# A field with finitely many elements [duplicate]

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 $$k=p^n$$

## marked as duplicate by Zev Chonoles, Amzoti, Micah, Dan Rust, Cameron BuieAug 10 '13 at 3:03

• See Theorem 1.5 here. This is essentially the argument given by A Wertheim below. It might help to read the theorem in a context, and learn more cool stuff :) – Prism Aug 10 '13 at 0:50
• Thanks for posting that, @Prism. I love Dr. Conrad's notes (and fortunately, I do believe he's an active member of MSE)! – Alex Wertheim Aug 10 '13 at 0:56
• – Prism Aug 10 '13 at 0:58
• @AWertheim: Me too! And yes, always looking forward to reading his posts here :) – Prism Aug 10 '13 at 0:59

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 $p<q$.

• Wow. The way I've usually seen this proved is the one I posted, but I really like this answer! +1 :) – Alex Wertheim Aug 9 '13 at 23:54
• @AWertheim Yes I also just thought about this argument now. – Amr Aug 9 '13 at 23:55
• New to me, too, and quite elegant. One remark: we don't need $p<q$ to know that $q$ cannot divide it. – Jonathan Y. Aug 10 '13 at 0:08
• That's a beautiful answer. – user66733 Aug 10 '13 at 0:37
• @user142526 Sure. I thought that the answer would be clearer. I don't remember why I thought this way. – Amr Aug 10 '13 at 5:44

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

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

• Could you expand on what you mean by using a dimension argument? – Makkers Aug 9 '13 at 23:56
• @Makkers certainly. Let $F'$ be a subfield of $F$. Suppose $F$ has dimension $n$ as a vector space over $F'$. What do elements of $F$ look like with respect to a basis in $F'$? Write every element of $F$ as a unique linear combination of basis elements in $F'$. Then we just count... – Alex Wertheim Aug 9 '13 at 23:58

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 $|F| = p^n$