Is $p\in k[z]$ irreducible iff $q(x,y)=y^{b\cdot\deg(p)}p(\frac{x^{a}}{y^{b}})\in k[x,y]$ is irreducible, where $a,b$ are coprime integers

Question

Let $a,b\in\mathbb{N}$ be coprime, $p\in k[z]$ a polynomial of degree $g$. Then the bivariate polynomial $q\in k[x,y]$, given by \begin{equation*} q(x,y)=y^{b\cdot g}p\left(\frac{x^{a}}{y^{b}}\right) \end{equation*} is irreducible if and only if $p$ is.

One direction is obvious: If $p$ is reducible and factors as \begin{equation*} p(z)=p_{1}(z)\cdot...\cdot p_{n}(z)\text{,} \end{equation*} and $g_{1},...,g_{n}$ are the respective degrees of $p_{1},...,p_{n}$, then $q$ factors as \begin{equation*} q(x,y)=\prod_{i=1}^{n}y^{b\cdot g_{i}}p_{i}\left(\frac{x^{a}}{y^{b}}\right)\text{,} \end{equation*} so $q$ is reducible.

One can try to prove the other direction by assuming that $q=\prod_{i=0}^{n}q_{i}$ is reducible. Then $p(z)=\prod_{i=0}^{n}q_{i}(z,1)$. But how do I know that none of the $q_{i}$s is constant?

Answer 1

Since you already proved one direction, I'll give you several hints for the other direction and hope you can reach to a solution (without much algebraic geometry). Let us assume that $p$ is nonconstant irreducible and $q$ isn't. I am assuming that $a,b$ are positive integers.

1. Since $a,b$ are coprime you may find $c,d\in \mathbb N$ such that $ac-bd = 1$
2. Show that $f\in k[x,y]$ is nonconstant iff $f(z^c,z^d) \in k[z]$ is nonconstant
3. If $q(x,y) = q_1(x,y)q_2(x,y)$ for nonconstant $q_1,q_2$ then what can you say about $q(z^c,z^d)$.

Answer 2

Let me attempt to answer my question myself, is this a valid solution?

Assume $q$ is irreducible. Endow $k[x,y]$ with the structure of a graded ring by setting $\deg(x)=b$, $\deg(y)=a$. Then $q$ is a homogeneous polynomial in $k[x,y]$, and its associated variety \begin{equation*} V(q)\subseteq\operatorname{Proj}(k[x,y]) \end{equation*} is irreducible. Let $D(y)$ be the open subset of $\operatorname{Proj}(k[x,y])$ where $y$ does not vanish, and $V(p)\subseteq\mathbb{A}^{1}_{k}$ the variety associated to $p$. Then $V(p)=V(q)\cap D(y)$, and is thus irreducible.