Let us show by induction on the integer $n \geq 0$ that every sequence $(u_1, \ldots, u_n)$ of distinct elements of $\mathscr{H}$ is free. The case $n = 0$ being trivial, we may henceforth suppose that $n \geq 1$; let $a_1, \ldots, a_n$ be elements of L such that

For $x,\ y$ in A we have whence By the induction hypothesis, the elements $u_1, \ldots, u_{n-1}$ of $\mathscr{H}$ are linearly independent, whence $\alpha_i [u_i(x) - u_n(x)] = 0$ for $1 \leq i \leq n-1$ and for all $x$ in A. Since the $u_i$ are distinct, this implies that $\alpha_i = 0$ for $i \neq n$, hence $\alpha_n u_n = 0$ and so $\alpha_n = \alpha_n u_n(1) = 0$ (on denoting by 1 the unit element of A). We have thus shown that $a_1, \ldots, a_{n-1}, \alpha_n$ are zero, and this proves the theorem.

Let A be the algebra of the monoid $\Gamma$ with coefficients in L and $(e_\gamma)_{\gamma \in \Gamma}$ the canonical basis of A over L (III, p. 446). For every L-linear mapping $u$ of A into L let us write $\tilde{u}(\gamma) = u(e_\gamma)$ (for $\gamma \in \Gamma$); then the mapping $u \mapsto 3$ is an isomorphism of vector L-spaces of $\mathrm{Hom}_L(A, L)$ onto $L^\Gamma$ which maps onto X the set of L-algebra homomorphisms of A into L. Now it suffices to apply Th. 1 with $K = L$.

COROLLARY 2 (Dedekind's theorem). — Let E and L be two extensions of K. The set of K-homomorphisms of E into L is free over L. If E is of finite degree over K, the number of K-homomorphisms of E into L is at most equal to $[E : K]$.

The last assertion follows from the first, taking account of Formula (2).

Let B be the set of elements of $L^n$ of the form $(u_1(x), \ldots, u_n(x))$ with $x \in A$. By Th. 1, there is no sequence $(\alpha_1, \ldots, \alpha_n)$ of elements not all zero in L such that $\sum_{i=1}^n \alpha_i u_i(x) = 0$ for all $x \in A$; therefore (II, p. 301, Th. 7) B generates the vector space $L^n$ over L. So there exist elements $a_1, \ldots, a_n$ of $A$ such that the matrix $(u_i(a_j))_{1 \leq i,j \leq n}$ is invertible.

Let us define the polynomial $g \in L[Y_1, \ldots, Y_n]$ by

Let $y_1, \ldots, y,$ be in $K$; writing $x = \sum_{i=1}^n y_i a_i$, we have whence $g(y_1, \ldots, y_n) = 0$ by the hypothesis on $f$. Since the field $K$ is infinite, we have $g = 0$ (IV, p. 18, Cor. 2); now the matrix $(u_i(a_j))$ has an inverse $(b_{ij})$ and we have whence $f = 0$.

Th. 2 has no analogue for finite fields. Thus let $K$ be a finite field with q elements, $A = L = K$ and $f(X) = X^q - X$. We have $x^q = x$ for all $x \in K$ (V, p. 93, Prop. 2); therefore if $u$ is the identity automorphism of $K$, we have $f(u(x)) = 0$ for all $x \in K$, even though $f$ is not zero.

The equivalence of a) and b) follows from Formula (6); on the other hand the n projections $K^n \to K$ are algebra homomorphisms, hence a) implies c). If c) holds, the algebra homomorphisms of A into $\mathbf{K}$ form a basis of the dual of A (V, p. 27, Th. 1), we denote them by $u_1, \ldots, u_n$; then $a \mapsto (u_i(a))$ is an isomorphism of A onto the algebra $K''$, whence a). We have thus established the equivalence of the conditions a), b) and c).

Suppose that b) holds and let M be an A-module; then the homotheties $(e_i)_M$ of ratio $e_i$ are projectors of $M$, and $M$ is a direct sum of the $e_i M$, which are sub-A-modules. We may thus suppose that there exists an index $i$ such that $(e_j)_M = 0$ for $j \neq i$. Hence every vector subspace of $M$ is a sub-A-module, whence d).

Conversely suppose that d) holds and consider the A-module A,. There exists then a basis $(f_i)$ of the vector K-space A such that $Af_i = Kf_i$ for $i = 1, \ldots, n$. After replacing each $f_i$ by a suitable scalar multiple, if necessary, we may suppose that $1 = f_i + \ldots + f_n$. If $i \neq j$, then $f_i f_j$ belongs to $Af_i \cap Af_j = Kf_i \cap Kf_j$ hence it is zero. Now $f_i = f_i f_1 + \ldots + f_i f_n = f_i^2$, whence b).

The vector space $\mathrm{Hom}_K(A, L)$ over L has dimension $[A : K]$, by Formula (2), and $\mathscr{H}$ is a free subset of $\mathrm{Hom}_K(A, L)$ by Th. 1 (V, p. 27). We thus have Card $\mathscr{H} \leq [A : K]$ with equality if and only if $\mathscr{H}$ is a basis of $\mathrm{Hom}_K(A, L)$. There exists an isomorphism of vector L-spaces, say $\pi : \mathrm{Hom}_K(A, L) \to A_{(L)}^*$, characterized by $u(x) = (\pi u)(1 \otimes x)$ for $x \in A$, and $\pi$ maps $\mathscr{H}$ onto the set $\mathscr{H}_L$ of L-algebra homomorphisms of $A(\cdot)$ into L. Finally the equivalence of a) and c) in Prop. 1 shows that the algebra $A_{(L)}$ over L is diagonalizable if and only if $\mathscr{H}_L$ generates the vector space $A_{(L)}^*$ over L. This completes the proof of the Corollary.

a) The algebra A is etale. b) There exists an extension of finite degree which diagonalizes A. c) The extension $\Omega$ of K diagonalizes A.

Suppose that A is etale. Let $n$ be the degree of A over K, let L be an extension of K which diagonalizes A and let $\mathscr{H}$ be the set of algebra homomorphisms of A into L. By the Cor. to Prop. 1 we have Card $\mathscr{H} = n$. On the other hand, for each $u \in \mathscr{H}$, we have $[u(A):K] \leq n$. By V, p. 18, Th. 2, the subextension L' of L generated by the images of elements of $\mathscr{H}$ is of finite degree over K. Since there exist $n$ distinct homomorphisms of A into L', the extension L' diagonalizes A, by the Cor. 1 of Prop. 1. This shows that a) implies b).

Since every extension of finite degree of K is isomorphic to a subextension of $\Omega$ (V, p. 20, Th. 1), b) implies c). Finally c) clearly implies a).

It suffices to show that an algebra $K^n$ has only a finite number of subalgebras and ideals, and that the subalgebras and quotient algebras of $K^n$ are diagonalizable. We denote by $(\varepsilon_1, \ldots, \varepsilon_n)$ the canonical basis of $K^n$.

Let A be a subalgebra of $K^n$ and let $v_1, \ldots, v_n$ be the restrictions to A of the n projections $K^n \to K$. Since the intersection of the kernels of the $v_i$ is clearly 0, the $v_i$ generate the vector K-space dual to A (II, p. 302, Cor. 1); hence the K-algebra A is diagonalizable (V, p. 29, Prop. 1).

For every subset I of $\{1, 2, \ldots, n\}$ put $\varepsilon_I = \sum_{i \in I} \varepsilon_i$. It is clear that the elements $\varepsilon_I$ are idempotents of $K^n$; we have $\varepsilon_I = 0$ if and only if I is empty, and $\varepsilon_I \varepsilon_J = \varepsilon_{I \cap J}$. By what has been said, every subalgebra A of $K^n$ is diagonalizable; by condition b) of Prop. 1 every subalgebra A of $K^n$ therefore admits a basis $(\varepsilon_{I_1}, \ldots, \varepsilon_{I_p})$, where $(I_1, \ldots, I_p)$ is a partition of $\{1, 2, \ldots, n\}$, and there are only a finite number of such subalgebras.

For every subset I of $(1, 2, \ldots, n)$ let $a_i$ be the vector subspace of $K^n$ having as basis the idempotents $\varepsilon_i$ for $i \in I$; then it is clear that $a_i$ is an ideal of $K^n$; moreover if $J = (1, 2, \ldots, n) - I$, then the residue classes $\overline{\varepsilon}_j$ of $\varepsilon_j$ mod $a_i$, for $j \in J$ form a basis of $K^n / a_i$. We have $\overline{\varepsilon}_j^2 = \overline{\varepsilon}_j$ and $\overline{\varepsilon}_j \overline{\varepsilon}_k = 0$ if $j \neq k$, hence the algebra $K^n / a_I$ is diagonalizable, by Prop. 1 of V, p. 29.

It remains to show that every ideal of $K^n$ is of the form $a_I$. Let I be the set of integers $i$ such that $1 \leq i \leq n$ and $\varepsilon_i \in a$, then $a_i \subset a$. Let $x = x_1 \varepsilon_1 + \cdots + x_n \varepsilon_n$ be an element of $a$ (with $x_1, \ldots, x_n$ in $K$) and let $i$ be in $(1, 2, \ldots, n) - I$. We have $x_i \varepsilon_i = x \varepsilon_i \in a_i$, and $\varepsilon_i \notin a$, whence $x_i = 0$. Thus $x = \sum_{i \in I} x_i \varepsilon_i$ and this shows that $x \in a_i$. We have now shown $a \subset a_I$, whence finally $a = a_i$.

Suppose that $A$ is etale; each of the algebras $A_1, \ldots, A_r$ is isomorphic to a quotient of $A$, and hence is etale by Prop. 3. Conversely, every extension of $K$ which diagonalizes $A_1, \ldots, A_r$ clearly diagonalizes $A$.

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).

Let $\Omega$ be an algebraic closure of $K$ and $A?$ the set of algebra homomorphisms of $A$ into $\mathbf{R}$. We have Card $A? = [A: K]$, and $A$ is etale if and only if $A$ is diagonalized by the extension $\Omega$ of $K$ (V, p. 30, Prop. 2). Thus Prop. 4 follows from the Cor. of V, p. 29.

We have $[C: K] = [A: K] \cdot [B: K]$ and the corresponding Formula (7) for the separable degrees. Further we have $[A: K], \leq [A: K]$ and corresponding formulae for $B$ and $C$. It follows that $[C: K] = [C: K]$, if and only if we have both $[A: K] = [A: K]$, and $[B: K] = [B: K]$; now it is enough to apply Prop. 4.

a) For a $K$-algebra $A$ to be etale, it is necessary and sufficient that the $K'$-algebra $A_{(K')}$ should be etale.

b) Let $A'$ be an algebra over $K'$, not reduced to 0. For $A'$ to be etale over $K$ it is necessary and sufficient that $A'$ should be etale over $K'$ and $K'$ etale over $K$.

We argue as for Cor. 1, applying this time (8) for a) and (9) for b).

A) Let $L$ be an algebraic closure of $K$ ($V$, p. 23, Th. 2). For $A$ to be etale it is necessary and sufficient that the algebra $A_{(L)}$ over $L$ should be diagonalizable ($V$, p. 30, Prop. 2). Further, the $A$-module $\Omega_L(A_{(L)})$ is isomorphic to $\Omega_K(A) \otimes_A A_{(L)}$ (III, p. 572, Prop. 20), hence to $\Omega_K(A) \otimes_K L$, by the associativity of the tensor product; therefore $\Omega_K(A) = 0$ is equivalent to $\Omega_L(A_{(L)}) = 0$. To prove Th. 3 it is therefore enough to consider the case where $K$ is algebraically closed and to show that the algebra $A$ is diagonalizable if and only if $\Omega_K(A) = 0$.

B) Suppose that $A$ is diagonalizable; then ($V$, p. 29, Prop. 1), the vector space $A$ is generated by the idempotents of $A$. The assertion $\Omega_K(A) = 0$ is thus a consequence of the following lemma:

From the relation $e = e^2$ we deduce $de = 2e \cdot de$; on multiplying by $e$ we obtain $e \cdot de = 2e \cdot de$, hence $e \cdot de = 0$, and so finally, $de = 2e \cdot de = 0$.

C) We first prove two lemmas:

The algebra $A/m$ is an extension of finite degree of the algebraically closed field $K$, whence $[A/m : K] = 1$. Hence for each $a \in A$ there exists a unique scalar $\lambda$ such that $a - \lambda \cdot 1 \in m$; write $D(a)$ for the residue class of $a - \lambda \cdot 1$ mod $m^2$. It is clear that $D$ is a $K$-derivation of $A$ into the $A$-module $m/m^2$. The universal property of $\Omega_K(A)$ (III, p. 569) and the hypothesis $\Omega_K(A) = 0$ now imply that $D = 0$, whence $m/m^2 = 0$ and so $m = m^2$.

Let $(a_1, ..., a_r)$ be a generating system of the ideal $a$; since $a = a^2$, there exist elements $x_{ij}$ in $a$ such that $a_i = \sum_{j=1}^r x_{ij} a_j$ for $1 \leq i \leq r$. Write $M$ for the square matrix of order $r$ whose elements are $\delta_{ij} - x_{ij}$ and let $D$ be its determinant. There exists (III, p. 532, Formula (26)) a square matrix $N$ of order $r$ with elements in $A$ such that $N \cdot M = D \cdot I$, whence immediately $Da_j = 0$ for $1 \leq j \leq r$ and so finally $Da = 0$. Now the matrix $M$ is congruent to $I$, mod $a$, hence $D \equiv I$ mod $a$. Put $e = 1 - D$; then $e \in a$ and $ex = x$ for all $x \in a$. It follows that $e$ is an idempotent and $a$ is equal to $Ae$.

With these lemmas established, let us show by induction on the degree of $A$ that $A$ is diagonalizable if $K$ is algebraically closed and $\Omega_K(A) = 0$. Let $m$ be a maximal ideal of $A$ (I, p. 104). By Lemmas 3 and 4, there exists an idempotent $e$ such that $m = Ae$; we have seen that $A/m$ is of degree 1 over $K$. Hence $A$ is a direct sum of the ideals $a = (1 - e)A$ and $m$ and we have $[a : K] = 1$, hence $A$ is isomorphic to $K \times A/a$. Since $\Omega_K(A/a)$ is isomorphic to a quotient of $\Omega_K(A)$ (III, p. 573, Prop. 22), it is zero and the induction hypothesis shows that $A/a$ is diagonalizable. This then shows $A$ to be diagonalizable.

The stated condition is clearly sufficient.

Conversely, assume that $A$ is reduced; arguing by induction on the degree of $A$ we see that it suffices to prove that if $A$ is not a field, there exist two non-zero algebras $A_1$ and $A_2$ such that $A$ is isomorphic to $A_1 \times A_2$, or also that there exists in $A$ an idempotent different from 0 and 1.

Suppose from now on that $A$ is reduced and is not a field. Among the ideals of $A$ different from 0 and $A$ let $a$ be an ideal whose dimension as vector $K$-space is minimal. For any $x \neq 0$ in $a$ we have $x^2 \neq 0$, because $A$ is reduced, whence $a' \neq \{0\}$. We have $a^2 \subset a$ and by the minimality of $a$ we find that $a' = a$. By Lemma 4 there exists an idempotent $e$ such that $a = Ae$, and we have $e \neq 0, e \neq 1$ because $a$ is distinct from 0 and $A$.

A) Let us first prove the equivalence of $a$, $b$ and $c$.

Suppose that $A$ is etale and let $L$ be an extension of $K$. Let $\Omega$ be an algebraically closed extension field of $L$ (V, p. 23, Th. 2). Then $L \otimes_K A$ is isomorphic to a subring of $\Omega \otimes_K A$ and the latter is isomorphic to a ring $\Omega^n$ by Prop. 2 (V, p. 30). Therefore the ring $L \otimes_K A$ is reduced.

We have thus shown that a) implies b), and c) is a particular case of b). Assume now that c) holds. For the K-algebra $A$ to be etale it is necessary and sufficient that the P-algebra $A_{(P)}$ should be etale (V, p. 32, Cor. 2). The algebra $A$ is therefore etale by the following lemma:

By Prop. 5 there exist extensions $L_1, \ldots, L_n$ of $P$ such that $B$ is isomorphic to the algebra $L, x \ldots x L_n$. Since a finite product of etale algebras is etale (V, p. 31, Cor.) it is enough to examine the case where $B$ is an extension of $P$. By Th. 3 (V, p. 33) it is enough to show that $dx = 0$ in $\Omega_P(B)$ for all $x \in B$.

Let $x \in B$; since $B$ is of finite degree over $K$, $x$ is algebraic over $K$ (V, p. 18, Prop. 2). Let $f$ be the minimal polynomial of $x$ and let $f'$ be the derivative off. The polynomial $f$ is non-constant. Suppose that $f' = 0$; by V, p. 9, Cor. the field $P$ is of characteristic $p \neq 0$ and we have $f \in P[X^p]$; since $P$ is perfect, we have $P[X^p] = P[X]^p$, but the irreducible polynomial $f$ cannot lie in $P[X]^p$.

We thus have $f' \neq 0$ and since the degree of $f'$ is strictly less than that off, we have $f'(x) \neq 0$. Now from $f(x) = 0$ we deduce $f(x) \cdot dx = 0$ in $\Omega_P(B)$, whence $dx = 0$, as we had to show.

*B) Suppose that $A$ is etale; by the equivalence of a) and b) the algebra $A$ is reduced, so there exist extensions $L_1, \ldots, L_n$ of $K$ such that $A$ is isomorphic to $L_1 \times \cdots \times L_n$ (Prop. 5). Since $A$ is etale, each of the extensions $L_i$ is an etale algebra (V, p. 31, Cor.), and hence by definition is a separable algebraic extension of $K$.

The implication $d) \Rightarrow a)$ follows from V, p. 31, Cor. \*