $\square$

We will generalise properties 4 and 5 in Further remarks and examples to arbitrary field extensions. In the separable algebraic case, it has a reinterpretation in terms of Galois theory: if $i \colon K \to \bar k$ and $j \colon L \to \bar k$ is a choice of embeddings, then we get closed subgroups $H_K$ and $H_L$ of $G_k = \operatorname{Gal}(\bar k/k)$. The condition on linear disjointness then means that for any open subgroups $U_K \supseteq H_K$ and $U_L \supseteq H_L$, we have $[G_k : U_K \cap U_L] = [G_k : U_K] \cdot [G_k : U_L]$.

Remark. One is tempted to believe that $K$ and $L$ are linearly disjoint if and only if $i(K) \cap j(L) = k$ for every pair of embeddings $i \colon K \to \bar k$ and $j \colon L \to \bar k$, but this is not true:

Example 1. Let $k \subseteq M$ be an $S_5$-extension; for example $k = \mathbf C(\sigma_1,\sigma_2,\sigma_3,\sigma_4.\sigma_5)$ and $M = \mathbf C(x_1, x_2, x_3, x_4,x_5)$, where $$\prod_{i=1}^5(X-x_i) = X^5 - \sigma_1 X^4 + \sigma_2 X^3 - \sigma_3 X^2 + \sigma_4 X - \sigma_5.$$ Let $K$ be the fixed field of $(12)$ and $L$ the fixed field of $(12345)$. Then $K \cap L$ is fixed by $(12)$ and $(12345)$, hence equals $k$. What's more, any conjugate of $K$ is the fixed field of a transposition, and any conjugate of $L$ is the fixed field of a $5$-cycle, hence their intersection is still $k$ since any pair of a transposition and a $5$-cycle generate $S_5$. However, $K$ and $L$ cannot be linearly disjoint, because $[K:k] = 60$ and $[L:k] = 24$, but they're both contained in a degree $120$ extension of $k$.

2. Reducedness

We will use repeatedly that if $k \subseteq K$ and $k \subseteq L$ are separable, then they are geometrically reduced, hence $K \otimes_k L$ is reduced by Tag 034N. In particular, if $\operatorname{char} k = 0$ there is nothing to discuss, so assume $\operatorname{char} k = p > 0$.

For any field $k$ of characteristic $p$, write $\Omega_k = \Omega_{k/\mathbf Z} = \Omega_{k/\mathbf F_p}$ for the module of absolute differentials. Although my criterion was originally based on the idea of $p$-bases (Tag 07P0), I was ultimately unable to prove it in that language. The proof uses the (naive) cotangent complex $\mathbf L_{B/A}$ (see Tags 00S0 and 08R6) instead, where again we write $\mathbf L_A$ for $\mathbf L_{A/\mathbf F_p}$.

Lemma 2. Let $k \subseteq \ell$ be a field extension in characteristic $p > 0$, and let $A$ be a successive extension $$\ell = A_0 \subseteq A_1 \subseteq \ldots \subseteq A_r = A$$ where $A_i = A_{i-1}[X_i]/(X_i^p - x_i)$ for some $x_i \in A_{i-1}$. Then the following are equivalent:

  1. $A$ is a field;
  2. $H_1\mathbf L_A = 0$;
  3. the boundary map $d \colon H_1\mathbf L_{A/k} \to \Omega_k \otimes_k A$ is injective.
  4. the boundary map $d \colon H_1\mathbf L_{A/\ell} \to \Omega_\ell \otimes_\ell A$ is injective.

Proof. For any field $F$ of characteristic $p$, the extension $\mathbf F_p \subseteq F$ is separable as $\mathbf F_p$ is perfect. Therefore, we have $H_1\mathbf L_F = 0$ (see for example Tag 07E5, but it also follows from Tag 08Q1, localisation, and the étale case of Tag 0D0M). Thus $H_1\mathbf L_A = 0$ if $A$ is a field, showing (1) $\Rightarrow$ (2).

If $A$ is not a field, then let $i$ be the largest index such that $A_i$ is a field, so by the above $H_1 \mathbf L_{A_i} = 0$. Then $A_{i+1}$ is not a field, meaning that $x_{i+1}$ is a $p$-th power in $A_i$. Then the long exact cotangent sequence (Tag 00S2) for $\mathbf F_p \subseteq A_i \subseteq A$ reads $$0 \to H_1 \mathbf L_A \to H_1 \mathbf L_{A/A_i} \stackrel d\to \Omega_{A_i} \underset {A_i}\otimes A \to \ldots.$$ Hence $H_1 \mathbf L_A \neq 0$ since $X_{i+1}^p - x_{i+1}$ maps to $-dx_{i+1} = 0$ under $d$, showing (2) $\Rightarrow$ (1).

Finally, the long exact cotangent sequence for $\mathbf F_p \subseteq k \subseteq A$ reads $$0 \to H_1\mathbf L_A \to H_1\mathbf L_{A/k} \stackrel d\to \Omega_k \underset k\otimes A \to \Omega_A \to \Omega_{A/k} \to 0.$$ This shows (2) $\Leftrightarrow$ (3), and replacing $k$ by $\ell$ gives (2) $\Leftrightarrow$ (4). $\square$

Proposition 3. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then the following are equivalent:

  1. $K \otimes_k L$ is reduced;
  2. $H_1 \mathbf L_{K \otimes_k L}$ is generically trivial;
  3. the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ (in $\Omega_k \otimes_k (K \otimes_k L)$) is generically trivial.

Proof. We can reduce to the case that $k \subseteq K$ and $k \subseteq L$ are finitely generated, since tensor product, formation of $\Omega$ and $\mathbf L$, and computation of kernel commute with filtered colimits, and nilpotents are defined over a finitely generated subring.

By 'baby Serre's criterion' (Tag 031R), we need to check that $K \otimes_k L$ is $S_1$ and $R_0$. But $S_1$ is automatic, for example by Tag 0339, so it suffices to check $R_0$. Thus, we see that $K \otimes_k L$ is reduced if and only if every localisation $A$ of $K \otimes_k L$ at a minimal prime $\mathfrak p$ is a field.

We claim that every such $A$ is of the form described in Lemma 2. Indeed, let $k \subseteq K' \subseteq K$ be a subextension such that $k \subseteq K'$ is separable and $K' \subseteq K$ is purely inseparable; for example by taking $K'$ to be the elements separable over a given transcendence basis. Then $K' \otimes_k L$ is reduced and $K' \otimes_k L \to K \otimes_k L$ is radicial and faithfully flat (in particular a bijection on irreducible components by Tag 01S4).

If $\mathfrak q$ is the preimage of $\mathfrak p$ in $K' \otimes_k L$ and $\ell$ is the localisation of $K' \otimes_k L$ at $\mathfrak q$, then $\ell$ is a field since $\mathfrak q$ is minimal and $K' \otimes_k L$ reduced. The map $\ell \to A$ is a map as described in Lemma 2, since $K' \subseteq K$ is of that form. Thus from Lemma 2 we see that $A$ is a field if and only if $H_1 \mathbf L_A = 0$. Since formation of $\mathbf L_A$ commutes with localisation by Tag 00S7, we see that $K \otimes_k L$ is reduced if and only if $H_1\mathbf L_{K \otimes_k L}$ is generically trivial, proving (1) $\Leftrightarrow$ (2).

For (2) $\Leftrightarrow$ (3), the exact sequences \begin{alignat*}{1} 0 & \to H_1\mathbf L_{K/k} & \stackrel d\to \Omega_k \underset k\otimes K & \to \Omega_K & \to \Omega_{K/k} & \to 0,\\ 0 & \to H_1\mathbf L_{L/k} & \stackrel d\to \Omega_k \underset k\otimes L & \to \Omega_L & \to \Omega_{L/k} & \to 0. \end{alignat*} for $\mathbf F_p \subseteq k \subseteq K$ and $\mathbf F_p \subseteq k \subseteq L$ identify $H_1\mathbf L_{K/k}$ and $H_1\mathbf L_{L/k}$ as the kernels of the maps on differentials. Choosing a presentation of $K \otimes_k L$ by tensoring presentations for $K$ and $L$ over $k$, we see that $$\mathbf L_{K \otimes_k L/k} \simeq \left( \mathbf L_{K/k} \underset k\otimes L \right) \oplus \left( \mathbf L_{L/k} \underset k\otimes K \right),$$ and the connecting map $H_1\mathbf L_{K \otimes_k L} \to \Omega_k \otimes_k (K \otimes_k L)$ is given by $(d,d)$ with respect to this decomposition. In particular, the connecting map for $K \otimes_k L$ is generically injective if and only if the images of the connecting maps for $K$ and $L$ have generically trivial intersection. $\square$

Remark. However, it is not true that $K \otimes_k L$ is a field if and only if there are no elements that become a $p$-th power in both; see Example 2 below (even in the finite purely inseparable case). In fact, $K \otimes_k K$ may be reduced even if $k \subseteq K$ is inseparable; see Example 3 below (a transcendental primary inseparable extension).

On the other hand, this last phenomenon does not happen for algebraic inseparable extensions $k \subseteq K$, since $k[x]/f \otimes_k k[x]/f \subseteq K \otimes_k K$ clearly picks up nilpotents if $f$ is inseparable. See also Example 4 at Further remarks and examples for an example of the interplay between the separable algebraic part and the purely inseparable part.

Example 2. Let $k = \mathbf F_p(x,y,z)$, let $K = k[u,v]/(u^{p^2}-z, v^p-xu^p-y)$ be the field adjoining a $p^2$-th root $u$ of $z$ and a $p^2$-th root $v$ of $x^pz+y^p$ (this example is due to Sweedler, as far as I can tell), and let $L = k(x^{1/p}, y^{1/p})$. Then $$K \underset k\otimes L = \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[v]}{(v^p-xz^{1/p}-y)} \cong \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[X]}{(X^p)},$$ since $xz^{1/p}+y$ is a $p$-th power in $\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})$. Thus, $K \otimes_k L$ is not reduced.

The elements in $k$ that become $p$-th powers in $K$ are given by $$k \cap K^p = \mathbf F_p(x^p,y^p,z).\label{Eq become p-th power}\tag{1}$$ Indeed, the inclusion $\supseteq$ is obvious. For the converse, consider the chain of field extensions $$K^{p^2} = k \cap K^{p^2} \subseteq k \cap K^p \subseteq k \cap K = k.$$ We have $[k:K^{p^2}] = [k:k^{p^2}]/[K^{p^2}:k^{p^2}] = (\operatorname{tr. deg.} k)^2/[K:k] = p^3$, and $x^p$ is in $k \cap K^p$ but not $k \cap K^{p^2}$ since $dx \neq 0 \in \Omega_K$ (see computation below). Thus, $[k:k \cap K^p] \leq p^2$, which gives equality in (\ref{Eq become p-th power}) by a dimension count.

By a similar (but easier) computation, we get $$k \cap L^p = \mathbf F_p(x,y,z^p).$$ Thus, the only elements in $k$ that become $p$-th powers in both $K$ and $L$ are already $p$-th powers in $k$ to begin with.

What's going on is that $\ker(\Omega_k \otimes_k K \to \Omega_K)$ is not defined over $k$, and a fortiori cannot be spanned by elements of the form $dx_i$ for $x_i \in k$. Indeed, $\Omega_k = k \cdot dx \oplus k \cdot dy \oplus k \cdot dz$, and one computes $$\Omega_K = \frac{K \cdot dx \oplus K \cdot dy \oplus K \cdot du \oplus K \cdot dv}{u^p\ dx + dy}.$$ The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $dz$ and $u^p\ dx + dy$, which is not defined over $k$.

Example 3. Let $k = \mathbf F_p(s,t)$, and let $K$ be the fraction field of $k[x,y]/(sx^p + ty^p - 1)$ (this example is due to Mac Lane, according to MO user nfdc23). Then $K$ is geometrically irreducible since $$\bar k[x,y]/(sx^p + ty^p - 1) = \bar k[x,y]/(ux+vy-1)^p,$$ where $u^p = s$ and $v^p = t$, which is a thickening of $\bar k[x,y]/(ux+vy-1) = \bar k[x]$. In particular, $k \subseteq K$ is primary, i.e. $k$ is algebraically closed in $K$.

The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $df = x^p\ ds + y^p\ dt$, where $f = sx^p + ty^p - 1$. Again this is not defined over $k$, and in this case not even over an algebraic extension of $k$.

If we take two copies of $K$, then the elements \begin{align*} df_1 &= x_1^p\ ds + y_1^p\ dt\\ df_2 &= x_2^p\ ds + y_2^p\ dt \end{align*} agree at the locus $V((x_1y_2-y_1x_2)^p)$, which is a strict closed subscheme. Thus the intersection of the kernels is generically trivial (since $K$ is the function field of a regular but not geometrically regular curve over $k$, so the diagonal is not an irreducible component), so $K \otimes_k K$ is reduced.

(In a previous version of this answer, I erroneously stated a version of the criterion for reducedness that did not take this type of example into account.)

On the other hand, $K \otimes_k K \otimes_k K$ cannot be reduced, because we get $3$ relations $df_1$, $df_2$, and $df_3$ involving only $2$ variables $ds$ and $dt$.

3. Zero-dimensionality

Lemma 4. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is $0$-dimensional if and only if one of $k \subseteq K$ and $k \subseteq L$ is algebraic.

Proof. Immediate by Grothendieck–Sharp.

This can actually be seen directly as well: clearly the tensor product is $0$-dimensional if one of $k \subseteq K$ and $k \subseteq L$ is algebraic. Conversely, if $K$ and $L$ both contain an element transcendental over $k$, then this gives embeddings $k \subseteq k(x) \subseteq K$ and $k \subseteq k(x) \subseteq L$. View $K$ and $L$ as extensions of $k(x)$ and choose $k(x)$-linear embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed extension $k(x) \subseteq \Omega$. Then $x \otimes 1 - 1 \otimes x$ is in the kernel $\mathfrak p$ of $i \otimes j \colon K \otimes_k L \to \Omega$.

The maps $k(x) \to K$ and $k(x) \to L$ are flat, hence the same goes for $k(x) \otimes_k k(x) \to K \otimes_k L$. The preimage of $\mathfrak p$ under this map contains $x \otimes 1 - 1 \otimes x$, hence is a nonzero ideal. But $k(x) \otimes_k k(x)$ is a domain, so going down for flat morphisms (see Tag 00HS) shows that $\mathfrak p$ cannot be minimal. Hence, $K \otimes_k L$ has positive dimension. $\square$

Further remarks and examples

Here is an interesting example of the interplay between the separable and purely inseparable parts, in light of statement 2 of the precise version of the theorem (note that already the separable part disqualifies this example from producing a field, but the point is only the reducedness statement now):

Example 4. Let $k = \mathbf F_p(\sigma_1,\sigma_2)$, and consider the extension $M = \mathbf F_p(x_1,x_2)$ given by $(X-x_1)(X-x_2) = X^2 - \sigma_1X + \sigma_2$. This is a $\mathbf Z/2$-Galois cover (even if $p = 2$, but feel free to assume $p > 2$). Let $K = \mathbf F_p(x_1^{1/p},x_2)$ and $L = \mathbf F_p(x_1,x_2^{1/p})$.

Then $K^{\operatorname{sep}} = L^{\operatorname{sep}} = M$, and $M \otimes_k M \cong M^{\mathbf Z/2}$ given by $m_1 \otimes m_2 \mapsto (m_1\sigma(m_2))_{\sigma \in \mathbf Z_2}$. That is, on the factor $M_0$ the inclusions are both given by $x_i \mapsto x_i$, and on the factor $M_1$ the inclusion from $L^{\operatorname{sep}}$ swaps $x_1$ and $x_2$.

Then $K \otimes_k L^{\operatorname{sep}} = M_0(x_1^{1/p}) \times M_1(x_1^{1/p})$, and $K^{\operatorname{sep}} \otimes_k L \cong M_0(x_2^{1/p}) \times M_1(x_1^{1/p})$ since the embedding $L^{\operatorname{sep}} \to M_1$ sends $x_2$ to $x_1$. We see that $K \otimes_k L$ is not reduced, because it contains $M_1(x_1^{1/p}) \otimes_{M_1} M_1(x_1^{1/p})$.

In terms of differentials, the relations $\sigma_1 = x_1+x_2$ and $\sigma_2 = x_1x_2$ give $d\sigma_1 = dx_1 + dx_2$ and $d\sigma_2 = x_1\ dx_2 + x_2\ dx_1$. The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is spanned by $$dx_1 = \frac{d\sigma_2 - x_1\ d\sigma_1}{x_2-x_1},$$ and the opposite happens for the kernel of $\Omega_k \otimes_k L \to \Omega_L$. These subspaces of $\Omega_k \otimes_k K$ and $\Omega_k \otimes_k L$ are not defined over $k$, so their intersection does not show up over $k$. In the factor $M_0$ of $K \otimes_k L$ they are linearly independent (note that indeed $M_0(x_1^{1/p}) \otimes_{M_0} M_0(x_2^{1/p})$ is a field), but in the other factor $M_1$ they give the same subspace because $x_2$ and $x_1$ are swapped in $L^{\operatorname{sep}} \to M_1$.

We end with the generalisation of Lemma 1 to arbitrary field extensions:

Lemma. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is a field if and only if for every pair of embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed field $\Omega$, the images $i(K)$ and $j(L)$ are linearly disjoint (i.e. $i \otimes j \colon K \otimes_k L \to \Omega$ is injective). If $K$ and $L$ are algebraic over $k$, it suffices to take $\Omega$ an algebraic closure of $k$, and it suffices to take one pair $(i,j)$ of embeddings.

Proof. Since $$\operatorname{Hom}_k(K,\Omega) \times \operatorname{Hom}_k(L,\Omega) = \operatorname{Hom}_k\left(K \underset k\otimes L, \Omega\right),$$ the result follows since a commutative ring $R$ is a field if and only if every map $R \to \Omega$ to a(n algebraically closed) field is injective.

This proves the first statement, and the final statement is Lemma 1. $\square$

  • 1
    $\begingroup$ In Example 1, it's not the case that any transposition and any 4-cycle generate $S_4$. E.g. $(1234)$ and $(13)$ generate a copy of $D_4$. So the corresponding conjugates of $K$ and $L$ should meet in a degree 3 extension of $k$. (So the tempting belief is still plausible, especially since $12\cdot 6 / 3 = 24$...) $\endgroup$ Jul 21, 2020 at 1:13
  • 1
    $\begingroup$ @benblumsmith oh you're right, you need prime order for that. Fixed. $\endgroup$ Jul 21, 2020 at 2:25
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    $\begingroup$ What a wonderfully detailed answer! Thanks a lot Remy. $\endgroup$ Jul 27, 2020 at 11:34

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.

  • 2
    $\begingroup$ 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. $\endgroup$ Nov 29, 2011 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.

  • $\begingroup$ (Will, you have actually stated the contrapositive of the Corollary in the original post (I nearly made the same mistake myself, awhile back)). $\endgroup$
    – B R
    Dec 15, 2011 at 0:11
  • $\begingroup$ Not true. My argument proves that things are fields if they satisfy certain conditions. $\endgroup$
    – Will Sawin
    Dec 15, 2011 at 1:28
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    $\begingroup$ 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). $\endgroup$
    – B R
    Dec 15, 2011 at 2:55
  • $\begingroup$ (The results are on pages 66 and 67) $\endgroup$
    – B R
    Dec 15, 2011 at 2:56
  • $\begingroup$ 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. $\endgroup$
    – name
    Mar 20, 2012 at 3:54

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