(1) cl(A $\otimes_KB$) $=$ cl(A) cl(B).

Let A, B, and C be K-algebras of finite degree, with respective classes $\alpha$, $\beta$, and $\gamma$. The K-algebras A and $\alpha$ are isomorphic, as are B and $\beta$. Therefore, the K-algebras $A\otimes_KB$ and $\alpha \otimes_K\beta$ are isomorphic, and we have cl(A$\otimes_KB$) $=$ cl($\alpha \otimes_K\beta$ ), which gives formula (1). It follows that $(\alpha \beta )\gamma$ is the class of the K-algebra $(A\otimes_KB)\otimes_KC$ and $\alpha (\beta \gamma )$ that of the K-algebra $A\otimes_K(B\otimes_KC)$. Now, these K-algebras are isomorphic (III, §4, No. 1, p. 461), so we have the equality $(\alpha \beta )\gamma =\alpha (\beta \gamma )$. Analogously, the relation $\alpha \varepsilon =\varepsilon \alpha =\alpha$ follows from the fact that the K-algebras $A\otimes_KK, K\otimes_KA$, and A are isomorphic, and the relation $\alpha \beta =\beta \alpha$ follows from the fact that the algebras $A\otimes_KB$ and $B\otimes_KA$ are isomorphic.

In the set $\mathscr{C}_K$, the relation “$\alpha$ and $\beta$ are Morita equivalent algebras” is an equivalence relation (VIII, p. 100). It is compatible with the law of composition on $\mathscr{C}_K$ by Proposition 13, d) of VIII, p. 111. We denote by $\mathscr{M}_K$ the quotient monoid of $\mathscr{C}_K$ by this equivalence relation and by $\varphi$ the canonical homomorphism from $\mathscr{C}_K$ to $\mathscr{M}_K$. For every K-algebra A of finite degree, we denote by [A] the image of cl(A) by $\varphi$. If A and B are K-algebras of finite degree, then we have [A] = [B] if and only if the K-algebras A and B are Morita equivalent; moreover, we have the relation

in the monoid $\mathscr{M}_K$.

By definition, we have [A] = [D] if and only if there exists an invertible $(A$, D)-bimodule, that is (VIII, p. 101, Theorem 1), a right vector space V of nonzero finite dimension over D endowed with an isomorphism of K-algebras $A\rightarrow$ End$_D(V)$. Lemma 1 follows.

Let A be a central simple algebra of finite degree over K. The algebra $A^o$ is central simple and of finite degree over K (VIII, p. 251). The algebra $A\otimes_KA^o$ is isomorphic to a matrix algebra $\mathbf{M}_n$(K), where $n^2=$ dim(A) (VIII, p. 252, Theorem 1). We therefore have $[A][A^o] = [A\otimes_KA^o] = [K]$ (Lemma 1), which proves that [A] is invertible, with inverse $[A^o]$.

Conversely, let A be a K-algebra of finite degree. If [A] is invertible in $\mathscr{M}_K$, then there exists a K-algebra B of finite degree such that [A][B] = [K]; by formula (2) and Lemma 1, this means that the K-algebra $A\otimes_KB$ is isomorphic to a matrix algebra $\mathbf{M}_n(K)$ with $n\geqslant 1$. By Remark 1 of VIII, p. 251, the algebra A is then central simple.

The existence of a $k$-linear bijection $\varphi$ with the property stated in the lemma follows from the compatibility of the tensor product with direct sums (II, §3, No. 7, p. 255, Proposition 7). The fact that $\varphi$ is an algebra homomorphism follows from the definition of a product matrix.

(i) We have [A] = [B] in the Brauer group Br(K).

(ii) There exists an integer $t\geqslant 1$ such that the K-algebra $A\otimes_KB^o$ is isomorphic to the matrix algebra $\mathbf{M}_t(K)$.

(iii) There exist strictly positive integers $r$ and $s$ such that the K-algebras $A\otimes_K\mathbf{M}_r(K)$ and $B\otimes_K\mathbf{M}_s(K)$ are isomorphic.

(iv) There exist a field D containing K and integers $m\geqslant 1$ and $n\geqslant 1$ such that A is isomorphic to $\mathbf{M}_m(D)$ and B to $\mathbf{M}_n(D)$.

(v) The K-algebras A and B are Morita equivalent.

Suppose that we have [A] = [B]. Since $[B^o]$ is the inverse of [B] in the Brauer group, we have $[K] = [B][B^o] = [A][B^o] = [A\otimes_KB^o]$. By Lemma 1, there exists an integer $t\geqslant 1$ such that the algebras $A\otimes_KB^o$ and $\mathbf{M}_t(K)$ are isomorphic. So (i) implies (ii).

Suppose that (ii) holds. Since $B^o\otimes_KB$ is isomorphic to a matrix algebra $\mathbf{M}_s(K)$ with $s\geqslant 1$ (VIII, p. 252, Theorem 1), the algebra $A\otimes_KB^o\otimes_KB$ is isomorphic to $A\otimes_K\mathbf{M}_s$(K), on the one hand, and to $\mathbf{M}_t(K)\otimes_KB$, on the other. So property (ii) implies property (iii).

Suppose that (iii) holds. By Wedderburn’s theorem (VIII, p. 120, Theorem 1), there exist integers $m\geqslant 1$ and $n\geqslant 1$ and fields D and $D'$ of finite degree over K with center K such that A is isomorphic to $\mathbf{M}_m(D)$ and B to $\mathbf{M}_n(D')$. Then the algebra $A\otimes_K\mathbf{M}_r(K)$ is isomorphic to $\mathbf{M}_{mr}$(D), and the algebra $B\otimes_K\mathbf{M}_{r'}(K)$ is isomorphic to $\mathbf{M}_{nr'}(D')$ (Lemma 2). By Corollary 2 of VIII, p. 121, the K-algebras D and $D'$ are isomorphic. So (iii) implies (iv).

Suppose that (iv) holds. The algebras A and D are Morita equivalent, as are the algebras B and D (VIII, p. 114, Example 2). Therefore, A and B are Morita equivalent, so (iv) implies (v).

Finally, the implication (v)$\Rightarrow$(i) is the definition of the set Br(K).

When the equivalent properties of the proposition hold, we say that the algebras A and B are similar.

This follows from the equivalence of properties (i) and (iv) of Proposition 3 and the fact that dim$_K(\mathbf{M}_n(D)) =$ dim$_K(D)\times n^2$.

This follows from Lemma 1 of VIII, p. 278 and Wedderburn’s theorem (VIII, p. 120, Theorem 1).

(i) The extension L is a splitting field for A.

(ii) There exists a central simple K-algebra of finite degree similar to A containing a maximal commutative subalgebra isomorphic to L.

Let us prove that (ii) implies (i); it suffices to consider the case when L is a maximal commutative subalgebra of A. Let $\psi : A\otimes_KA^o\rightarrow$ End$_K(A)$ be the canonical isomorphism that sends $a\otimes a'$ to the K-linear mapping $x\mapsto axa'$ (VIII, p. 252, Theorem 1). We view A as a right vector space over L; then $\psi$ sends $A\otimes_KL$ into the subspace End$_L(A)$ of End$_K(A)$ and induces, by restriction, an injective homomorphism of L-algebras $\psi ': A\otimes_KL\rightarrow$ End$_L(A)$. Set $n= [L : K]$. By VIII, p. 262, Proposition 3, we have $[A : L] =n$, and therefore $[A\otimes_KL : L] = [A : K] =n^2$ and [End$_L(A) : L] =n^2$. Consequently, $\psi '$ is an isomorphism, which proves (i).

The converse follows from Lemma 3 below.

We identify A with a subring of $A_{(L)}$. We view V as a K-vector space. The ring C is the commutant of $\varphi (A)$ in End$_K(V)$. It is a central simple K-algebra of finite degree, and the homomorphism $a\otimes c\mapsto ac$ from $A\otimes_KC$ to End$_K(V)$ is an isomorphism (VIII, p. 260, Theorem 6, a)). Consequently, the K-algebras A and $C^o$ are similar (VIII, p. 279, Proposition 3).

Let $L_V$ be the ring of homotheties of the L-vector space V; it is the commutant of End$_L(V)$ in End$_K(V)$ (VIII, p. 82, Corollary 1). Now, the K-algebra End$_L(V)$ is generated by $\varphi (A)$ and $L_V$; therefore, in End$_K$(V), we have

$L_V=$ End$_L(V)'=\varphi (A)'\cap L'_V= C\cap L'_V$,

where for any subset B of End$_K$(V), the commutant of B in End$_K(V)$ is denoted by $B'$. So $L_V$ is a maximal commutative subalgebra of C (VIII, p. 261, Lemma 3), and therefore also of $C^o$. The mapping $\lambda \mapsto \lambda_V$ is a K-algebra isomorphism from L to $L_V$. This proves the lemma.

Suppose that there exists a morphism $\varphi$ from L to A. Let M be a maximal semisimple commutative subalgebra containing $\varphi (L)$. By Proposition 3 of VIII, p. 262, we have $[A : K] = [M : K]^2$, hence [M : K] = [L : K] and $M =\varphi (L)$. By Proposition 5, the field L splits A. Conversely, suppose that the field L splits A; then it is isomorphic to a maximal commutative subalgebra of a central simple K-algebra B similar to A (Proposition 5). We have $[B : K] = [L : K]^2$ (VIII, p. 262, Proposition 3), hence [B : K] = [A : K]. Consequently, B is isomorphic to A (VIII, p. 280, Corollary). Corollary 1 follows.

Denote the reduced degree of D (VIII, p. 253) by $r$; by definition, we have $[D : K] =r^2$. By Proposition 5, there exists a central simple K-algebra B similar to D of which L is a maximal commutative subalgebra. Since B is isomorphic to a matrix algebra $\mathbf{M}_n(D)$ (VIII, p. 278, Lemma 1), we have $[B : K] =n^2r^2$ and consequently $[L : K] =nr$ (VIII, p. 262, Proposition 3).