Subfields of finite fields 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.
 A: If $p^t-1\mid p^s-1$, that is, $p^s-1=k(p^t-1)$, then by dividing both sides by $p-1$, we have $1+\cdots+p^{t-1}+p^t+\cdots+p^{s-1}=k(1+\cdots+p^{t-1})$. Let $V=1+\cdots+p^{t-1}$. Then, $V+p^t+\cdots+p^{s-1}=kV$. So, $V+p^t(1+\cdots+p^{s-1-t})=kV.$ In other words $V+p^t(V+p^t+\cdots+p^{s-1-t})=kV.$ 
Again if we can, $V+p^t(V+p^t(V+p^t+\cdots+p^{s-1-2t})=kV$, and so on. The left hand side is divisible by $V$ only if for some $j$, $s-1-jt=s-1$, that is $s=jt.$
A: 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.
A: Suppose $p^t-1|p^s-1$ but $t$ doesn't divide $s$. Then, write $s=ta+b$, where $0<b<t$. Now, $p^s-1=p^{ta+b}-1={(p^t)^a\cdot p^b}-1\equiv p^b-1(\mod p^t-1)$ which cannot be $0$. Contradiction. Hence, $t$ divides $s$.
A: (Subfield Criterion)
Let Fq be the finite field with q = p^n
elements. Then every subfield of Fq has order p^m, where m is
a positive divisor of n. Conversely, if m is a positive divisor of n, then there is exactly one subfield
of Fq with p^m
 elements.
