What are some examples of algebraically closed fields? Wikipedia lists exactly two: $\mathbb{C}$ and the (complex) algebraic numbers.

EDIT: scrolling to the bottom of the Wikipedia article, they mention that every field has an essentially unique "algebraic closure", which is algebraically closed, and that proving this fact in full generality requires the Axiom of Choice.

But that leaves open some questions: are there any algebraically closed fields that are used often enough that they have names? When do you need AC to show $\mathbb{F}$ has an algebraic closures, and when don't you need AC?

I could not find a similar question here, and Google was not helpful.

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    $\begingroup$ Pick a field (for example finite fields) and take its algebraic closure. $\endgroup$
    – anon
    Jan 5, 2014 at 5:00
  • 2
    $\begingroup$ See [this example][1] on the trusty MathOverflow. [1]: mathoverflow.net/questions/25344/… $\endgroup$
    – Igor Rivin
    Jan 5, 2014 at 5:04
  • $\begingroup$ Can we consider $\cup_{k\geq{0}}F_{p^{2^k}}$? And $F_{q}$ denote the finite field with number $q$. $\endgroup$
    – gaoxinge
    Jan 5, 2014 at 5:07
  • $\begingroup$ @anon : thanks. I added to my question. $\endgroup$ Jan 5, 2014 at 5:09
  • $\begingroup$ @gaoxinge That would not be algebraically closed. (For example $x^{p^{\large 3}}-x$ does not split in it.) The algebraic closure of $\Bbb F_p$ is the whole union $\bigcup_{n\ge1}\Bbb F_{p^n}$. $\endgroup$
    – anon
    Jan 5, 2014 at 5:10

1 Answer 1


In general, an algebraic closure of a field $K$ is denoted by $\overline{K}$. Typical examples arising in number theory are $K=\mathbb{Q}$, $K=\mathbb{F}_p(t)$, $K=\mathbb{Q}_p$. Usually one needs the axiom of choice in order to prove the existence of algebraic closures. There are (at least) two exceptions: For $K=\mathbb{R}$ we have $\overline{K}=\mathbb{C}$. Then, for every subfield $K \subseteq \mathbb{R}$, we may realize $\overline{K}$ as the subfield of $\mathbb{C}$ which consists of complex numbers which are algebraic over $K$. It also exists without AC. We may also replace $\mathbb{R}$ by a real closed field, one only has to adjoin $\sqrt{-1}$. For $K=\mathbb{F}_q$, a finite field, we have for every $n$ an extension $\mathbb{F}_{q^n}$ of degree $n$ and every divisibility relation $n|m$ induces a canonical $\mathbb{F}_q$-homomorphism $\mathbb{F}_{q^n} \to \mathbb{F}_{q^m}$. It follows that we may consider the colimit $\mathbb{F}_{q^{\infty}} := \varinjlim_{n} \mathbb{F}_{q^n}$ (often this is written as a union, which is not quite correct). This turns out to be an algebraic closure of $\mathbb{F}_q$.

Let me also share a quite nice construction of an algebraic closure: Consider the (infinite) tensor product $A$ of all the $K$-algebras $K[x]/(f)$, where $f \in K[x] \setminus \{0\}$. By linear algebra it is non-zero, hence has a maximal ideal $\mathfrak{m}$ (Zorn's Lemma!). Then $K' := A/\mathfrak{m}$ is a field extension of $K$, and by construction every $f \in K[x] \setminus \{0\}$ has a root in $K'$. It is a nontrivial result that this already is the algebraic closure; but even if we don't use this, we can just repeat this process $K \hookrightarrow K' \hookrightarrow K'' \hookrightarrow K''' \hookrightarrow \dotsc$ and observe that the colimit $\overline{K}$ is an algebraic closure of $K$. A similar reasoning can be obtained to show that every two algebraic closures are isomorphic (but not in a canonical way).

  • 1
    $\begingroup$ Just an observation: The field $\mathbb{F}_{q^{\infty}}$ can indeed be realised as "honest" union $\bigcup_{n=1}^{\infty}\mathbb{F}_{q^{n!}}$, where $\mathbb{F}_{q^{(n+1)!}}$ is constructed recursively as the field extension of $\mathbb{F}_{q^{n!}}$ of degree $(n+1)$. $\endgroup$ Jan 5, 2014 at 11:41
  • $\begingroup$ It is not a union, since these are not subsets, but rather these fields are linked by homomorphisms (which should not be confused with inclusions). It doesn't matter if we take the poset $(\mathbb{N},|)$ or $(\mathbb{N},\leq)$. $\endgroup$ Jan 5, 2014 at 11:49
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    $\begingroup$ Since the field $\mathbb{F}_{q^k}$ is usually defined only up to isomorphism, I don't see a reason why one can't choose the representatives of elements of $\mathbb{F}_{q^{n!}}$ such that the morphisms are set-theoretic inclusions and then take the set-theoretic union. Or am I missing something here? $\endgroup$ Jan 5, 2014 at 12:08
  • $\begingroup$ You can do this, but this is not natural. $\endgroup$ Jan 5, 2014 at 12:12
  • 1
    $\begingroup$ 1. Yes. 2. The field of $p$-adic rational numbers. You may skip this since it is just an additional example. 3. A colimit is a kind of union where the involved sets/groups/etc. are linked by transition maps. Details can be found everywhere, for example at Wikipedia en.wikipedia.org/wiki/Direct_limit. $\endgroup$ Jan 6, 2014 at 2:25

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