a) Every direct sum of absolutely semisimple A-modules is an absolutely semisimple A-module. Every submodule and quotient module of an absolutely semisimple module is absolutely semisimple.

b) Let M be an A-module and let L be an extension of the field K. The A-module M is absolutely semisimple if and only if the $A_{(L)}$-module $M_{(L)}$ is absolutely semisimple.

Assertion a) follows from the analogous assertion for semisimple modules (VIII, p. 56, Corollaries 1 and 3).

Suppose that the A-module M is absolutely semisimple, and let $L'$ be an extension of L. As an $A_{(L')}$-module, $L'\otimes_LM_{(L)}$ is isomorphic to $M_{(L')}$ (II, §5, No. 1, p. 273, Proposition 2); it is therefore a semisimple module. This proves that $M_{(L)}$ is absolutely semisimple.

Conversely, suppose that $M_{(L)}$ is absolutely semisimple. Let $L'$ be an extension of K. There exists a composite extension $(\Omega , u, v)$ of L and $L'$ (V, §2, No. 4, p. 13, Corollary); we identify L and $L'$ with subextensions of Ω. The $A_{(\Omega )}$-module $M_{(\Omega )}$ is isomorphic to $(M_{(L)})_{(\Omega )}$; it is therefore semisimple. But $M_{(\Omega )}$ is also isomorphic to $(M_{(L')})_{(\Omega )}$, and Proposition 8, a) of VIII, p. 222 implies that $M_{(L')}$ is semisimple. So M is absolutely semisimple.

(i) The A-module M is absolutely semisimple.

(ii) There exists an extension P of K that is a perfect field such that the $A_{(P)}$-module $M_{(P)}$ is semisimple.

(iii) The A-module M is semisimple, and the center of its commutant is an étale algebra over the field K.

It is clear that (i) implies (ii). Assume that (ii) holds; then the $A_{(P)}$-module $M_{(P)}$ is absolutely semisimple by Corollary 1, a) of VIII, p. 222. By Proposition 1, b), M is then absolutely semisimple. Therefore, (ii) implies (i).

Suppose that M is semisimple. Let Z be the center of the commutant of M; it is a commutative algebra of finite degree over the field K. If L is an extension of K, then it follows from Proposition 8, b) of VIII, p. 222 that the $A_{(L)}$-module $M_{(L)}$ is semisimple if and only if the ring $Z_{(L)}$ is reduced. The equivalence of (i) and (iii) therefore follows from Theorem 4 of V, §6, No. 7, p. 34.

If $p= 1$, then D = Z. We therefore assume $p >1$.

Let us give a proof by contradiction and suppose that D is not commutative. Let $a$ be an element of D- Z and $q$ a power of $p$ such that $a^q$ belongs to Z. Denote the identity mapping on D by I and the inner automorphism $x\mapsto axa^{-1}$ from D to D associated with $a$ by $\sigma$; we have $\sigma^q= I$ because $a^q$ belongs to Z. We have $\sigma -I\not= 0$ because $a$ does not belong to Z, and we have $(\sigma -I)^q=\sigma^q-I = 0$ because Z has characteristic $p$. Let $f$ be the greatest natural number such that $(\sigma -I)^f\not= 0$; we have $f\geqslant 1$. Choose an element $c$ of D such that $(\sigma -I)^f(c)\not= 0$, and set

By construction, we have $y\not= 0$ and $\sigma (y) =y$; if we set $z=y^{-1}x$, then it follows that

and therefore $\sigma (z^{p^j}) = 1 +z^{p^j}$ for every natural number $j$. Choose an integer $m\geqslant 0$ such that $z^{p^m}$ belongs to the center Z of D; we have

This contradiction implies Lemma 1.

Denote the characteristic exponent of K by $p$. Let $a$ be an element of D. The ring $K[a]$ is an algebraic extension of K (V, §3, No. 1, p. 17, Corollary 1). Since the field K is separably closed, it follows from V, §7, No. 7, p. 44, Proposition 13 that the algebra $K[a]$ is a $p$-radical extension of K. Hence there exists an integer $m\geqslant 0$ such that $a^{p^m}$ belongs to K. By Lemma 1, the field D is consequently commutative.

By the structure theorem for semisimple algebras (VIII, p. 135, Theorem 1), A is isomorphic to an algebra $\prod^r_{i=1}\mathbf{M}_{n_i}(D_i)$, where $r$ is an integer $\geqslant 0,n_1, . . . , n_r$ are strictly positive integers, and $D_1, . . . ,D_r$ are K-algebras of finite degree that are fields. Since the field K is separably closed, each field $D_i$ is commutative by Proposition 3; the corollary follows.

(i) The K-algebra A is absolutely semisimple.

(ii) The algebra A has finite degree over K, and there exists an extension P of K that is a perfect field, for which the P-algebra $A_{(P)}$ is semisimple.

(iii) The K-algebra A is semisimple and has finite degree over K, and its center is an étale K-algebra.

(iv) There exists a finite family $(n_i,D_i)_{i\in I}$, where $n_i$ is a strictly positive integer and $D_i$ a K-algebra of finite degree that is a field, such that the center $Z_i$ of $D_i$ is a separable extension of K for every $i\in I$ and A is isomorphic to the product of the matrix rings $\mathbf{M}_{n_i}(D_i)$.

(v) There exist an extension L of K and a finite family of integers $(n_i)_{i\in I}$ such that the L-algebra $A_{(L)}$ is isomorphic to the algebra $\prod_{i\in I}\mathbf{M}_{n_i}(L)$.

(vi) There exist a Galois extension L of K of finite degree and a finite family of integers $(n_i)$ such that $A_{(L)}$ is isomorphic to the product of the matrix algebras $\mathbf{M}_{n_i}(L)$.

We first prove the implications (v) $\Rightarrow$ (iv) $\Rightarrow$ (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i). Denote the center of A by Z.

If property (v) holds, then the L-algebra $A_{(L)}$ is semisimple and has finite degree over L, and its center (isomorphic to $Z_{(L)}$, III, p. 41, Corollary) is isomorphic to $L^r$ for some integer $r\geqslant 0$. By Corollary 2, a) of VIII, p. 222, the algebra A is semisimple and has finite degree over K. It is therefore isomorphic to a finite product of rings $\prod_{i\in I}\mathbf{M}_{n_i}(D_i)$ with $n_i\geqslant 1$ for every $i\in I$, where the field $D_i$ is an algebra of finite degree over K. The center $Z_i$ of $D_i$ is a commutative field that is an extension of K, and Z is isomorphic to $\prod_{i\in I}Z_i$. Therefore, $Z_{(L)}$ is isomorphic, on the one hand, to $\prod_{i\in I}(Z_i)_{(L)}$ and, on the other hand, to $L^r$. In other words, the algebra $\prod_{i\in I}Z_i$ is étale over the field K (V, §6, No. 3, p. 28, Definition 1), and each of the extensions $Z_i$ is separable over K (V, §6, No. 4, p. 31, Corollary, and V, §7, No. 1, p. 42). So (v) implies (iv).

If property (iv) holds, then it is clear that A is a semisimple algebra and has finite degree over K. Its center Z is isomorphic to the product $\prod_{i\in I}Z_i$ of separable extensions of K of finite degree; it is therefore an étale algebra (loc. cit.). So (iv) implies (iii).

The implications (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i) follow from Proposition 2 of VIII, p. 230 applied to the A-module $A_s$.

Let $x$ be an element of D - L; since L is algebraically closed, the extension $L'= L(x)$ of L is not algebraic and $x$ is transcendent over L. The ring $B = L'\otimes_LL'$ is then an integral domain by Proposition 5 of V, §17, No. 4, p. 141.

The element $y=x\otimes 1-1\otimes x$ of B is not zero, but if $\varphi$ is the homomorphism from B to $L'$ that sends $\xi \otimes \eta$ to $\xi \eta$, then we have $\varphi (y) = 0$, so $y$ is not invertible in B. We view the ring $C = D\otimes_LL'$ as a right module over its subring B; it is a free module because D is a right vector space over its subfield $L'$. Since $y$ is a nonzero and noninvertible element of the integral domain B, right multiplication by $y$ in C is a mapping $R_y$ that is injective but not bijective. Now, $R_y$ is an endomorphism of the left C-module $C_s$; consequently (VIII, p. 28, Corollary 1), the ring C is not left Artinian.

Let us now prove that (i) implies (v). This is a consequence of the following lemma.

The L-algebra $A_{(L)}$ is semisimple; it is therefore isomorphic to a product of finitely many algebras of the form $\mathbf{M}_{n_i}(D_i)$, where $D_i$ is a field containing L in its center and $n_i$ an integer $\geqslant 1$ (VIII, p. 137, Remark 1).

Let $L'$ be an extension of L. Since the K-algebra A is absolutely semisimple, the ring $A_{(L')}$ is semisimple and therefore left Artinian. Now, the ring $A_{(L')}$ is isomorphic to $L'\otimes_LA_{(L)}$, hence to $\prod_{i\in I}\mathbf{M}_{n_i}(L'\otimes_LD_i)$; by Proposition 5 of VIII, p. 7, each of the rings $\mathbf{M}_{n_i}(L'\otimes_LD_i)$ is therefore left Artinian.

Let $n\geqslant 1$ be an integer and B a ring. Let $(\mathfrak{b}_r)_{r\geqslant 0}$ be a decreasing sequence of left ideals of B; denote by $\mathfrak{c}_r$ the set of square matrices of order $n$ with entries in $\mathfrak{b}_r$. Then $(\mathfrak{c}_r)_{r\geqslant 0}$ is a decreasing sequence of left ideals of $\mathbf{M}_n(B)$. In particular, if the ring $\mathbf{M}_n(B)$ is left Artinian, then so is B.

By the above, for every $i\in I$ and every extension $L'$ of L, the ring $D_i\otimes_LL'$ is left Artinian. By Lemma 2, we have $D_i= L$ for every $i\in I$, which implies Lemma 3.

We will use the following lemma to prove the implication (i)$\Rightarrow$(vi).

Let $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ be finite generating sets for the algebras A and B, respectively. Let $u$ be an isomorphism from $A_{(K')}$ to $B_{(K')}$ and $v$ the inverse isomorphism; there exists a subextension L of $K'$, finitely generated over K, such that we have $u(1\otimes e_i)\in B_{(L)}$ for every $i\in I$ and $v(1\otimes f_j)\in A_{(L)}$ for every $j\in J$. Consequently, $u$ sends $A_{(L)}$ into $B_{(L)}$, and $v$ sends $B_{(L)}$ into $A_{(L)}$. The induced mappings $u': A_{(L)}\rightarrow B_{(L)}$ and $v': B_{(L)}\rightarrow A_{(L)}$ are ring homomorphisms; they are inverse bijections.

Let us complete the proof of the implication (i)$\Rightarrow$(vi). Let $K'$ be a separable closure of K (V, §7, No. 8, p. 45). Then $A_{(K')}$ is an absolutely semisimple algebra over $K'$. By the implication (i)$\Rightarrow$(iv), the $K'$-algebra $A_{(K')}$ is isomorphic to a product $\mathbf{M}_{n_1}(D_1)\times \cdots \times \mathbf{M}_{n_r}(D_r)$, where $D_i$ is a $K'$-algebra of finite degree that is a field whose center $Z_i$ is a separable extension of $K'$. Since $K'$ is separably closed, we have $Z_i= K'$. By Proposition 3 of VIII, p. 231, the field $D_i$ is commutative. We therefore have $D_i= K'$. We denote the K-algebra $\mathbf{M}_{n_1}(K)\times \cdots \times \mathbf{M}_{n_r}(K)$ by B. The $K'$-algebras $A_{(K')}$ and $B_{(K')}$ are isomorphic by the above. Every subextension of $K'$ that is finitely generated over K is separable and has finite degree over K, hence is contained in a subextension L of $K'$ that is Galois and of finite degree over K (V, §10, No. 1, p. 57, Proposition 2). The implication therefore follows from Lemma 4.

The implication (vi)$\Rightarrow$(v) is immediate.

a) If $A_2$ is semisimple, then so is $A_1\otimes_KA_2$.

b) If $A_2$ is absolutely semisimple, then so is $A_1\otimes_KA_2$.

Denote the center of $A_1$ by $Z_1$ and that of $A_2$ by $Z_2$. The center Z of $A_1\otimes_KA_2$ is equal to $Z_1\otimes_KZ_2$ by the corollary of III, §4, No. 4, p. 468. Suppose that $A_2$ is semisimple; then $Z_2$ is a reduced algebra (VIII, p. 137, Propositions 2 and 3). By Theorem 1, $Z_1$ is an étale and therefore separable K-algebra. By Proposition 5 of V, §15, No. 2, p. 120, the ring $Z = Z_1\otimes_KZ_2$ is reduced; since $A_1$ has finite degree over K (Theorem 1), it follows from Proposition 7 of VIII, p. 221 that the ring $A_1\otimes_KA_2$ is semisimple.

Now suppose that $A_2$ is absolutely semisimple. Let L be an extension of K. Then the algebra $A_{1(L)}$ is absolutely semisimple, and the algebra $A_{2(L)}$ is semisimple. Therefore, by a), the algebra $A_1\otimes_KA_{2(L)}$ is semisimple.

By Theorem 1, A is isomorphic to an algebra of the form $\prod^r_{i=1}\mathbf{M}_{n_i}(D_i)$ for an integer $r\geqslant 0$, integers $n_1, . . . , n_r$, and K-algebras of finite degree $D_1, . . . ,D_r$ that are fields and whose centers are separable extensions of K and therefore equal to K. By Proposition 3 of VIII, p. 231, we have $D_i= K$ for $i\in [1, r]$.

a) Let M be a semisimple A-module. The A-module M is absolutely semisimple if and only if every simple module belonging to the support of M is.

b) Let S be a simple A-module, and let D be its commutant. The following properties are equivalent:

(i) The A-module S is absolutely semisimple.

(ii) The K-algebra D is absolutely semisimple.

(iii) The K-algebra D is a field, has finite degree over K, and its

center is a separable extension of K.

Assertion a) follows from Proposition 1, a) of VIII, p. 229. Let S and D be as in b), and let L be an extension of K. The $A_{(L)}$-module $S_{(L)}$ is semisimple if and only if the ring $D_{(L)}$ is (VIII, p. 222, Proposition 8, c)). This proves the equivalence of (i) and (ii), and that of (ii) and (iii) follows from Theorem 1 because D is a field.

The module $M_1$ is the direct sum of absolutely semisimple simple $A_1$-modules (Proposition 4). It therefore suffices to prove the assertion in the case when the modules $M_1$ and $M_2$ are simple. Denote their commutants by $D_1$ and $D_2$. The K-algebra $D_1$ is absolutely semisimple (loc. cit.); by Corollary 1 of VIII, p. 234, the K-algebra $D_1\otimes_KD_2$ is semisimple. It then follows from Corollary 1 of VIII, p. 215 that the $(A_1\otimes_KA_2$)-module $M_1\otimes_KM_2$ is semisimple.

(i) The B-module A is projective.

(ii) There exists an element $e$ of the $(A,A)$-bimodule B satisfying the following two conditions: $\varepsilon (e) = 1$ and $ae=ea$ for every $a\in A$.

The mapping $\varepsilon : B\rightarrow A$ is surjective because we have $\varepsilon (a\otimes 1) =a$ for every $a\in A$; it is a homomorphism of $(A$, A)-modules, hence a B-linear mapping. If the B-module A is projective, then there exists a section $s$ of $\varepsilon$ (II, §2, No. 2, p. 231, Proposition 4); it is a homomorphism of $(A$, A)-bimodules from A to B. If we set $e=s$(1), then we have $\varepsilon (e) =\varepsilon (s(1)) = 1$ and $ae=s(a) =ea$ for every $a\in A$. So (i) implies (ii).

Conversely, let $e$ be an element of B satisfying the conditions in item (ii). Define a mapping $s$ from A to B by the formula

It is a homomorphism of $(A$, A)-modules, and we have $\varepsilon \circ s= 1_A$; in other words, $s$ is a B-linear section of the surjective mapping $\varepsilon$. Consequently, the B-module A is isomorphic to the direct factor submodule $s(A)$ of $B_s$ (II, §1, No. 9, p. 211, Proposition 15) and is therefore projective (II, §2, No. 2, p. 231, Proposition 4). This proves that (ii) implies (i).

(i) The K-algebra A is absolutely semisimple.

(ii) The K-algebra $B = A\otimes_KA^o$ is semisimple.

(iii) The B-module A is projective.

(iv) There exists an element $e$ of the $(A,A)$-bimodule B satisfying $\varepsilon (e) = 1$ and $ae=ea$ for every $a\in A$.

Suppose that the algebra A is absolutely semisimple, hence semisimple. Then the algebra $A^o$ is semisimple (VIII, p. 137, Proposition 2), and it follows from Corollary 1 of VIII, p. 234 that $B = A\otimes_KA^o$ is a semisimple K-algebra. Therefore, (i) implies (ii).

Since every module over a semisimple ring is projective (VIII, p. 138, Proposition 4), (ii) implies (iii).

The equivalence of (iii) and (iv) follows from Proposition 5. To complete the proof, let us show that (iv) implies (i). Let $e=\sum^r_{i=1}a_i\otimes a'_i$ be an element of B satisfying the conditions in item (ii) of Proposition 5. Let L be an extension of the field K; we must prove that the ring $A_{(L)}$ is semisimple or, equivalently, that every $A_{(L)}$-module is semisimple (VIII, p. 138, Proposition 4). Let M be an $A_{(L)}$-module, and let N be a submodule of M; we view M as a left $(A$, L)-bimodule and N as a sub-bimodule (III, §4, No. 3, p. 466). Since L is a field, there exists an L-linear projector $u$ in M with image N. Since the homotheties $a_M$ associated with the elements $a$ of A are L-linear, there exists a unique group homomorphism from $A\otimes_KA^o$ to End$_L(M)$ that sends an element $a\otimes a'$ to the L-linear mapping $x\mapsto au(a'x)$. We denote the image of $e$ by this homomorphism by $v$; it follows from Remark 2 that $v$ is an $A_{(L)}$-linear projector with image N. The kernel of $v$ is an $A_{(L)}$-submodule of M, supplementary to N. By Corollary 2 of VIII, p. 56, the $A_{(L)}$-module M is semisimple.

We then have

for every integer $n\geqslant 1$ and every $f\in C^n(A,P)$.

Let $x\in B_n$; we have

so, by formula (6),

But we have

by property (ii) of Proposition 5 of VIII, p. 236 and, for $i\geqslant 2$,

which gives

and therefore formula (17).

We can finish the proof of Proposition 6 using Lemma 5. Let $n$ be an integer $\geqslant 1$ and $f$ an element of $C^n(A,P)$. By construction, we have

and, consequently, by formulas (9) and (18),

where the last equality follows from (17). Since $\alpha^n$ is bijective, the proposition follows.

We must prove that for every integer $n\geqslant 1$, every element $f$ of $C^n(A,P)$ with $\partial^n(f) = 0$ is of the form $\partial^{n-1}(g)$ for an element $g$ of $C^{n-1}(A,P)$. By Proposition 5, this is an immediate consequence of Proposition 6.

This is a translation of the equality $H^1(A,P) = 0$.