# $n\mid d \iff a^d-1\mid a^n-1$ and prove field with $p^d$ elements is a subfield of a field with $p^n$ elements $\iff$ $d\mid n$

Prove $$n\mid d \iff a^d-1\mid a^n-1$$ and prove field with $$p^d$$ elements is a subfield of a field with $$p^n$$ elements $$\iff$$ $$d\mid n$$

So for the first part i'm betting there is some elementary algebra identities that would help a lot, and then the second part will follow from the first part because the splitting field of $$p^n-1$$ over $$\mathbb{Q}$$ is a field with $$p^n$$ elements but if anyone wants to talk about what's going on here in more depth that'd be cool. Thanks!

• I suppose your intial hypothesis is $d\mid n$, not $n\mid d$. – Bernard Nov 1 '18 at 23:21

Here $$a^n-1=(a^d-1)(a^{n-d}+ a^{n-2d}+ \cdots +a^d+1)=(a^d-1)k$$ so $$a^d-1$$ divides $$a^n-1$$ whenever $$d$$ divides $$n$$.

$$\Longleftarrow$$:

Let $$F$$ be a field with $$p^n$$ elements and assume $$d$$ divides $$n$$. Now consider $$F_1=\{x \in F: x^{p^d}=x\}$$ Check $$F_1$$ is a sub field of $$F$$! We have $$\vert F_1 \vert$$ is atmost $$p^d$$, since $$x^{p^d}=x$$ has atmost $$p^d$$ roots in $$F$$. Since multiplicative group of $$F$$ is cyclic, we have $$\langle F^*\rangle=\langle a\rangle$$. Using this to see $$a^k \in F_1$$ an so $$F_1$$ is a subfield of order $$p^d$$. Moreover $$F_1$$ is unique(?)

$$\Longrightarrow$$: Suppose $$F_1$$ is a sub field of order $$p^d$$ in a field $$F$$ of order $$p^n$$. Now $$n=[F: \Bbb{F}_p]=[F:F_1][F_1:\Bbb{F}_p]=[F:F_1] \cdot d$$ so $$d$$ divides $$n$$

• What is $k$ and why is $a^k \in F_1$? – Math is hard Nov 2 '18 at 18:57
• I already mention $k$ in the first equation, namely, $k=a^{n-d}+\cdots+1$. That is we know $a^d-1$ divides $a^n-1$, so $a^n-1=(a^d-1).k, k \in \Bbb{Z}$ . I mention this $k$ – Chinnapparaj R Nov 3 '18 at 2:39
• how does the fact that you wrote $a^d-1$ as a divisor of $a^n-1$ depend on the fact that $d$ divides $n$? – Math is hard Nov 3 '18 at 16:46

Hint:

You have the high-school identity: $$a^k-1=(a-1)(a^{k-1}+a^{k-2}+\dots+a+1).$$