Lemma 3

Let A be an absolutely semisimple K-algebra and L an algebraically closed extension of K. Then the algebra $A_{(L)}$ is isomorphic to a product of finitely many matrix algebras over L.

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).