Let $L,E$ be extensions (you may assume finite) of a field $k$, and let $\sigma: L\to E$ be a $k$-isomorphism, then there exists $\phi\in \text{Aut}(\overline k/k)$ such that $\phi|_L=\sigma$. I am stuck trying to prove this.
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May be a duplicate of this? I don't want to cast the first vote to close as a dupe, when it is not clear cut. Particularly as A) I answered the other one (easier to find those), B) dupehammer privilege means that mine would also be the last vote needed. – Jyrki Lahtonen Apr 17 '24 at 03:53
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Yes, I forgot to add that, thanks! – mrod1605 Apr 17 '24 at 03:53
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1I see how the other post extend the isomorphism to a morphism from $\overline k$ to itself, but I don’t see how you can make it an isomorphism. Maybe if I look carefully through the proof it implies what I want. – mrod1605 Apr 17 '24 at 03:56
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That. Is a good question. The resulting morphism is, of course, injective. For surjectivity you need to use the facts that A) the morphism must permute the roots of every irreducible polynomial with coefficients in $k$, and B) every element of $\overline{k}$ is a root of some such polynomial. I think that works, but I'm not sure I have had enough morning coffee. – Jyrki Lahtonen Apr 17 '24 at 04:07
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Good luck with your coffee! It’s 1 am here so I’m gonna get some sleep but I’ll try thinking what you said tomorrow! Thank you – mrod1605 Apr 17 '24 at 04:13