# 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.

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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
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
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
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

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 '14 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 '14 at 10:13

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 $s=jt.$

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amazing idea and proof – Fardad Pouran Dec 28 '14 at 5:53
Really amazing: $s-1-jt=s-1\Rightarrow jt=0$ not $s=jt$. – user26857 Aug 3 at 10:34

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 $s$.

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