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

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

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 $\mathbb Q$.

• Simple example, thank you! – Seirios Apr 18 '13 at 9:22
• @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 $\mathbb{Q}$? – LJR Apr 18 '13 at 10:24
• yes, the algebraic closure of $\mathbb Q$ (which is called the field of algebraic numbers) contains $\mathbb Q(i)$. – Ittay Weiss Apr 18 '13 at 10:31
• @IttayWeiss, thank you very much. – LJR Apr 18 '13 at 11:04

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 $\kappa$.

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.

• thank you very much. What is the transcendence degree of $\mathbb{C}/\mathbb{Q}$? – LJR Apr 18 '13 at 9:32
• The cardinality of a transcendence basis of $\mathbb{C}/\mathbb{Q}$ equals the cardinality of $\mathbb{R}$. – Hagen Knaf Apr 18 '13 at 10:27