# The proof that a finite field has a prime power order

I don't seem to grasp the proof. First we construct a vector space over a subfield with prime order $p$ where $p$ is the characteristic of the field . As the field is finite , the vector space will be finite-dimensional so there is a finite basis of $n$ elements that spans the space and every vector in that space can be uniquely expressed as a linear combination of the basis , so the field has order $p^n$. My question is : why can't we just take any subfield , for example the field itself , as the underlying field and conclude that our field $F$ has $|F|^n$ elements?

• If you view $F$ as an $F$-vector space, its dimension is $1$, and you get $\lvert F\rvert = \lvert F\rvert^1$, which isn't very informative. You must use a subfield where you know the number of elements. The prime field of $F$ is such a subfield. Jan 9 '14 at 21:17

The dimension of a vector space depends on the field you're talking about, so thinking its dimension over the prime subfield and its dimension over itself are the same is a mistake! (Or a typo: sorry if that's the case :) )

Choosing $F$ itself would not be very helpful, since $F$ is one dimensional over $F$, and thus $n=1$, so that your result is $|F|=|F|^1$.

You can use other subfields: if $K$ is a subfield of $F$ and $F$ is $n$ dimensional over $K$, then $|F|=|K|^n$. And then if $E$ is a subfield of $K$ over which $K$ is $m$ dimensional, $|K|=|E|^m$, and $|F|=|E|^{mn}$. It's all consistent.

The nice part about using the prime subfield is that we're guaranteed every finite field contains a copy of one of the prime order fields. It's just a nice, canonical choice.

• Can I conclude that every proper subfield of a finite field has the same prime order ? Jan 9 '14 at 21:33
• @user1978522 The same prime divides it yes, but of course it won't be the same order. A proper subset of a finite set cannot have the same order (as you probably know.) Jan 9 '14 at 21:34
• No. For example $\mathbb{F}_16$ has a unique proper subfield strictly larger than $\mathbb{F}_2$, namely $\mathbb{F}_4$. In general, $\mathbb{F}_{p^n}$ is always a subfield of $\mathbb{F}_{p^m}$ provided $n$ divides $m$. Jan 9 '14 at 21:34

If you know group theory, there's another way to look at it: every nonzero element of this field has additive order $p$. But by Cauchy's Theorem, if a prime divides the order of a group (in this case the additive group), then there is an element whose order is that prime. Thus, since every nonzero element has that same prime order, only 1 prime can divide the order of the field, so it must be of prime power order.