"Two algebraically closed fields with char $0$ and same cardinality $\kappa> \aleph_0$ are isomorphic." Any reference for this result?

by Roland   Last Updated December 12, 2017 05:20 AM

I already know how to prove the result, however I need a reliable source I can cite so I can omit the proof. I'm having a terrible hard time trying to find it throughout Google. Can anyone please help me?

Answers 2

I do not know a reference, but probably, an algebra textbook like Hungerford contains sufficient information for you to work out this problem, or maybe cite. Here is the sketch of my approach for future reference in this site, if it is useful.

Let $K$ and $L$ be two algebraically closed field of the same characteristic. Write $E$ and $F$ for the algebraic closures of the prime fields of $K$ and $L$, respectively. Of course, $E$ and $F$ are isomorphic. Since $K$ and $L$ have the same cardinality $\kappa>\aleph_0=|E|=|F|$, the transcendence degrees of $K$ over $E$ and $L$ over $F$ are the same. Thus, any transcendence bases $B$ of $K$ over $E$ and $C$ of $L$ over $F$ have the same cardinality $\kappa$.

Now, we pick an arbitrary field isomorphism $\sigma:E\to F$. Extend $\sigma$ to $E(B)$ and $F(C)$, and the resulting extension $\tilde{\sigma}$ is still a field isomorphism. Finally, since $K$ is the algebraic closure of $E(B)$ and $L$ is the algebraic closure of $F(C)$, the isomorphism $\tilde{\sigma}$ extends to a field isomorphism $\bar{\sigma}:K\to L$.

December 12, 2017 05:00 AM

If you want just some published statement to cite, this statement is the case $p=0$ of Proposition 2.2.5 in David Marker's Model Theory: An Introduction.

(Marker doesn't actually give a detailed proof, though; he instead cites section X.1 of the 1971 edition of Lang's Algebra for the fact that two algebraically closed fields are isomorphic iff they have the same characteristic and transcendence degree and then very briefly argues that transcendence degree and cardinality are the same in the uncountable case. Somewhat infuriatingly, this result from Lang's Algebra was not included in the 2005 edition, or at least I am unable to find it.)

Eric Wofsey
Eric Wofsey
December 12, 2017 05:11 AM

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