Do finite algebraically closed fields exist? 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?
Thanks.
 A: 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
$$f(x)=1+\prod_{\alpha\in K}(x-\alpha)=1+(x^{p^n}-x).$$
A: 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.
A: 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. 
A: 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.
