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The following question struck me as pretty interesting: Let $\Bbb F$ be a field of characteristic $p$ (a prime, of course). I'm then asked to show that $|\mathbb{F}| = p^n$ for some $n\geq 1$.

Here's my intuition. Certainly we know that the prime subfield of $\mathbb{F}$ has order $p$. Now if there's an element (treating $\mathbb{F}$ now as a vector space over itself) independent from it, we have the $Span\{1,a_1\}$ as the usual set of linear combinations of $1$ and $a_1$. And any element of a field of characteristic $p$ added to itself $p$ times is $0$, so now we have $p^2$ possible linear combinations. And so on, arguing inductively. Is this argument kosher? Or does more need to be said to make it rigorous?

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The question is answered… and elsewhere on this site. – Gerry Myerson Sep 28 '12 at 6:36
@GerryMyerson Might you point me to the elsewhere? That answer uses language a bit over my head unfortunately...Apologies for the reposting... – AsinglePANCAKE Sep 28 '12 at 6:38…? which you can find by looking at the list headed Linked on the right side of the page in my earlier comment. Look around, check out some of the links, you won't break anything. – Gerry Myerson Sep 28 '12 at 7:34
@GerryMyerson Except perhaps the patience of veteran stack-exchangers! Many thanks! – AsinglePANCAKE Sep 28 '12 at 7:38
You mean, a finite field. – lhf Sep 28 '12 at 14:58

3 Answers 3

up vote 3 down vote accepted

In priciple, yes. $F$ is a vector space over $\mathbb F_p$ and hence in the finite case it is in bijection with some $\mathbb F_p^n.$ Of course $|\mathbb F_p^n|=|\mathbb F_p|^n=p^n$.

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I'll try to dumb this down to the very basics.

You already observed that $F$ contains $\mathbb F_p$ as a subfield. The essential point is that by only considering multiplication by elements of $\mathbb F_p$, your $F$ becomes a vector space over $\mathbb F_p$. You surely know that one can choose in any vector space a basis; since $F$ is finite, the basis is of course also finite. So if $f_1,f_2,\ldots,f_n$ is a basis of $F$, then by definition every $x\in F$ can be written uniquely as $x=x_1f_1+\cdots+x_nf_n$ with $x_1,\ldots,x_n\in\mathbb F_p$. Since each $x_i$ can independently take $p$ different values, there are $p^n$ values that can be obtained for $x$, all distinct and they fill up $F$.

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Ah yes, that's the answer I was poorly articulating above. Thank ya! – AsinglePANCAKE Sep 28 '12 at 7:06

I did not manage to follow your argument however there is a very simple argument here:

Since you already noted that the prime field of $F$ is $\mathbb{F}_{p}$ (up to isomorphism) all you have to recall is that $F$ is a vector space over its prime field hence $$|F|=|\mathbb{F}_{p}|^{dim_{\mathbb{F_p}}(F)}=p^{dim_{\mathbb{F_p}}(F)}$$

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If you don't mind, might you elaborate a bit on the string of equalities? They don't resonate with my n00b-self as obvious... – AsinglePANCAKE Sep 28 '12 at 6:41
Do you know that up to isomorphism there is only one vector space of dimesnion $n$ over a given field ? – Belgi Sep 28 '12 at 6:43
Umm, this makes sense, but I don't think I've seen it before. – AsinglePANCAKE Sep 28 '12 at 6:44

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