If $f(x^2)$ is irreducible, then $f(x)$ is also irreducible

Question

Let $f(x)$ be a monic polynomial over $\mathbb{Q}$ of degree $d>1$. Show that if $f(x^2)$ is irreducible, then $f(x)$ is also irreducible.

If we approach this by contrapositive, then we assume that $f(x)$ is reducible. So $f(x)=g(x)h(x)$, where the degrees of $g$ and $h$ are greater than 1.

Hence, $f(x^2)=g(x^2)h(x^2)$. It follows that $f(x^2)$ is also reducible.

I'm not sure about my statement that $f(x^2)=g(x^2)h(x^2)$. Is it correct? If not, how would I approach this problem?

Thanks!

Answer 1

If $f(x)=g(x)h(x)$, then $f(x^2)=g(x^2)h(x^2)$ is true (you are just changing the input from $x$ to $x^2$). Since $g(x)$ and $h(x)$ are non-constants, $g(x^2)$ and $h(x^2)$ must also be non-constants, and since the only units in $\mathbb{Q}[x]$ are the non-zero constants, $f(x^2)$ is reducible.