Skip to main content
1 of 9

This is the most complete treatment I could come up with; see Conclusions below for the final criterion (which has some computational practicality but is also a little limited). All morphisms below will be $k$-linear.

Generalities

Lemma 1. Let $k \subseteq K' \subseteq K$ and $k \subseteq L' \subseteq L$ be field extensions. If $K \otimes_k L$ is a field, then so is $K' \otimes_k L'$.

Proof. By Grothendieck–Sharp, either $K$ or $L$ is algebraic over $k$, hence either $K'$ or $L'$ is algebraic over $k$. Again by Grothendieck–Sharp, we get $$\dim\left(K' \underset k\otimes L'\right) = 0.$$ Since $K'\otimes_k L'\subseteq K\otimes_k L$ is a domain, it must be a field. $\square$

Corollary 2. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is a field if and only if $K' \otimes_k L'$ is a field for all finitely generated subfields $k \subseteq K' \subseteq K$ and $k \subseteq L' \subseteq L$.

Proof. One direction follows from the lemma, and the other because $$K \underset k\otimes L = \left(\underset{\substack{\longrightarrow \\ k \subseteq K' \subseteq K \\ \text{finitely generated}}}{\operatorname{colim}} K'\right) \underset k\otimes \left( \underset{\substack{\longrightarrow \\ k \subseteq L' \subseteq L \\ \text{finitely generated}}}{\operatorname{colim}} L'\right) = \underset{\substack{\longrightarrow \\ k \subseteq K' \subseteq K \\ k \subseteq L' \subseteq L \\ \text{finitely generated}}}{\operatorname{colim}} K' \underset k\otimes L',$$ and a filtered colimit of fields is a field. $\square$

We will also use the following well-known lemma.

Lemma 3. Let $k \subseteq K$ and $k \subseteq L$ be extensions such that $K \otimes_k L$ is a field. Then $k \subseteq K$ is separable algebraic, purely inseparable, or regular if and only if $L \subseteq K \otimes_k L$ is.

Proof. All three statements reduce to the finitely generated case. The result follows since a finitely generated field extension is separable algebraic if and only if it is étale, purely inseparable if and only if it is radicial, and regular if and only if it is geometrically integral, and these properties are stable under base change and descend under faithfully flat morphisms. $\square$

Now 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 decompositions of $k \subseteq K$ and $k \subseteq L$ into a separable algebraic, a purely inseparable algebraic, and a regular extension. Consider the commutative diagram Diagram of fields where each rectangle is a pushout.

Key reduction

By Corollary 2, if $K \otimes_k L$ is a field, the same goes for all other terms in the diagram. Conversely, all squares except the three squares going down the middle have the property that if the bottom three terms in it are fields, then so is the top, by the sufficient condition of the OP. Thus, following the central three vertical squares, we reduce to studying the case where $k \subseteq K$ and $k \subseteq L$ are either both separable algebraic, both purely inseparable, or both regular.

By Lemma 3, all arrows in the diagram are either étale, radicial, or geometrically integral, only depending on their location in the diagram. If all entries are fields, this translates to separable algebraic, purely inseparable algebraic, and regular field extensions.

Case I: separable algebraic extensions

This is actually the hardest part to say something explicit about. The following lemma is virtually devoid of content.

Lemma 4. Let $k \subseteq K$ and $k \subseteq L$ be separable algebraic extensions, and let $\bar k$ be a separable algebraic closure of $k$. Then the following are equivalent:

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

Proof. If $i(K)$ and $j(L)$ are linearly disjoint for one choice of $i$ and $j$, then $i \otimes j \colon K \otimes_k L \to \bar k$ is injective, so $K \otimes_k L$ is a field. The converse is also clear. $\square$

We will generalise this in Conclusions 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 $[G_k : U_K \cap U_L] = [G_k : U_K] \cdot [G_k : U_L]$.

Case II: purely inseparable algebraic extensions

This is a little more concrete. 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. We will make use of some ideas around $p$-bases (Tag 07P0) as well as (naive) cotangent complexes (Tag 00S0).

Lemma 5. Let $k$ be a field of characteristic $p > 0$, and let $A$ be a successive extension $$k = A_0 \subseteq A_1 \subseteq \ldots \subseteq A_r = A$$ where $A_i = A_{i-1}[X_i]/(X_i^p - x_i)$ for some $x_i \in A_{i-1}$. Then the $dx_i$ span the kernel of $\Omega_k \otimes_k A \to \Omega_A$, and $A$ is a field if and only if the $dx_i$ are linearly independent in $\Omega_k \otimes_k A$.

Proof. Since $\mathbf F_p \subseteq k$ is a separable field extension as $\mathbf F_p$ is perfect, we have $H_1\mathbf L_k = 0$ (see for example Tag 07E5, but it also follows from Tag 08Q1, localisation, and étale base change). Thus, the long exact cotangent sequence (Tag 00S2) for $\mathbf F_p \subseteq k \subseteq A$ reads $$0 \to H_1\mathbf L_A \to H_1\mathbf L_{A/k} \stackrel d\to \Omega_k \underset k\otimes A \to \Omega_A \to \Omega_{A/k} \to 0.$$ Moreover, we have $H_1\mathbf L_{A/k} = \bigoplus_i A(X_i^p-x_i)$ with connecting map taking $X_i^p-x_i$ to $d(X_i^p-x_i) = dx_i$. This proves the first statement. For the second, it suffices to show that $H_1\mathbf L_A = 0$ if and only if $A$ is a field. If $A$ is a field we already saw that $H_1\mathbf L_A = 0$. If $A$ is not a field, then let $i$ be the largest index such that $A_i$ is a field. Then $A_{i+1}$ is not a field, meaning that $x_{i+1}$ is a $p$-th power in $A_i$. Then the exact sequence $$0 \to H_1 \mathbf L_A \to H_1 \mathbf L_{A/A_i} \stackrel d\to \Omega_{A_i} \underset {A_i}\otimes A \to \ldots$$ shows that $H_1 \mathbf L_{A/A_i} \neq 0$ since $x_i$ maps to $0$ under $d$. $\square$

Proposition 6. Let $k \subseteq K$ and $k \subseteq L$ be purely inseparable extensions. Then $K \otimes_k L$ is a field if and only if $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $\ker(\Omega_k \otimes_k L \to \Omega_L) \otimes_k K$ intersect trivially (in $\Omega_k \otimes_k (K \otimes_k L)$).

Proof. We can reduce to the case that $k \subseteq K$ and $k \subseteq L$ are finitely generated, since tensor product as well as formation of $\Omega$ commutes with filtered colimits, and by Corollary 2. Say that $K$ can be obtained as $$k = K_0 \subseteq K_1 \subseteq \ldots \subseteq K_r = K,$$ where $K_i = K_{i-1}[X_i]/(X_i^p-x_i)$ for some $x_i \in K_{i-1}$. By Lemma 5, the $dx_i$ are a basis of the kernel of $\Omega_k \otimes_k K \to \Omega_K$. Moreover, $K \otimes_k L$ can be obtained from $L$ by $$L = K_0 \underset k\otimes L \subseteq \ldots \subseteq K_r \underset k\otimes L,$$ and $K_i \otimes_k L = (K_{i-1} \otimes_k L)[X_i]/(X_i^p - x_i)$ for all $i$. Again by Lemma 5, we see that $K \otimes_k L$ is a field if and only if the $dx_i$ remain linearly independent in $\Omega_L \otimes_k (K \otimes_k L)$, i.e. if and only if the intersection $$\left(\ker\left(\Omega_k \underset k\otimes K \right) \underset k\otimes L\right) \cap \left(\ker\left(\Omega_k \underset k\otimes L \right) \underset k \otimes L\right)$$ is trivial. $\square$

Case III: regular extensions

If $k \subseteq K$ and $k \subseteq L$ are regular field extensions, then Grothendieck–Sharp implies that $K \otimes_k L$ is a field if and only if one of $k \subseteq K$ and $k \subseteq L$ is a trivial extension.

This can actually be seen directly as well: if $K$ and $L$ both contain an element not in $k$, then these have to be transcendental over $k$ by regularity. 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 $i \otimes j \colon K \otimes_k L \to \Omega$ is not injective, since $x \otimes 1$ and $1 \otimes x$ both map to $x \in k(x) \subseteq \Omega$. Hence, $K \otimes_k L$ has a nonzero ideal.

Conclusions

Putting everything together, we obtain the following criterion:

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 $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $\ker(\Omega_k \otimes_k L \to \Omega_L) \otimes_k K$ intersect trivially.
  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.

Proof. The first statement follows from Lemma 4, noting that purely inseparable and regular extensions are geometrically irreducible, so do not affect irreducibility.

For the second statement, note that the separable extensions $k \subseteq K^{\operatorname{sep}}$, $k \subseteq L^{\operatorname{sep}}$, $K^{\operatorname{alg}} \subseteq K$, and $L^{\operatorname{alg}} \subseteq L$ do not affect reducedness. Moreover, for a separable extension $k \subseteq \ell$ the map $\Omega_k \otimes_k \ell \to \Omega_\ell$ is injective (for example by the cotangent computation in the proof of Lemma 5), and an isomorphism if $k \subseteq \ell$ is moreover algebraic. This gives $$\ker\left(\Omega_k \underset k\otimes K^{\operatorname{alg}} \to \Omega_{K^{\operatorname{alg}}} \right) \underset{K^{\operatorname{alg}}}\otimes K = \ker\left(\Omega_k \underset k\otimes K \to \Omega_K\right),$$ and similarly for $L$. Since both subspaces in question are defined over $K^{\operatorname{alg}} \otimes_k L^{\operatorname{alg}}$, we reduce to the same statement with $K$ and $L$ replaced by $K^{\operatorname{alg}}$ and $L^{\operatorname{alg}}$. We also get $\Omega_k \otimes_k K^{\operatorname{sep}} = \Omega_{K^{\operatorname{sep}}}$, so $$\ker\left(\Omega_k \underset k\otimes K^{\operatorname{alg}} \to \Omega_{K^{\operatorname{alg}}}\right) = \ker\left(\Omega_{K^{\operatorname{sep}}} \underset{K^{\operatorname{sep}}}\otimes K^{\operatorname{alg}} \to \Omega_{K^{\operatorname{alg}}}\right),$$ and similarly for $L$. Base change along $k \subseteq L^{\operatorname{alg}}$ may give multiple irreducible components on $K^{\operatorname{alg}} \otimes_k L^{\operatorname{alg}}$, but two submodules of $V \otimes_k (K^{\operatorname{alg}} \otimes_k L^{\operatorname{alg}})$ for some $k$-vector space $V$ intersect trivially if and only if they do so on each component of $K^{\operatorname{alg}} \otimes_k L^{\operatorname{alg}}$. The components of $K^{\operatorname{alg}} \otimes_k L^{\operatorname{alg}}$ are in bijection with those of $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$, and now our result follows by Proposition 6 applied to each component of $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ separately.

The third statement was explained by the OP, using Grothendieck–Sharp, and the final statement follows since a commutative ring is a field if and only if it is irreducible and reduced of dimension $0$. $\square$

We end with the generalisation of Lemma 4 to arbitrary field extensions:

Corollary. 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. If $K$ and $L$ are algebraic over $k$, it suffices to take $\Omega$ an algebraic closure of $k$.

Proof. If $K \otimes_k L$ is a field, then the map $i \otimes j \colon K \otimes_k L \to \Omega$ is forced to be injective. Conversely, we will construct embeddings where this fails whenever one of the three criteria of the theorem fails.

If criterion 1 fails, then choose embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$ into an algebraic closure $\bar k$ of $k$ for which $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ are not linearly disjoint. All we need to do is extend to embeddings of $K$ and $L$ into a big algebraically closed field $\Omega$.

The purely inseparable part always extends uniquely, so we automatically get $i \colon K^{\operatorname{alg}} \to \bar k$ and $j \colon L^{\operatorname{alg}} \to \bar k$. Finally, the further extensions to $K$ and $L$ are regular, hence geometrically integral, so $\bar k \subseteq K \otimes_{K^{\operatorname{alg}}} \bar k$ and $\bar k \subseteq L \otimes_{L^{\operatorname{alg}}} \bar k$ are still field extensions. Then choosing a big algebraically closed overfield $\Omega$ of both does the trick.

If criterion 2 fails, then $K \otimes_k L$ is not reduced, so the map $i \otimes j \colon K \otimes_k L \to \Omega$ can never be injective.

So far, pretty much any embedding sufficed, but for the case of criterion 3 we need to choose the correct one. This is also easy: if both $k \subseteq K$ and $k \subseteq L$ are transcendental, choose purely transcendental subfields $k(x) \in K$ and $k(x) \in L$. Viewing both as extensions of $k(x)$ instead of $k$, we can construct $k(x)$-linear embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into some algebraically closed field $\Omega$. Then the map $i \otimes j \colon K \otimes_k L \to \Omega$ is not injective, because both $x \otimes 1$ and $1 \otimes x$ are sent to $x \in k(x) \subseteq \Omega$.

This proves the first statement, and the final statement is Lemma 4. $\square$