Exercise 11.2.12

Proof.

(a)

Let $\alpha =[i]+⟨f⟩$ be the coset of the constant polynomial $[i]\in {𝔽}_p$. Then $(T-1_R)(\alpha )={\alpha }^p-\alpha =({[i]}^p-[i])+⟨f⟩=[0]+⟨f⟩=\bar{0}$, so $\alpha \in \ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R)$. $${\mathrm{vect}}_{{𝔽}_p}([1]+⟨f⟩)=\{\alpha \in R|\exists <!--\; FUNCTION\; APPLICATION\; -->[i]\in {𝔽}_p,\alpha =[i]+⟨f⟩\}$$

is a one-dimensional subspace of the kernel of $T-1_R$, corresponding to the constant polynomials in ${𝔽}_p[x]$.

(b)

By part (a), the rank of $T-1_R$ is not $n$, and by hypothesis this rank is not $n-1$, so the rank of $T-1_R$ is less than $n-1$, so the kernel of $T-1_R$ has dimension at least 2.

Therefore there exists a polynomial $h$, with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)<n$, such that $\bar{h}=h+⟨f⟩\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R)$ which is not in the one-dimensional subspace of part (a), thus $h$ is not a constant polynomial. Since $h\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R)$, $h^p+⟨f⟩=h+⟨f⟩$, thus $f\mid h^p-h$.

Conclusion: there is a nonconstant polynomial $h$ of degree $<n$ such that $f\mid h^p-h$.

(c)

In $k[x]$, we know that $$x^p-x=\underset{a\in {𝔽}_p}{\prod }(x-a).$$

The formal composition with $h(x)$ gives

$$h^p-h=\underset{a\in {𝔽}_p}{\prod }(h-a).$$

(d)

We know that if $f_1,\dots ,f_s$ are polynomials in ${𝔽}_p[x]$ such that $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f_i,f_j)=1$ if $i\ne j$, then $$\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f_1\cdots f_s)=\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f_1)\cdots \gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f_s).$$

Take $f_a=h-a\in {𝔽}_p[x]$. Then $a\ne b\Rightarrow \gcd <!--\; FUNCTION\; APPLICATION\; -->(f_a,f_b)=1$, since

$$\frac{1}{b-a}(h-a)-\frac{1}{b-a}(h-b)=1.$$

Therefore, since $f\mid h^p-h$,

$$\begin{array}{llll}\hfill f& =\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h^p-h)& \hfill & \\ \hfill & =\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,\underset{a\in {𝔽}_p}{\prod }(h-a))& \hfill & \\ \hfill & =\underset{a\in {𝔽}_p}{\prod }\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a)& \hfill & \end{array}$$

(e)

Since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)<n$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h-a)<n$, therefore $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a))<n$. Moreover, since $f\ne 0$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a))\ge 0$, so $f$ cannot divide $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a)$ when $a\in {𝔽}_p$.

Therefore the product ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{a\in {𝔽}_p}\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a)$ has more than one nonconstant factor (if all factors except one are constant, then $f\mid \gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a)$ for some $a\in {𝔽}_p$, which is impossible).

So $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,h-a)$ is a nonconstant factor of $f$ (and non associate to $f$) for at least two $a\in {𝔽}_p$.