Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Nope. And what you've written is a correct proof that no such fields exist. –  Qiaochu Yuan Aug 8 '11 at 22:42
1  
This question makes me wonder whether the field with one element (whatever that means...) is algebraically closed. –  Mark Aug 9 '11 at 14:33
    
@Mark: apparently $\mathbb{F}_1$ has extensions called $\mathbb{F}_{1^n}$ for all $n$, whatever that means. –  Qiaochu Yuan Dec 1 '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. –  you-sir-33433 Oct 30 '13 at 9:33
add comment

4 Answers

up vote 20 down vote accepted

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 $$f(x)=1+\prod_{\alpha\in K}(x-\alpha)=1+(x^{p^n}-x).$$

share|improve this answer
    
There is also a nice similarity to the proof of the infinitude of primes, as Bill Dubuque points out in his answer below. –  Chris Leary Dec 1 '11 at 21:21
add comment

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.

share|improve this answer
1  
The "genesis" has a more obvious explanation, no? It comes from the absolutely immediate observation that over a finite field a polynomial may well represent the zero function! –  Mariano Suárez-Alvarez Aug 9 '11 at 18:00
1  
@Mar Sure, one could view it that way too. But Euclid's idea is no more complex and works more generally. –  Bill Dubuque Aug 9 '11 at 18:20
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.