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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?2011-12-13
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    Good point, I'll edit.2011-12-13
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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.2011-12-13
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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 $r$.2011-12-13
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    I feel silly now. I even wrote down the tower law, but then didn't pursue it. Thanks!2011-12-13
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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 $r$.2011-12-13

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