Does every algebraically closed field contain the field of complex numbers?

Question

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

Answer 1

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 $\mathbb Q$.

Answer 2

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 $\kappa$.

Answer 3

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.