Possible Cardinality of a Field

Question

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?

Answer 1

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 $|\mathbb F_p^n|=|\mathbb F_p|^n=p^n$.

Answer 2

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 $$|F|=|\mathbb{F}_{p}|^{dim_{\mathbb{F_p}}(F)}=p^{dim_{\mathbb{F_p}}(F)}$$

Answer 3

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 $F$.