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?

  • 9
    $\begingroup$ 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 $\;\overline{\Bbb F_p}\;$ ... $\endgroup$
    – DonAntonio
    Jul 13, 2016 at 19:44
  • 4
    $\begingroup$ math.stackexchange.com/questions/627662/… and mathoverflow.net/questions/25344/… have some good answers. $\endgroup$ Jul 13, 2016 at 19:44
  • 2
    $\begingroup$ Might I remark that $\mathbb C$ actually IS a non-trivial example of an algebraically closed field. Mathematicians needed years to prove it. $\endgroup$
    – MooS
    Jul 14, 2016 at 11:28

4 Answers 4


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 $K((X))$.


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


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 $\Bbb C$.

  • 1
    $\begingroup$ What do you mean by complete? With respect to Euclidean metric? $\endgroup$ Jul 13, 2016 at 22:55
  • 1
    $\begingroup$ @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. $\endgroup$ 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .