Proposition 3

Let $K$ be a field.

a) Every subalgebra of a separable $K$-algebra is separable.

b) Every direct limit of separable $K$-algebras is separable.

c) Every product of separable $K$-algebras is separable.

d) Let $\mathbf{A}$ be a $K$-algebra and $K'$ an extension of $K$. For $\mathbf{A}$ to be separable it is necessary and sufficient for the $K'$-algebra $\mathbf{A}_{(K')}$ obtained from $\mathbf{A}$ by extension of scalars to be separable.

Let $L$ be an extension of $K$. Let $\mathbf{A}$ be a separable algebra over $K$ and $B$ a subalgebra of $\mathbf{A}$; then the ring $L \otimes_K \mathbf{A}$ is reduced, and $L \otimes_K B$ is isomorphic to a subring of $L \otimes_K \mathbf{A}$, hence is reduced. Thus $B$ is separable and a) is proved. In the same way $b$) may be proved, using the canonical isomorphism of $L \otimes_K \varprojlim A_i$ with $\varprojlim L \otimes_K A_i$ (II, p. 290, Prop. 7) and $c$) is proved by remarking that $L \otimes_K \left( \prod_{i \in I} A_i \right)$ is isomorphic to a subring of $\prod_{i \in I} (L \otimes_K A_i)$ (II, p. 306, Prop. 15).

We shall use the notation of $d$). For every extension $L'$ of $K'$ the rings $L' \otimes_{K'} A_{(K')}$ and $L' \otimes_K A$ are isomorphic (II, p. 278, Prop. 2). We conclude that if $A$ is a separable $K$-algebra, then $A_{(K')}$ is a separable $K'$-algebra. Conversely suppose that $A_{(K')}$ is a separable $K'$-algebra ; the preceding remark shows that $L' \otimes_K A$ is reduced for every extension $L'$ of $K$ containing $K'$ as subextension. Let $L$ be an extension of $K$; by the Scholium (V, p. 13) there exists an extension $L'$ of $K$ containing $K'$ as a subextension and a $K$-homomorphism of $L$ into $L'$; the ring $L \otimes_K A$ is thus isomorphic to a subring of $L' \otimes_K A$ and hence reduced. This proves that $A$ is a separable $K$-algebra.