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I need to cite and read the proof of the following:

Theorem: For every characteristic $p\geq0$ and uncountable cardinal $k$, there is up to field isomorphism exactly one algebraically closed field of characteristic $p$ and cardinality $k$.

Question: Where can I find it and does the theorem have a name?

EDIT: This result was first published in 1910 by Ernst Steinitz on the paper Algebraische theorie der körper, Journal für die reine und angewandte Mathematik, 137:167–309.

I need a not-that-old reference in English (probably it will be a book since the result was already published in a paper).

Chilote
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2 Answers2

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This is not a difficult result. An algebraically closed field $F$ is determined, up to isomorphism, by the transcendence degree over the prime field ( that can be $\mathbb{Q}$ or $\mathbb{F}$). If the transcendence degree is uncountable, the cardinality of $F$ equals to the transcendence degree. (not so if the transcendence degree is finite or countable, then $F$ is countable, so the cardinality does not determine the transcendence degree in this case).

orangeskid
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    I am afraid that you are using the theorem I am looking for: An algebraically closed field F is determined, up to isomorphism, by the transcendence degree over the prime field. Do you know a couple of sources that prove this result and that are different than Steinitz's 1910 paper that is in German? – Chilote May 04 '18 at 12:10
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    @Chilote See e.g. Hungerford's book "Algebra", Chapter IV, Section 1. – dragoon Mar 04 '19 at 06:46
  • @dragoon you mean Chapter VI, not IV. – Brauer Suzuki Jun 30 '22 at 19:37
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I shall add the proof the fact claimed by @orangeskid above :

Let $E$ be an algebraically closed field and $F$ its prime field. Then the cardinality of $E$ equals the transcendence degree $\mathrm{trdeg}(E|F)$, if the cardinality of $E$ is uncountable.

Proof: Let $\mathcal{B}$ be a transcendental basis of $E|F$. Then $\mathrm{card}(E) = \mathrm{card}(F(\mathcal{B}))$, since $F(\mathcal{B})$ is infinite and $E|F(\mathcal{B})$ is algebraic. Here we use the following lemma:

Lemma: For any algebraic extension $L|K$, if $\mathrm{card}(K)=\infty$, then $\mathrm{card}(L)=\mathrm{card}(K)$. If $\mathrm{card}(K)<\infty$, then $\mathrm{card}(L) \leq \aleph_0$.

Moreover, we claim that $\mathrm{card}(F(\mathcal{B}))=\mathrm{card}(\mathcal{B})$. In fact, $$ \mathrm{card}(\mathcal{B}) \leq \mathrm{card}(F(\mathcal{B})) = \mathrm{card}(F[\mathcal{B}]) \leq \sum_{n \geq 0} \mathrm{card}(\{P \in F[\mathcal{B}]: \deg P = n\}) \leq \sum_{n \geq 0} \mathrm{card}(\mathcal{B}) = \mathrm{card}(\mathcal{B}), $$ so the claim holds. Therefore, we see that $\mathrm{card}(\mathcal{B}) = \mathrm{trdeg}(E|F) = \mathrm{card}(E)$, as desired.

The lemma is proved in the same manner as the long equation above.

Sorry if there are any possible mistakes and misunderstandings.

Hetong Xu
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  • But this doesn't say anything about the structure of the field, only its cardinality. The whole point is to show that it is uniquely determined up to isomorphism by this information. – Tobias Kildetoft Oct 30 '21 at 19:03
  • @TobiasKildetoft: Thank you for your comment! Yes, the main point here is to show that it is uniquely determinded up to isomorphism. This relies on the fact that orangeskid said before:An algebraically closed field $E$ is determined, up to isomorphism, by the transcendence degree over the prime field $F$. I'm not proving this fact, which is closely related to the field structure, since dragoon had provided a reference. I'm here just to add details on that when $E$ is uncountable, the transedental degree of $E$ over $F$ equals the cardinality of $E$....... – Hetong Xu Oct 31 '21 at 02:04
  • ...... Then if $E$ and $E^{\prime}$ share the same cardinality, they have the same transedental degree over their common prime field $F$, then invoking orangeskid's quoted claim, we are done. Hope that I haven't got that wrong. Thank you! – Hetong Xu Oct 31 '21 at 02:05