# Is there a field with $n$ elements for all $n \in \mathbb{N}$? [duplicate]

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 $n \in \mathbb{N}$.

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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 1 '13 at 7:06
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 1 '13 at 7:07
@joriki Okay, I'll make separate posts. –  chubbycantorset Mar 1 '13 at 7:10
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 $k = \operatorname{dim}_P (F)$.