Proposition 9

— Let K be a field, $\overline{K}$ an algebraic closure of K and L an extension of K. Then the following conditions are equivalent:

a) L is separable over K and K is relatively algebraically closed in L.

b) L is a regular extension of K.

c) The ring $\overline{K} \otimes_K L$ is an integral domain.

d) Let $\overline{L}$ be an algebraic closure of L. Then L is linearly disjoint over K from the relative algebraic closure of K in $\overline{L}$.

Moreover, when these conditions hold, then $\overline{K} \otimes_K L$ is a field.

a) $\Rightarrow$ b): Let M be an extension of K. Under the hypotheses of a), the ring $M \otimes_K L$ is an integral domain, by V, p. 140, Cor.

b) $\Rightarrow$ c): This follows from Def. 2.

c) $\Rightarrow$ d): With the notation of d) we can identify $\overline{K}$ with the relative algebraic closure of K in $\overline{L}$ (V, p. 22, Ex. 2). Suppose that the ring $A = \overline{K} \otimes_K L$ is an integral domain. Let E be a subextension of K of finite degree over K; the subring

E $\otimes_K$ L of A is an integral domain and hence is an algebra of finite degree over L; by the Cor. of V, p. 10 it is a field. Since $\bar{K}$ is the union the increasing directed set of extensions E of the above type, A is a field (V, p. 11, Prop. 3). The canonical homomorphism of A into $\bar{L}$ mapping $x \otimes y$ to $xy$ (for $x \in \bar{K}$ and $y \in L$) is therefore injective, so L and $\bar{K}$ are linearly disjoint over K.

$d) \Rightarrow a)$: Under the hypotheses of d) we have $L \cap \bar{K} = K$, so K is relatively algebraically closed in L; further if p is the characteristic exponent of K, the field $L$ is linearly disjoint from $K^{p^{-\infty}}$ over $K$, hence L is separable over K (V, p. 123, Cor. 1).