Lemma 1

— Let $\Omega$ be an algebraic closure of $K$; then we have $h(L) \leq h(\Omega)$ for every extension $L$ of $K$, with equality when $L$ is algebraically closed.

Let $L'$ be the algebraic closure of $K$ in $L$. For every homomorphism $u$ of $A$ into $L$ we have $[u(A) : K] \leq n$, hence $u(A) \subset L'$ by V, p. 18, Prop. 2; we thus have $h(L') = h(L)$. Since the extension $L'$ of $K$ is isomorphic to a subextension of $\Omega$ (V, p. 20, Th. 1), we have $h(L') \leq h(\Omega)$. If $L$ is algebraically closed, then $L'$ is an algebraic closure of $K$; the extensions $L'$ and $\Omega$ of $K$ are then isomorphic (V, p. 23, Th. 2) and so $h(L') = h(\Omega)$; the Lemma follows directly from this.

By Lemma 1, the number $h(L)$ has the same value for all algebraically closed extensions $L$ of $K$; this number will be denoted by $[A : K]$, and called the separable degree of $A$.

Let $A$ and $B$ be two commutative algebras of finite degree over $K$. We shall establish the formula

Let $L$ be an algebraically closed extension of $K$ and denote by $\mathscr{H}(A)$ the set of algebra homomorphisms of $A$ into $L$, and similarly define $\mathscr{H}(B)$ and $\mathscr{H}(A \otimes_K B)$. By definition we have Card $\mathscr{H}(A) = [A : K]$, and corresponding formulae for [B : K], and [A \otimes_K B : K]. Moreover (III, p. 465, Formula (6)), the formula $(u * v)(a \otimes b) = u(a)v(b)$ defines a bijection $(u, v) \mapsto u * v$ of $\mathscr{H}(A) \times \mathscr{H}(B)$ onto $\mathscr{H}(A \otimes_K B)$, whence Formula (7) follows.

Let $K'$ be an extension of $K$; we shall prove the formula

For take $L$ to be an algebraic closure of $K'$. The formula $\tilde{u}(x) = u(1 \otimes x)$ (for $x \in A$) defines a bijection $u \mapsto \tilde{u}$ between the set of $K'$-homomorphisms of $A_{(K')}$ into $L$ and the set of $K$-homomorphisms of $A$ into $L$, whence (8).

Finally, suppose that $K'$ is an extension of finite degree of $K$; if $A'$ is a commutative $K'$-algebra of finite degree, then it is also a commutative $K$-algebra of finite degree and we have $[A': K] = [A': K'] \cdot [K': K]$ (V. p. 10, Th. 1). We shall prove the formula

For let $S$ (resp. $T$) be the set of $K$-homomorphisms of $K'$ (resp. $A'$) into an algebraic closure $L$ of $K$; for every $\sigma \in S$ we denote by $T_\sigma$ the set of elements $f$ of $T$ such that $f(a \cdot 1) = \sigma(a)$ for all $a \in K'$. Then the family $(T_\sigma)_\sigma$ is a partition of $T$ and we have Card $S = [K': K]$; now for each $\sigma \in S$ the set $T_\sigma$ consists of $K'$-homomorphisms of $A'$ into the algebraically closed extension $(L, \sigma)$ of $K'$, hence Card $T_\sigma = [A': K']_s$, and so we have proved (9).