Algebra which becomes a field after a base change

Question

Let $A$ be an associative algebra over a field $k$, and let $L/k$ be any field extension.

Assume that $A \otimes_k L$ is a field (not necessarily a finite extension of $L$ !). Can we deduce that $A$ is a field?

I know that $A$ is a (commutative, unital) integral domain, since $A \cong A \otimes_k k$ embeds in $A \otimes_k L$. But how to go further?

Maybe if $L/k$ is Galois, we can use $A \cong (A \otimes_k L)^{Gal(L/k)}$, after checking that this is indeed a $k$-algebra isomorphism ?

Answer 1

Yes, $A$ is a field. We don't need to assume that the field extension is Galois or finite.

Lemma: If $k \to L$ is a field extension, then the base change / extension of scalars functor $(-) \otimes_k L$ (which we'll write $(-)_L$ to keep things short) is faithful (induces injections $\text{Hom}_k(V, W) \to \text{Hom}_L(V_L, W_L)$) and conservative (reflects isomorphisms: if $T : V \to W$ is a linear map such that $T_L : V_L \to W_L$ is an isomorphism, then $T$ is an isomorphism).

Proof. Faithfulness can be proven pretty explicitly by choosing bases if necessary. As for conservativity, if $V$ and $W$ are finite-dimensional we can argue using determinants; in general we can use the fact that faithful functors reflect epimorphisms and monomorphisms, so if $T_L$ is an isomorphism then $T$ is both an epimorphism and a monomorphism and hence an isomorphism (note that we really need to be working in a nice category like an abelian category here; this is false in general categories). $\Box$

Now we can argue as follows. An element $a \in A$ is invertible iff left multiplication $A \xrightarrow{a} A$ is an isomorphism (of $k$-modules). If $a \in A$ is nonzero then $a_L \in A_L$ is nonzero by faithfulness, and hence invertible since $A_L$ is a field by hypothesis. By conservativity, $a$ is invertible. Hence $A$ is a field.

This argument applies more generally to any faithfully flat ring homomorphism $R \to S$ replacing $k \to L$: these are also faithful and conservative, and so also don't make anything invertible that wasn't invertible already.