# Non-trivial example of algebraically closed fields

I'm beginning an introductory course on Galois Theory and we've just started to talk about algebraic closed fields and extensions.

The typical example of algebraically closed fields is $\mathbb{C}$ and the typical non-examples are $\mathbb{R}, \mathbb{Q}$ and arbitrary finite fields.

I'm trying to find some explicit, non-typical example of algebraically closed fields, but it seems like a complicated task. Any ideas?

• What do you mean "explicit"? A very impotant, nice example of alg. closed field different from $\;\Bbb C\;$ is the algebraic closure of the rationals $\;\overline{\Bbb Q};$ . You can also take the alg. closures of the $\;p\,-$ adic fields and etc., or the alg. closures of the finite fields of positive characteristic $\;\overline{\Bbb F_p}\;$ ... Commented Jul 13, 2016 at 19:44
• math.stackexchange.com/questions/627662/… and mathoverflow.net/questions/25344/… have some good answers. Commented Jul 13, 2016 at 19:44
• Might I remark that $\mathbb C$ actually IS a non-trivial example of an algebraically closed field. Mathematicians needed years to prove it.
– MooS
Commented Jul 14, 2016 at 11:28

Another concrete example is given by Puiseux's theorem:

If $K$ is an algebraically closed field of characteristic $0$, the field $K\langle\langle X\rangle\rangle$ of Puiseux's series is an algebraic closure of the field of formal power series $K((X))$.

Note:

$K\langle\langle X\rangle\rangle=\displaystyle\bigcup_{n\ge1}K((X^{1/n}))$

You can start with $\Bbb Q$ and take its algebraic closure $\bar{\Bbb Q}\subsetneq\Bbb C$ and you get an algebraically closed subfield of $\Bbb C$ that's much much smaller than $\Bbb C$ (countable versus uncountable). Then you can add any transcendental to it like $\pi$ and you can take the algebraic closure of that $\overline{\bar{\Bbb Q}(\pi)}$. So you can produce infinitely many algebraically closed subsets of $\Bbb C$ in this way. What makes $\Bbb C$ special is not just that it's algebraically closed but that it's also complete.

Other examples are the p-adic fields which have complete and algebraically closed extensions which are very different from $\Bbb C$.

• What do you mean by complete? With respect to Euclidean metric? Commented Jul 13, 2016 at 22:55
• @MarcoFlores Yes, strictly speaking it means every Cauchy sequence converges. Informally it means there are no holes. Like even if you add to $\Bbb Q$ the roots of all polynomials with rational coefficients, the space still has holes, like $\pi$ and $e$. You can write sequences of rational numbers that bunch up around $\pi$ but have nowhere to converge to if $\pi$ is not there. Commented Jul 13, 2016 at 23:30

In On Numbers and Games, Conway defined a field structure on the set of all ordinals, and he calls the result $\mathbf{On}_2$. It is an algebraically closed field of characteristic two, if you are willing to ignore the fact that it's really too big to be a set.

It is also possible to "cut" $\mathbf{On}_2$, that is, to only consider ordinals smaller than a given limit and to get some algebraically closed fields. For example, the ordinals smaller than $\omega^{\omega^\omega}$ give the algebraic closure of $\mathbb F_2$, cf. this Lenstra's article.

Here's an introduction to this construction.

One interesting thing to note is that the first-order theory of algebraically closed fields of characteristic $$0$$ is $$\kappa$$-categorical for $$\kappa > \aleph_0$$. That means that there other algebraically closed field of characteristic $$0$$ of the same cardinality as $$\mathbb{C}$$! Maybe not the most helpful response, but I think quite interesting.