Theorem 4

— Let $\Omega$ be an algebraically closed extension of a field K and L a subextension of $\Omega$. Then the following conditions are equivalent: a) L is separable over K. b) The intersection of the kernels of the 0-algebra homomorphisms of $\Omega \otimes_K L$ into $\Omega$ is reduced to 0. c) For any elements $a_1, \ldots, a_n$ of L linearly independent over K there exist K-automorphisms $\sigma_1, \ldots, \sigma_n$ of $\Omega$ such that $\det(\sigma_i(a_j)) \neq 0$. d) Let V be a vector sub-K-space of L of finite dimension. Every K-linear mapping of V into $\Omega$ is a linear combination (with coefficients in R) of the restrictions to V of K-automorphisms of $\Omega$.

$d) \Rightarrow c)$: Let $a_1, \ldots, a_n$ be elements of L linearly independent over K and let V be the vector sub-K-space of L generated by $a_1, \ldots, a_n$. The mapping $f \mapsto (f(a_1), \ldots, f(a_n))$ is an R-linear bijection of $\mathrm{Hom}_K(V, R)$ onto $\Omega^n$. Suppose that d) holds, then there exist K-automorphisms $a_1, \ldots, a_n$ of $\Omega$ such that the elements $(\sigma_i(a_1), \ldots, \sigma_i(a_n))$ of $\Omega^n$ (for $1 \leq i \leq n$) form a basis of $\Omega^n$. We thus have $\det(\sigma_i(a_j)) \neq 0$, so d) implies c).

c) $\Rightarrow$ b): Assume that c) holds and let x be in $\Omega \otimes_K L$. We write x in the form $\sum_{j=1}^n x_j \otimes a_j$, with $x_1, \ldots, x_n$ in $\Omega$ and the elements $a_1, \ldots, a_n$ of L linearly independent over K. Choose K-automorphisms $\sigma_1, \ldots, \sigma_n$ of $\Omega$ such that $\det \sigma_i(a_j) \neq 0$; let $\chi_1, \ldots, \chi_n$ be the R-homomorphisms of $\Omega \otimes_K L$ into $\Omega$ such that $\chi_i(a \otimes b) = a \cdot \sigma_i(b)$ for $a \in \Omega$ and $b \in L$. Suppose that $\chi_i(x) = 0$ for $1 \leq i \leq n$, in other words, that $\sum_{j=1}^n x_j \cdot \sigma_i(a_j) = 0$ for $1 \leq i \leq n$. Since we have assumed that the matrix $(\sigma_i(a_j))$ has a non-zero determinant, we have $x_i = 0$ for $1 \leq i \leq n$ and so $x = 0$.

b) $\Rightarrow$ a): Since every extension of a field of characteristic 0 is separable (V, p. 122, Th. 1) it suffices to examine the case where K is of characteristic $p \neq 0$. Let X be the set of all R-algebra homomorphisms of $\Omega \otimes_K L$ into $\Omega$ and f the homomorphism of $\Omega \otimes_K L$ into $\Omega^X$ defined by $f(u) = (\chi(u))_{\chi \in X}$ for $u \in \Omega \otimes_K L$. Condition b) means that f is injective and so implies that the ring $\Omega \otimes_K L$ is reduced. Hence condition b) of Th. 2 (V, p. 122) is satisfied with $K' = \Omega$, so L is separable over K.

a) $\Rightarrow$ d): Suppose that L is separable over K. Let V be a vector sub-K-space of finite dimension of L, $V_{(\Omega)} = \Omega \otimes_K V$ the vector $\Omega$-space derived from V by extension of scalars and $f_0$ the linear form on $V_{(\Omega)}$ such that $f_0(x \otimes y) = xy$ for $x \in \Omega, y \in V$. Denote by G the group of K-automorphisms of $\Omega$; for $\sigma \in G$ we put $\sigma_V = \sigma \otimes \mathrm{Id}_V$ and $g_\sigma = \sigma \circ f_0 \circ \sigma_V^{-1}$.

For each $\sigma \in G$ the mapping $g_\sigma$ of $V_{(\Omega)}$ into $\Omega$ is $\Omega$-linear and maps $x \otimes y$ to $x \cdot \sigma(y)$ for $x \in \Omega, y \in V$. The kernel $N_\sigma$ of $g_\sigma$ is therefore a vector subspace of $V_{(\Omega)}$, and the same holds for $N = \cap_{\sigma \in G} N_\sigma$. If p is the characteristic exponent of K, the field of invariants of G in $\Omega$ is equal to $K^{p^{-\infty}}$ (V, p. 112, Prop. 10). We clearly have $\sigma_V(N) = N$ for all $\sigma \in G$; therefore (V, p. 63, Cor.) the vector $\Omega$-space N is generated by $N_0 = N \cap (K^{p^{-\infty}} \otimes V)$. Since L is separable over K, the fields $K^{p^{-\infty}}$ and L are linearly disjoint over K (V, p. 123, Cor. 1); we have $K^{p^{-\infty}} \otimes_K V \subset K^{p^{-\infty}} \otimes_K L$ and $f_0(x \otimes y) = xy$ for $x \in \Omega$ and $y \in V$. It follows that the restriction of $f_0$ to $K^{p^{-\infty}} \otimes_K V$ is injective. Now $f_0 = g_1$ is zero on N and a fortiori on $N_0 \subset K^{p^{-\infty}} \otimes_K V$. We thus have $N_0 = 0$, whence $N = 0$. Since V is of finite dimension over K, $V_{(\Omega)}$ is of finite dimension over $\Omega$; the intersection of the kernels of the linear forms $g_\sigma$ is zero, hence (II, p. 301, Th. 7) the family $(g_\sigma)_{\sigma \in G}$ generates the dual of $V_{(\Omega)}$. Now let $u$ be a K-linear mapping of V into $\Omega$; let $\tilde{u}$ be the linear form on $V_{(\Omega)}$ which maps $x \otimes y$ to $xu(y)$ for $x \in \Omega$ and $y \in V$. By what has been said there exist elements $\sigma_1, \ldots, \sigma_n$ of G and $\lambda_1, \ldots, \lambda_n$ of $\Omega$ such that $\tilde{u} = \sum_{i=1}^n \lambda_i g_{\sigma_i}$, whence $u(y) = \sum_{i=1}^n \lambda_i \sigma_i(y)$ for all $y \in V$. So we have shown that a) implies d).