Let $K$ be an algebraically closed field ($\operatorname{char}K=p$). Denote $${\mathbb F}_{p^n}=\{x\in K\mid x^{p^n}-x=0\}.$$ It's easy to prove that ${\mathbb F}_{p^n}$ consists of exactly $p^n$ elements.

But if $|K|<p^n$, we have collision with previous statement (because ${\mathbb F}_{p^n}$ is subfield of $K$).

So, are there any finite algebraically closed fields? And if they exist, where have I made a mistake?


Rodrigo de Azevedo
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    Nope. And what you've written is a correct proof that no such fields exist. – Qiaochu Yuan Aug 08 '11 at 22:42
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    This question makes me wonder whether the field with one element (whatever that means...) is algebraically closed. – Mark Aug 09 '11 at 14:33
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    @Mark: apparently $\mathbb{F}_1$ has extensions called $\mathbb{F}_{1^n}$ for all $n$, whatever that means. – Qiaochu Yuan Dec 01 '11 at 19:22
  • @MarkSchwarzmann As Qiaochu says, $\mathbb{F}_1$ is not algebraically closed. And this is the popular opinion; I think the only claim I have seen that it is algebraically closed was from someone who was also claiming that $\mathbb{F}_1 = \mathbb{C}$. But then there is an interesting question: does the algebraic closure of $\mathbb{F}_1$ have a finite number of elements? And I think that the popular answer is no; $\mathbb{F}_{1^n}$ is supposed to have $n$ elements. – Slade Oct 30 '13 at 09:33
  • F_1 is not a field, not even a domain. In a field K\{0} is a multiplicative group, so it has a 1, so it cannot be empty. In a field all elements BUT 0 are invertible (0 cannot be). In a domain 0 is a prime ideal (and prime is proper "by definition", the complement must be a multiplicative subset, in particular it contains 1). – Marcus Barão Camarão Aug 22 '19 at 14:32
  • This is as hard as explaining why 1 is not a prime number: prime numbers have EXACTLY 2 positive divisors. If 1 were considered a prime, you wouldn't have uniqueness of factorization (fundamental theorem of arithmetics). Primes are the set of (INDEPENDENT!) multiplicative generators of positive integers, 1 being the empty product. – Marcus Barão Camarão Aug 22 '19 at 14:32

4 Answers4


No, there do not exist any finite algebraically closed fields. For suppose $K$ is a finite field; then the polynomial $$f(x)=1+\prod_{\alpha\in K}(x-\alpha)\in K[x]$$ cannot have any roots in $K$ (because $f(\alpha)=1$ for any $\alpha\in K$), so $K$ cannot be algebraically closed.

Note that for $K=\mathbb{F}_{p^n}$, the polynomial is $$f(x)=1+\prod_{\alpha\in K}(x-\alpha)=1+(x^{p^n}-x).$$

Marc van Leeuwen
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Zev Chonoles
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Hint $\rm\:F[x]\:$ has infinitely many primes for every field $\rm\,F\,$ -- by mimicking Euclid's proof for integers. In particular, if $\rm\,F\,$ is algebraically closed, there are infinitely many nonassociate primes $\rm\ x - a_i\:$ therefore there are infinitely many elements $\rm\:a_i\in F\:.\:$

Remark $\ $ This explains the genesis of the polynomial employed in Zev's answer.

Bill Dubuque
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As others have said, there cannot be any finite algebraically closed fields (and if there were, algebraic geometry would be a rather different subject than it is;-). In fact there cannot even be any finite field $K$ over which all quadratic polynomials have roots, by the following simple counting argument.

Let $q=|K|$, then there are $q$ monic degree $1$ polynomials $X-a$, and similarly $q^2$ monic degree $2$ polynomials $X^2+c_1X+c_0$ in $K[X]$. By commutativity there are only $\frac{q^2+q}2$ distinct products of two degree $1$ polynomials, which leaves $q^2-\frac{q^2+q}2=\binom{q}2$ irreducible monic quadratic polynomials. (Even without using unique factorization, one gets at least so many monic irreducible polynomials.)

There are in fact formulas in terms of $q$ for the number of (monic) irreducible polynomials over $K$ of any degree, obtained by the inclusion–exclusion principle.

These formulas show that, if the mythological field with $1$ element were to exist, it would be algebraically closed.

Added: It turns out that finding the number monic irreducible polynomials over $K$ of a given degree using only inclusion–exclusion (and nothing about finite fields) gets rather hairy. Rather, one can use the existence of finite field $K'$ of order $q^n$ to find the formula. All elements of $K'$ have a minimal polynomial of degree $d$ dividing $n$, since they are contained in a subfield of order $q^d$, and inversely all $d$ roots of such an irreducible polynomial are distinct and lie in $K'$. Then if $c_d(q)$ counts the irreducible polynomials of degree $d$ over $K$, one has $\sum_{d|n}dc_d(q)=q^n$. From this a Möbius inversion argument (which is a form of inclusion–exclusion) gives $$ c_n(q)=\frac{\sum_{d|n}\mu(n/d)q^d }n $$ where $\mu$ is the classical Möbius function.

Marc van Leeuwen
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As an alternative approach, suppose we have a field $K$ such that $\overline{K}$, the algebraic closure of $K$, is finite (and we'll also assume that $\vert K \vert >1$). It is clear that $K$ must then be finite, so $K=\mathbb{F}_p^n$ for some prime $p$ and some $n\in \mathbb{N}$.

However, for $i \vert j$, we have $\mathbb{F}_{p^i}$ isomorphic to a subfield of $\mathbb{F}_{p^j}$. Thus, $\overline{K}=\overline{\mathbb{F}_{p^n}}=\bigcup\limits_{n\vert m} \mathbb{F}_{p^m}$, which is infinite.

Therefore the algebraic closure of any (non-trivial) field is infinite.