There is exactly one algebraically closed field with prescribed characteristic and cardinality

Question

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).

Answer 1

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).

Answer 2

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.