<s>[INST] Proving a polynomial divides other polynomials Given $\mathbb{F}_p[u]$ over $\mathbb{F}_p$ with $p$ prime, how can I prove that $u^2-cu+d$ irreducible is a factor of both $u^p+u-c$ and $u^{p+1}-d$?
 [/INST] When $p=2,$ the claim is trivial, so assume $p$ is an odd prime. In the finite field $\mathbb F_{p^2}=\mathbb F_p[u]/(u^2-cu+d)$ (it is a field by the irreducibility of $u^2-cu+d$,) we want to show $u^p+u-c=0$ and $u^{p+1}-d=0.$
Assuming the first equation, we immediately have that $u^{p+1}-d=u^{p+1}+u^2-cu=u(u^p+u-c)=0,$ so all that is left to do is show $u^p+u-c=0.$
Now, in $\mathbb F_p[u]/(u^2-cu+d),$ we can express $u^p=au+b,$ for some $a,b\in\mathbb F_p.$ Then, taking both sides to the $p$-th power, we have $u^{p^2}=u=(au+b)^p=au^p+b=a(au+b)+b=a^2u+(a+1)b.$ Thus, $a^2=1$ and $(a+1)b=0.$
If $b=0,$ we would have $u^p=au,$ or $u^{p-1}=a.$ Then, $(u^{p-1})^p=u^{p^2-p}=u^{-p}=a,$ so $u=a^{-2},$ a contradiction.
Hence, we have that $b\neq 0,$ so since $(a+1)b=0,$ we have that $a=-1.$ Now, since $u^2-cu+d=0,$ $(u^2-cu+d)^p=cu^{2p}-cu^p+d=(2c-2b)u+(b^2-cb)=0,$ showing $b=c.$
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