# Are algebraically closed fields of characteristic 0 always uncountable?

I am working on a model theory problem that has very little to do with field theory.

I basically want to know if and why every algebraically closed field of characteristic 0 is uncountable. I cannot think of a reason why this is true, but given the problem I am working on, I suspect it must be. I know many abstract algebra theorems assume uncountability. I also know that the cardinality of the algebraic closure of a field F is $$max\{\aleph_0, |F|\}$$. Is it true that algebraically closed fields of characteristic 0 are uncountable?

• No, the field of algebraic numbers is countable. – Angina Seng Nov 10 '19 at 9:03

## 1 Answer

$$\mathbb Q$$ has characteristic $$0$$ and is countable by a famous spiral argument. As you correctly state, the cardinality of the algebraic closure of a field $$F$$ is $$\max\{\aleph_0, |F|\}$$, so the cardinality of the algebraic closure of $$\mathbb Q$$ is $$\aleph_0$$.

• Wouldn't the correct statement be that the cardinality of the algebraic closure $\bar{k}$ of a field $k$ is at most $\mathrm{max}(\aleph_0,|k|)$?  (Though of course this is not a problem here since we just get $|\mathbb{Q}|=\aleph_0\leq|\bar{\mathbb{Q}}|\leq\mathrm{max}(\aleph_0,|\mathbb{Q}|)=\aleph_0$ from the inclusion $\mathbb{Q}\hookrightarrow\bar{\mathbb{Q}}$) – Théo Dec 11 '19 at 21:43
• Well, this is of course also correct, but since there is no finite algebraically closed field (there are irreducible polynomials of any degree over any $\mathbb F_p$) we have that equality always holds. Or was it something else you were doubting? – rawbacon Dec 11 '19 at 21:59
• I got confused and mixed that up with another result (Tag 09GK of the Stacks Project, which I now see is about the cardinality of the extension field of an algebraic extension, rather than about algebraically closed fields), but I understand it now. Sorry for bothering you with this point and many thanks for your explanation! – Théo Dec 11 '19 at 22:41