$y^{q-1}-f(x)$ irreducible over $\overline{F_{q}}[x]$ When $(\deg f,q-1)=1$

Question

I am trying to solve the following problem, and I’d like to ask for some help. Let $q$ be a prime power, $f(x)\in F_{q}[x]$ a polynomial of degree $d$, such that $(d,q-1)=1$. I’d like to show that $G(x,y)=y^{q-1}-f(x)$ is irreducible over $\overline{F_{q}}[x,y]$.

I only see one direction to work with, and it is not yielding enough. That is, suppose that $G(x,y)=\sum a_iy^{i} \sum b_j y^{j}$ where $a_i,b_j$ are polynomials in $x$. Then, any root of $f(x)$ must be a root of all $a_i,b_j$ (except the leading coefficients, which we assume are just 1), for if we plug this root in we will get a decomposition of $y^{q-1}$. I feel this is a nice direction, as the things I have in mind are $y^{2}-f^{2}$ as counterexamples, but it is not enough. I’ve also tried plugging in some $y’s$, and use the fact that $y^{q-1}=y$ for nonzero $y$ in $F_{q}$, and I’ve also tried some tricks with differentiating, e.g the $y$ derivative of $yG(x,y)$ does not depend on $x$, but none of these tricks yielded anything conrecte or substantial.

I am quite stuck and can’t find a reference online. Even a slight hint would be greatly appreciated! I feel like this shouldn’t be hard, and I am missing something simple…

Answer 1

Hopefully this works.

Let us suppose that $y^{q-1}-f(x)$ is reducible over $\overline{\mathbb F_q}(x)$ and let $h(y)$ be an irreducible factor. Consider the algebraic extension of of $\overline{\mathbb F_q}(x)$ given by:$$L:=\overline{\mathbb F_q(x)}[y]/(h(y)).$$ If $\alpha\in L$ is a root of $h(y)$, then note that all the roots of $h$ are of the form: $$\alpha\cdot \zeta,$$ where $\zeta$ is a $q-1$-root of unity. Moreover, all the roots of $h(y)$ have this form and we can identify the Galois group of $L/\overline{\mathbb F_q}(x)$ with a subgroup $H$ of the $q-1$-roots of unity in $\overline{\mathbb{F_q}}$. We conclude that: $$\prod_{\zeta\in H}\zeta\cdot \alpha\in \overline{\mathbb{F_q}}(x).$$ Note that the subgroup $H$ has to be be strictly contained in the group of $q-1$-roots of unity, otherwise $h(y)=y^{q-1}-f(x)$. Now let $\beta:=\prod_{\zeta\in H}\zeta\cdot \alpha$ be the above product and let $T$ be the group of roots of unity of order $(q-1)/\#H$. Then we have that: $$\prod_{\omega\in T}\omega \beta=f(x).$$ It follows that $\deg(f(x))=\deg(\beta)\cdot \#T$, and since $\#T|q-1$, we get a contradiction.

Answer 2

There's a general result for irreducibility of polynomials in $F[x,y]$, where $F$ is any field, which specializes nicely to your question. Let $$f(x,y) = f_0y^d + f_1(x)y^{d-1} + \dots + f_d(x) \in F[x,y]$$ where $f_i(x) \in F[x]$ and $f_0\neq 0$. Set $$\psi(f) = \max_{1\leq i\leq d} \frac{1}{i}\deg_x(f_i)$$ where $\deg_x$ means the degree as a polynomial in $F[x]$. If $\psi(f) = \frac{m}{d}$ with $m$ and $d$ coprime, then $f(x,y)$ is not just irreducible, but absolutely irreducible. This is the statement that specializes to your question.

As for the proof, I think you can find it in Schmidt's book Equations Over Finite Fields. Unfortunately I can't find my copy, but I remember the proof looking roughly like:

Step (1): Whenever $f(x,y) = g(x,y)h(x,y)$ is a factorization, derive an easy formula for $\psi(f)$ in terms of $\psi(g)$ and $\psi(h)$.

Step (2): Assume $\psi(f) = \frac{m}{d}$ with $(m, d) = 1$ but $f(x,y) = g(x,y)h(x,y)$, then arrive at a contradiction by using the fact that $\deg_y(g) < d$, $\deg_y(h) < d$, and appealing to the simple formula from step 1 and the definition of $\psi$.