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Hi I would like to prove that for algebraic closed field $F_1,F_2$ if:

  1. They have the same characteristic $p$
  2. They have the same cardinal and is uncountable

Then they are isomorphic.

I have proved that with the same characteristic $F_1,F_2$ has isomorphic sub-field, donate $F'$, and $F_1,F_2$ are vector space over $F'$

I think the next step is to expend the isomorphic mapping from the sub-field to the whole field but I stucked in this and cannot find a way.

Alex Kruckman
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Shore
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    It's possible for $F_1$ and $F_2$ to be isomorphic but for no isomorphism to preserve $F$. Take the algebraic closure of $F=F_1=\mathbb{Q}$ adjoin countably many elements and $F_2$ be the algebraic closure of $F_1(X)$. They both have the same characteristic and transcendence degree, so are isomorphic, but $F=F_1 \subset F_2$ so the isomorphism cannot restrict to the identity on $F$. – Zoe Allen May 28 '24 at 02:08
  • @zoeallen Thanks a lot! In that case how should I prove that if $F_1,F_2$ has the same characteristic and cardinal, they are isomorphic???? – Shore May 28 '24 at 02:15
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    You need to pick a transcendence basis and build the isomorphism from a bijection of the transcendence bases. – Zoe Allen May 28 '24 at 02:17
  • @zoeallen could you tell where I can find the detailed prove?? Thanks a lot – Shore May 28 '24 at 02:19
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    https://math.stackexchange.com/q/3623568/1107685 covers the finite degree case, and the infinite degree case is much the same. I don't have a link for the infinite case. – Zoe Allen May 28 '24 at 02:21
  • @zoeallen thanks a lot! I've read this post, but this is the other way around which trying to prove that if two algebraic closed field is isomorphic, then the isomorphic mapping preserves the transcendence base – Shore May 28 '24 at 02:29
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    Here is another duplicate: https://math.stackexchange.com/q/2562735/7062 – Alex Kruckman May 28 '24 at 02:46

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