Lemma 9.26.11. Let $K/k$ be a finitely generated field extension. The algebraic closure of $k$ in $K$ is finite over $k$.

Proof. Let $x_1, \ldots , x_ r \in K$ be a transcendence basis for $K$ over $k$. Then $n = [K : k(x_1, \ldots , x_ r)] < \infty$. Suppose that $k \subset k' \subset K$ with $k'/k$ finite. In this case $[k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] = [k' : k] < \infty$, see Lemma 9.26.10. Hence

$[k' : k] = [k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] \leq [K : k(x_1, \ldots , x_ r)] = n.$

In other words, the degrees of finite subextensions are bounded and the lemma follows. $\square$

Comment #4976 by Laurent Moret-Bailly on

I believe the equality $[k'(x_1,\dots,x_r):k(x_1,\dots,x_r)]=[k':k]$ needs some justification.

Comment #6317 by Peng DU on

I think the last "<" can be "\leq".

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