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If $F$ is a finite field, show that $|F|=p^r$, for $p$ prime and $r>0$ integer.

I know that $p$ is zero in this field since $p$ is characteristic. So the kernel of this will be any multiple of of $p$. But am confused how to go from here. Don't fields have the requirement that they need to be prime. So wouldn't $p$ divide $p^r$ so it could not be a field?? Any help would be appreciated. Thank you

ervx
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Jacob Rodgers
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  • The ideal $\langle x^2 + x + 1 \rangle \leq \Bbb F_2[x]$ is maximal, so the quotient $\Bbb F_2[x] / \langle x^2 + x + 1 \rangle$ is a field of order $4$. – Travis Willse May 11 '16 at 15:52

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If $F$ is a finite field, then there is a prime $p$ such that $F$ contains $\mathbb{F}_p$. Since $F$ is finite, the dimension of $F$ as an $\mathbb{F}_p$-vector space must be finite, so there are elements $x_1,\dots,x_r\in F$ which form a basis for $F$ as an $\mathbb{F}_p$-vector space.

Then every element of $F$ can be written uniquely in the form $n_1x_1+\dots+n_rx_r$ with $n_i\in \{0,1,\dots,p-1\}$, hence $|F|=p^r$.

carmichael561
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