Separable extension iff $\Omega_{E/F}=0$?

Question

I know if $E/F$ is a finitely generated algebraic extension of fields, then $E/F$ is separable iff the Kahler differential $\Omega_{E/F}=0$.

Question: I wonder if we only assume that $E/F$ is algebraic without finitely generated, and $\Omega_{E/F}=0$, then is $E/F$ separable?

Edit: This question doesn't solve my question, since the answer there uses the assumption that $E/F$ is finitely generated.

Does this answer your question? Zero module of differentials implies finite extension? - Lemma 1. — Dietrich Burde, Jan 14 '21 at 11:43
@DietrichBurde. No. The answer in the question you refer uses the assumption that $E/F$ is finitely generated. — , Jan 14 '21 at 11:44

score 2 · Accepted Answer · answered Jan 14 '21 at 11:54

Kahler differentials commute with colimits, and any algebraic extension is the colimit of its finite subextensions (more generally, any field extension is the colimit of its finitely generated subextensions). Applying the result in the finite case, we have your desired result as the colimit of a diagram of zero modules is zero.

Separable extension iff $\Omega_{E/F}=0$?

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