Lemma 2

Let L be an algebraically closed field and D a field containing L in its center. If D is distinct from L, then there exists an extension $L'$ of L such that the ring $D\otimes_LL'$ is not left Artinian.

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.