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Recently I have been looking into the module of Kaehler differentials and found that in the case of finite field extensions, $L/K$ is separable if and only if $\Omega_{L/K}^1=0$. There is also a way to extend this to genereal algebraic field extensions as seen from this post by considering that algebraic field extensions are colimits of finite extensions.

It would be great if someone could point me towards a book stating / proving this fact as I have seen multiple other answers such as this and this (filtered colimit in this case) using this fact. I would like to learn more about it (also to learn more about interactions between field theory and category theory in general).

SuperPannda
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  • Two things, first you can check out the algebraic closure of $\mathbb{F}_p$ which is a very clear colimit, but if you want to see that every algebraic extension is a colimit, check out the proof of Zorn lemma (acknowledging the choice axiom) it basically uses filtered colimits in posets, now if you link it to the proof of existence of algebraic closure, it is very clear that these colimits are indeed filtered colimits of field extensions which are decomposition fields of some irreducible polynomials. Your poset is here given by algebraic extensions of your base field. – julio_es_sui_glace Apr 06 '24 at 12:14
  • Maybe this was not what you were asking about? – julio_es_sui_glace Apr 06 '24 at 12:18
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    @julio_es_sui_glace The fact that algebraic extensions are colimits of finite extensions is easier than all that! The key point is that if $K\subseteq L$ is an algebraic extension and $a_1,\dots,a_n\in L$, then $K(a_1,\dots,a_n)/K$ is a finite extension. It follows that the subfields of $L$ which are finite over $K$ form a directed system under inclusions whose union is $L$, so $L$ is the directed colimit of this system. – Alex Kruckman Apr 06 '24 at 12:32
  • @AlexKruckman yes it is indeed a simpler way to see it. – julio_es_sui_glace Apr 06 '24 at 12:51
  • Thank you both for your answers, both of you made it clear to me that ultimately it is because all the sub-extensions under the algebraic extension that are finite form a directed system so that the directed colimit gives the algebraic extension – SuperPannda Apr 06 '24 at 12:52

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