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I need some help proving the following:

If $F$ is a finite field then $|F|=p^n$ for prime $p$ and some integer $n$.

By contradiction, suppose $|F| =pq$ for primes $p$ and $q$. Then by Cauchy's theorem there exists $a, b\in F$ such that the order of $a$ is $p$ and the order of $b$ is $q$. I am not quite sure where to go from here, but I think we need to contradict that $F$ is a finite field. The only thing that comes to mind is to show zero-divisors.

  1. Where can I go from here?
  2. I thought every finite field had to have prime order? $p^n$ is not prime!

Thanks for all the help!

CodeKingPlusPlus
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    No, not every finite field has prime order. They all have prime power order. – anon Dec 16 '13 at 09:00
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    Yes it does have to do with zero divisors. Let $p>0$ be the additive order of 1. If $p=rs$ is not prime, say for example $p=6$, then you get $(1+1)(1+1+1)=0$, contradiction. So $p$ is prime, and then all elements of the field must have additive order $p$. – Derek Holt Dec 16 '13 at 09:03
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    I really like Amr's proof here: http://math.stackexchange.com/questions/463954/a-field-with-finitely-many-elements – Prism Dec 16 '13 at 09:06

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Think of $F$ as a vector space over its smallest subfield containing $1$ (which must be ...?).

anon
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