A field with finitely many elements

Question

Let $\Bbb F$ be a field with $k$ elements, where $k$ is a finite number. Prove that there exists a prime $p$ and a positive integer $n$ such that $$k=p^n$$

Answer 1

This is equivalent to showing that $k$ is only divisible by one prime only. Suppose that $k$ is divisible by 2 distinct primes $p,q$ (with $p<q$) . Considering $F$ as a group under addition, we use Cauchy's theorem to deduce that there exists elements $x,y\in F$ such that $|x|=p,|y|=q$. Clearly, $x,y\not=0$.

Since $x+x+...+x=0\,$ (addition is done $p$-times), therefore by multiplying by $yx^{-1}$ we get $y+y+...+y=0$ (addition is done $p$-times). Therefore $|y|=q$ divides $p$, this contradicts the fact that $p<q$.

Answer 2

Hint: first show that the characteristic* of any field is prime. You'll want to show $F$ is a vector space over any subfield, and then use a dimension argument to count elements.

*If you don't know what the characteristic of a field is, it is the least integer $r$ such that

$$\underbrace{1_{f} + 1_{f} + \cdots 1_{f}}_{r \text{ times}} = 0$$

Answer 3

Well, there is an easy way to see this:

Consider the ring homomophism $\varphi: \mathbb{Z} \to F$ given by $\varphi(n)=n\cdot 1_F$ where $1_F$ is the multiplicative identity in $F$. What is the kernel of $\varphi$?

Well, From the first isomorphism theorem we know that $\mathbb{Z}/\ker(\varphi) \cong \operatorname{im}(\varphi)$. Since $\operatorname{im}(\varphi)$ is a subset of the field $F$, it must be an integral domain. Hence, $\ker(\varphi)$ must be a prime ideal of $\mathbb{Z}$ and because $|\operatorname{im}(\varphi)| < \infty$ since $F$ is finite we conclude that $\ker(\varphi)$ is a non-zero prime ideal of $\mathbb{Z}$. So, it's of the form $p\mathbb{Z}$ for a prime number p. But then, it would be a maximal ideal of $\mathbb{Z}$ and since $\mathbb{Z}/\ker(\varphi) \cong \operatorname{im}(\varphi)$ we conclude that $E = \operatorname{im}(\varphi)$ must be a subfield of $F$.

In other words, $F$ is a finite extension of $E=\operatorname{im}(\varphi)$ so we must have $\dim_E(F)=n$ for some n. So, everything in $F$ can be written as a linear combination of elements in $E$ and it's clear by a simple counting argument that we'll have $p^n$ such linear combinations. Therefore $|F| = p^n$