Lemma 4

Let A and B be algebras over the field K that have finite generating sets, and let $K'$ be an extension of K. If the $K'$-algebras $A_{(K')}$ and $B_{(K')}$ are isomorphic, then there exists a subextension L of $K'$, finitely generated over K, such that the L-algebras $A_{(L)}$ and $B_{(L)}$ are isomorphic.

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.