Lemma 1

Let E be a field and K a subfield of the center of E not equal to E. If the field K is algebraically closed, then we have $[E : K]\geqslant$ Card(K).

Let $x$ be an element of E - K and L the (commutative) subfield of E generated by $K\cup \{x\}$. Since K is algebraically closed, $x$ is transcendent over K. By VII, §2, No. 3, p. 10, Theorem 2 and p. 11, the elements $(x-\alpha )^{-1}$ of L, where $\alpha$ runs through K, are linearly independent over K. We therefore have $[E : K]\geqslant [L : K]\geqslant$ Card(K).