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We know that if a finite field $F$ has characteristic $p$ (prime), then $F$ has cardinality $p^r$ where $r = [F:\mathbb{F}_p]$.

I'm now trying to say something about the possible cardinalities of subfields of $F$. I can see that there is a subfield of cardinality $p^s$ for each $s$ that divides $r$, given by the fixed field of the group generated by $\phi^s$, where $\phi$ is the Frobenius automorphism.

Now suppose $K$ is a subfield of $F$. Then (since both are additive groups), Lagrange gives us that $|K|$ divides $|F|$, so $|K| = p^t$ for some $1 \leq t \leq r $ (alternatively, $K$ contains $\mathbb{F}_p$ and so is a vector space over $\mathbb{F}_p$ and is thus isomorphic to $\mathbb{F}_p^t$, where $t = [K:\mathbb{F}_p]$). By considering the multiplicative group of units of $K$ and $F$ respectively, we get that $ p^t - 1$ divides $p^r -1$. I want to make the leap to $t|r$, but I'm failing to see why this needs to be true. Any help would be appreciated.

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Does it have to be finite just because the characteristic is finite? What about a transcendental extension? –  Arthur Dec 13 '11 at 13:02
    
Good point, I'll edit. –  Jonathan Dec 13 '11 at 13:04
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If you show that a finite field is Galois over its prime field, and identify its Galois group $G$, you can translate the problem to finding the subgroups of $G$. Since $G$ is quite simple, this is easier. –  Mariano Suárez-Alvarez Dec 13 '11 at 13:09
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Consider $K$ a subfield of $F$ so that we have $\mathbb{F}_p \subseteq K \subseteq F$. Then $F$ is not only a vector space over $\mathbb{F}_p$ but also a vector space over $K$. In fact, we have $[F:\mathbb{F}_p] = [F:K][K:\mathbb{F}_p]$. So if $|K|=p^t$, then $[K:\mathbb{F}_p]=t$ and thus $t$ divides $r$. –  Bill Cook Dec 13 '11 at 13:09
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A question that gets closed pretty quickly on math.SE is "Prove $\text{gcd}(x^a-1,x^b-1)=x^{\text{gcd}(a,b)}-1$". Since $p^t-1\mid p^r-1$, $\text{gcd}(p^t-1,p^r-1)=p^t-1=p^{\text{gcd}(t,r)}-1$ and so $\text{gcd}(t,r) = t$ showing that $t$ divides $r$. –  Dilip Sarwate Dec 13 '11 at 14:24
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1 Answer 1

You know more than that a finite field $F$ of characteristic $p$ has cardinality $q=p^r$ for some number $r\geq1$ ($r=\dim_{\Bbb F_p}F$). Namely, you know that for every number of the form $q=p^r$ there is a unique, up to isomorphism, field $\Bbb F_q$ with $q$ elements and moreover $\Bbb F_q$ can be realized as the set of the roots of the polynomial $X^q-X$ in some algebraic closure of $\Bbb F_p$.

If $\Bbb F_q\supset K\supseteq\Bbb F_p$ is a subextension with $K=\Bbb F_{q'}$, $q=p^r$ and $q=p^s$ a dimensional argument ($\Bbb F_q$ is also a $K$-vector space) shows that $s\mid r$.

But the condition is also sufficient, because the roots of the polynomial $X^{p^s}-X$ aro roots also of $X^{p^r}-X$.

Thus $\Bbb F_{p^r}$ contains a field with $p^s$ elements if and only if $s\mid r$, and such subfield is unique.

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@ Andera Mori : How can you say that if $s | r$, then there exist a field $K$ such that $F_p \subseteq K \subset F_q$. Please clear my doubts. Thank you –  user120386 Mar 29 at 2:25
    
As I said, the polynomial $X^{p^s}-X$ divides $X^{p^r}-X$, thus the roots of the former (elements of ${\Bbb F}_{p^s}$) are also roots of the latter (elements of ${\Bbb F}_{p^r}$) –  Andrea Mori Mar 29 at 10:13
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