Mathematical Reflections 2012, Issue 4 - Problem U236

Problem:
Let $f(X)$ be an irreducible polynomial in $\mathbb{Z}[X]$. Prove that $f(XY)$ is irreducible in $\mathbb{Z}[X,Y]$.

Proposed by Mircea Becheanu.

Solution:

If $f(X)$ is a constant polynomial, the statement is trivial. So, assume that $\deg f(X) > 0$. Since $f(X)$ is irreducible, then must be primitive. Assume that $f(XY)$ is reducible in $\mathbb{Z}[X,Y]$. Then, there are two non constant polynomials $g(X,Y), h(X,Y) \in \mathbb{Z}[X,Y]=\mathbb{Z}[X][Y]$ such that $f(XY)=g(X,Y)h(X,Y)$. It cannot be $f(XY)=g(X)h(Y)$, otherwise $f(0)=g(0)h(Y)$, so $h(Y)$ must be constant. Then both $g(X,Y)$ and $h(X,Y)$ must contain both $X$ and $Y$. But $$f(X)=g(X,1)h(X,1)=\overline{g}(X)\overline{h}(X), \qquad \overline{g}(X), \overline{h}(X) \in \mathbb{Z}[X]$$ is the product of two non constant polynomials, i.e. $f(X)$ is reducible, contradiction.