1

I don't think this is true, but I'm not sure. I certainly know of finite fields with 2,4 and 8 elements, and of course $p^n$ elements where $p$ is prime, for all $n \in \mathbb{N}$.

chubbycantorset
  • 2,065
  • 1
  • 20
  • 33
  • 2
    Nope. Hint: in a finite field, consider the subfield generated by $1$. This is called the prime subfield. Any field is a vector space over its prime subfield... – Qiaochu Yuan Mar 01 '13 at 07:06
  • 1
    Please ask one question per post. If you believe that several questions are sufficiently related to warrant asking them in one post, please point out where you see the connection. – joriki Mar 01 '13 at 07:07
  • @joriki Okay, I'll make separate posts. – chubbycantorset Mar 01 '13 at 07:10

1 Answers1

1

Let $F$ be a field of order $n$ and $P$ its prime subfield. Then $P\cong \mathbb Z_p$ where $p = \lvert P \rvert$. Thus $p$ is prime. So now $F$ is a finite $P$ vector space, and thus $n = \lvert F \rvert = \lvert P \rvert ^k = p^k$, where $k = \operatorname{dim}_P (F)$.

Stefan
  • 3,451