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Let $p$ be a prime integer, and let $q=p^r$ and $q'=p^k$. For which values of $r$ and $k$ does $x^{q'}-x$ divide $x^q-x$ in $\mathbb{Z}[x]$?

From Artin's Algebra, Chapter 15, problem 7.12 from the section on finite fields.

I believe it's those $k$ and $r$ such that $k|r$, but I'm having a hard time rationalizing it. I believe I need to show $\mathbb{F}_{q'}\subset\mathbb{F}_q$, but not sure how to show this either.

EthanAlvaree
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  • The factor $x$ poses no problem here, so you are left with a special case of: when does $x^m-1$ divide $x^n-1$ in $\Bbb{Z}[x]$? If $n=am+r$ with $0\le r<m$, then show that $(x^n-1)-(x^r-1)$ is always divisible by $x^m-1$. This will lead you to a useful reduction a la Euclid. Then apply the same to $x=p$. – Jyrki Lahtonen Mar 31 '14 at 05:15

3 Answers3

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Expanding on Mathemanic's answer for the forward direction,

Suppose $x^{q'}-x \mid x^q - x$ in $\mathbb{Z}[x]$. Then, reducing both polynomials mod $p$ preserves this property, so we have that $x^{q'}-x \mid x^q - x$ in $\mathbb{F}_p[x]$.

Now, any root of $x^{q'}-x$ must also be a root of $x^q - x$ by this divisibility relation. But, the roots of $x^{q'} - x$ are exactly the elements of $\mathbb{F}_{q'}[x]$, so we have that $\mathbb{F}_{q'}[x] \subset \mathbb{F}_q[x]$. Thus, $k \mid r$.

B. Mehta
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Consider a field $\Bbb F$ containing both $\Bbb F_{q,q'}$. Let $F$ the Froebenius morphism $x\in F \to x^p\in F$.

Hence $\Bbb F_q = \{x\in F: x^{p^r}=x\} = \ker F^r, \Bbb F_{q'} = \ker F^k$. Let $r = mk + s$, with $s<k$.

$$ (X^{p^k}-X) | (X^{p^r}-X) \iff \ker F^k\subset \ker F^r $$ If $s=0$, RHS is true.

If $\ker F^k\subset \ker F^r$: let $x\in F, F^k x = x$, such as $k = \min\{l>0:F^lx=x\}$. $$ x = F^rx = F^s(F^k)^mx = F^sx $$hence $s=0$ and $k|r$.

mookid
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  • Thanks! But we've never encountered the Froebenius sadly. I tried putting together a proof as an answer to this question but just need to fill in some of the gaps (explain some of the steps). Thanks again for your help! – EthanAlvaree Mar 31 '14 at 03:38
  • well, you just need to prove that this is a fiel homomorphism, this is a consequence of Pascal formula. Then I do not use any if its properties, I believe (this is just an easier way to write things) – mookid Mar 31 '14 at 03:44
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For the forward direction: $x^{q'}-x|x^q-x$ in $\mathbb{Z}[x]$ $\implies x^{q'}-x|x^q-x$ in $\mathbb{F}_p[x]$. Therefore because the roots of $x^{q'}-x$ are all roots of $x^q-x$, and the roots of this are the elements in $\mathbb{F}_{q'}[x]$, then $\mathbb{F}_{q'}[x]\subset \mathbb{F}_q[x]$, which implies that $k|r$.

For the backward direction, $k|r \implies r=kt, t\in\mathbb{Z}$. So $p^k-1|p^{kt}-1$, which implies $x^{p^k-1}-1|x^{p^{kt}-1}-1$, and multiplying by $x$ gives $x^{p^k}-x|x^{p^{tk}}-x$. Therefore we have $x^{q'}-x|x^q-x$.

Just need some reasoning for some of the steps (especially in the backward direction)...

EthanAlvaree
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  • And I should add that part of this relies on this lemma: http://math.stackexchange.com/questions/128007/xa-1-divides-xb-1-if-and-only-if-a-divides-b – EthanAlvaree Mar 31 '14 at 04:17