Theorem 1

Let A be a K-algebra of finite degree. The following properties are equivalent:

(i) The algebra A is central simple.

(ii) The algebra A is central and without radical.

(iii) The canonical homomorphism from the K-algebra $A\otimes_KA^o$ to the K-algebra End$_K(A)$ that sends $a\otimes a'$ to the K-linear mapping $x\mapsto axa'$ from A to A is bijective.

(iv) There exist an extension L of the field K and an integer $n\geqslant 1$ such that the L-algebras $A_{(L)}$ and $\mathbf{M}_n(L)$ are isomorphic.

(v) For every separable closure $K'$ of K, there exists an integer $n\geqslant 1$ such that the $K'$-algebras $A_{(K')}$ and $\mathbf{M}_n(K')$ are isomorphic.

(vi) There exist a Galois extension L of the field K of finite degree and an integer $n\geqslant 1$ such that the L-algebras $A_{(L)}$ and $\mathbf{M}_n(L)$ are isomorphic.

(vii) There exist a K-algebra of finite degree D that is a field with center K and an integer $n\geqslant 1$ such that the algebra A is isomorphic to the algebra $\mathbf{M}_n(D)$.

A ring is simple if and only if it is semisimple and its center is a field (VIII, p. 143, Corollary of Proposition 10). Since A is an algebra of finite degree over the field K, it is a left Artinian ring; it is therefore semisimple if and only if it is without radical (VIII, p. 154, Proposition 4). The equivalence of (i) and (ii) follows.

Set $E = A\otimes_KA^o$ and F = End$_K(A)$; denote by $\varphi$ the canonical homomorphism from E to F defined by the relation $\varphi (a\otimes a')(x) =axa'$ for $x, a, a'$ in A. If the algebra A is central simple, then so are $A^o$ and, therefore, E (Remark 1), so $\varphi$ is injective. Now, we have $[E : K] = [A : K]^2= [F : K]$, so $\varphi$ is bijective. Conversely, suppose that $\varphi$ is bijective; since the algebra F is central simple (because it is isomorphic to a matrix algebra $\mathbf{M}_m$(K)), so is E, and therefore so is A (Remark 1). We have thus proved the equivalence of (i) and (iii).

By Remark 4, assertion (vii) implies assertion (i). The converse implication follows from Corollary 3 of VIII, p. 122 and Corollary 2 of VIII, p. 83.

It is clear that (vi) implies (iv), and (iv) implies (i) by Remark 2.

It remains to prove the implications (i)$\Rightarrow$(v)$\Rightarrow$(vi). Suppose that A is central simple, and let $K'$ be a separable closure of K (V, §7, No. 8, p. 45) . Then $A_{(K')}$ is a central simple algebra of finite degree over $K'$ (VIII, p. 251, Remark 2). By the corollary of VIII, p. 231, there consequently exist an integer $n\geqslant 1$ and an isomorphism of $K'$-algebras from $A_{(K')}$ to $\mathbf{M}_n(K')$; observe that the $K'$-algebras $\mathbf{M}_n(K')$ and $\mathbf{M}_n(K)_{(K')}$ are isomorphic. By Lemma 4 of VIII, p. 234, there exists a subextension L of $K'$, finitely generated over K, such that the L-algebras $A_{(L)}$ and $\mathbf{M}_n(K)_{(L)}$ are isomorphic. Then L is separable and of finite degree over K, so 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 $L'$-algebras $A_{(L')}$ and $\mathbf{M}_n(L')$ are then isomorphic.