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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 $u^{p+1}-d$?

frobeniusguy
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  • what is $u$ in $\Bbb{F}_{p}[u]$? – Anurag A Mar 19 '20 at 21:18
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    Let $a$ and $b$ be the roots of $u^2-cu+d$ in an extension of $\Bbb F_p$. Show that $a^{p}$ and $b^{p}$ are also roots of $u^2-cu+d$. Use this result to verify that $a^{p+1}=d=b^{p+1}$. Finally note that $u^{p+1}-d=u(u^p+u-c)-(u^2-cu+d)$. – Batominovski Mar 19 '20 at 21:59

2 Answers2

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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 $b=c.$

Kenta S
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Kenta's argument is fine. I want to add a dash of Galois theory to the mix for I think it shows why the argument works (and how it can be easily discovered).

It is given that the quadratic $f(u)=u^2-cu+d$ is irreducible over the prime field. Therefore its zeros are in the quadratic extension field $K=\Bbb{F}_{p^2}$. Assume that the zeros are $\alpha,\beta$, so we have the factorization $$ f(u)=(u-\alpha)(u-\beta). $$ By Vieta relations (or simply expanding the right hand side) we arrive at the equations $$ \begin{aligned} c&=\alpha+\beta,\ \text{and}\\ d&=\alpha\beta. \end{aligned} $$ By basic Galois theory the untrivial automorphism of $K$ is the Frobenius automorphism $F(x)=x^p$. Irreducibility of $f$ means that its zeros are Galois conjugates of each other, so $F(\alpha)=\beta$ and $F(\beta)=\alpha$ (neither can be a fixed point of $F$ for those are the elements of the prime field). So $\alpha^p=\beta$, $\beta^p=\alpha$, and the equations above become $$ \begin{aligned} c&=\alpha+\alpha^p,\ \text{and}\\ d&=\alpha^{p+1} \end{aligned} $$ and the same equations with $\alpha$ replaced by $\beta$.

But these last equations imply that $\alpha$ and $\beta$ are (distinct) zeros of both $u^p+u-c$ and $u^{p+1}-d$. This implies that $f(u)=(u-\alpha)(u-\beta)$ is a factor of both these polynomials.

So, in a sense, a one-step argument if you are familiar with the somewhat simple Galois theory of $K$.

Jyrki Lahtonen
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  • Thank you, this argument is very insightful! – Kenta S Mar 20 '20 at 06:16
  • That follows from the calculations in Kenta's answer. All the common zeros of $u^p+u-c$ and $u^{p+1}-d$ are simple so it suffices to check that $\alpha$ and $\beta$ are the only common zeros. If $\gamma$ is a zero of both, then $$\gamma^p=c-\gamma$$ and it follows that $\gamma^{p+1}=c\gamma-\gamma^2$. On the other hand $\gamma^{p+1}=d$, and putting these two together gives $$\gamma^2-c\gamma+d=0.$$ Meaning that $\gamma$ must be equal to $\alpha$ or $\beta$. – Jyrki Lahtonen Mar 21 '20 at 04:38
  • The zeros of both polynomials are simple already. It is rather easy to show that neither has common factors with its derivative. – Jyrki Lahtonen Mar 21 '20 at 09:44