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Does every algebraically closed field contain the field of complex numbers? Thank you very much.

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No. The field of complex numbers has characteristic $0$. Every field $F$ has an algebraic closure, which must have the same characteristic as $F$. So any algebraic closure of a field of non-zero characteristic can't contain any isomorphic copy of the field of complex numbers. In particular, the algebraic closure of $\mathbb Z_2$ does not contain any isomorphic copy of the field of complex numbers.

Even the algebraic closure of a field of characteristic $0$ does not need to contain an isomorphic copy of $\mathbb C$. For instance, the algebraic closure of $\mathbb Q$.

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    $\begingroup$ Simple example, thank you! $\endgroup$
    – Seirios
    Commented Apr 18, 2013 at 9:22
  • $\begingroup$ @Ittay, thank you very much. I have a question about the algebraic closure of $\mathbb{Q}$. I think that $x^2+1\in \mathbb{Q}$. So the algebraic closure of $\mathbb{Q}$ should contain $\sqrt{-1}$ and hence the algebraic closure of $\mathbb{Q}$ contains $\mathbb{Q}(i)$. Is this true? What is the algebraic closure of $\mathbb{Q}$? $\endgroup$
    – LJR
    Commented Apr 18, 2013 at 10:24
  • $\begingroup$ yes, the algebraic closure of $\mathbb Q$ (which is called the field of algebraic numbers) contains $\mathbb Q(i)$. $\endgroup$ Commented Apr 18, 2013 at 10:31
  • $\begingroup$ @IttayWeiss, thank you very much. $\endgroup$
    – LJR
    Commented Apr 18, 2013 at 11:04
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The following is close to a positive answer to your question.

Let $\kappa$ be any uncountable cardinal. Then any two algebraically closed fields of characteristic $0$ and cardinality $\kappa$ are isomorphic.

In particular, any algebraically closed fields of characteristic $0$ and cardinality $c$ (the continuum) is isomorphic to the complex numbers.

And any algebraically closed field of characteristic $0$ that is big enough (cardinality $\ge c$) contains a copy of the complex numbers.

Note that we need $\kappa$ to be uncountable. There are non-isomorphic countable algebraically closed fields of characteristic $0$.

But any algebraically closed field of characteristic $0$ contains an isomorphic copy of the field of algebraic numbers.

Naturally, if the field $K$ has characteristic $p\ne 0$, there cannot be an embedding of $\mathbb{C}$ in $K$. And there are algebraically closed fields of characteristic $p$ and cardinality $\kappa$ for every prime $p$ and every infinite cardinal $\kappa$.

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An algebraically closed field $K$ contains a subfield isomorphic to the field of complex numbers if and only if $K$ has characteristic $0$ and the transcendence degree of $K/\mathbb{Q}$ is bigger or equal to the transcendence degree of $\mathbb{C}/\mathbb{Q}$. This is a consequence of Steinitz' classification of algebraically closed fields.

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  • $\begingroup$ thank you very much. What is the transcendence degree of $\mathbb{C}/\mathbb{Q}$? $\endgroup$
    – LJR
    Commented Apr 18, 2013 at 9:32
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    $\begingroup$ The cardinality of a transcendence basis of $\mathbb{C}/\mathbb{Q}$ equals the cardinality of $\mathbb{R}$. $\endgroup$
    – Hagen Knaf
    Commented Apr 18, 2013 at 10:27

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