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The Lecture Notes (and lecture as well) to Math E-222 states the following (I've reworded a little bit):

Theorem: Let $F$ be a finite field. Then $|F|=p^n$ for some prime $p$.

Proof: Consider the canonical map $f:\mathbb{Z}\to F$. Since $\mathbb{Z}$ is infinite and $F$ finite, $\text{ker}f\not=(0)$. So $\text{ker}f=(p)$ and $\mathbb{Z}/p\mathbb{Z}\cong im(f)$ by the first isomorphism theorem. This gives F the structure of a vectorspace over the field $\mathbb{Z}/p\mathbb{Z}$ which has finite dimension, say $n$. Therefore $|F|=p^n$.

Why does F have the structure of a vectorspace? At first I thought to use Lagrange's theorem, but I think this would only show that $|F|=pn$.

I think what this is ultimately trying to show is that $F\cong F[x]/(p, x^n)$. In the case where $\text{im}(f) = F$, then it is clearly true (with restriction $x=0$), but I'm not sure what to say about those elements which are not in the image of f.

I thought that maybe we could consider the cosets of $\text{im}(f)$, but this has lead me nowhere.

Am I on the right track? Not even close? Missing something obvious?

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If anyone knows the actual name of the theorem, let me know and/or retitle it. He called it "A theorem of Galois", and so that's how I've titled it here. –  Xodarap Jul 3 '11 at 22:03
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$\text{im}(f)$ acts by left multiplication on $F$ and this gives the vector space structure. The axioms are straightforward to verify here. More generally, given any ring $R$ and subring $S$ of it, left multiplication turns $R$ into a left $S$-module. –  Qiaochu Yuan Jul 3 '11 at 22:05
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Every field is a vector space over its prime field. The field operations restricted to the prime field as one factor essentially give the axioms for a vector space. I suggest you look at the basic axioms for a vector space and as a specific question about any which seem difficult. –  Mark Bennet Jul 3 '11 at 22:07
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@Xodrap: please observe that it is not true that $F\cong F[x]/(p, x^n)$, as you claim. Indeed, the right hand side is not even a field: for example, the class $\xi$ of element $x\in F[x]$ in the quotient is a non-zero nilpotent element. –  Mariano Suárez-Alvarez Jul 3 '11 at 22:12
    
@Mariano: thanks, good to know that I was unable to prove it because it is false. –  Xodarap Jul 3 '11 at 22:17

1 Answer 1

up vote 4 down vote accepted

Note that $im(f)$ is contained in $F$ so that $F$ contains a subfield that is (isomorphic to)$\mathbb{Z}/ p\mathbb{Z}$. In general, if you have two fields $K \subset L$, $L$ can be viewed as a $K-$ vector space with the obvious scalar multiplication: if $k \in K$ and $l \in L$, then $k \cdot l = kl$ where $kl$ is just the product of $k$ and $l$. So in this case $F$ contains a subfield isomorphic to $\mathbb{Z}/ p\mathbb{Z}$, so we may view $F$ as a vector space over that subfield, i.e. a vector space over $\mathbb{Z}/ p\mathbb{Z}$.

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