# If $f(x) \in \mathbb{Q}[x]$ is irreducible, then is $f(x^2)$irreducible?

I found a set of practice questions, one of which asks whether or not $f(x) \in \mathbb{Q}[x]$ irreducible implies $f(x^2)$ irreducible.

Is this true? I'm having trouble thinking of a counterexample. Is there an irreducibility criterion that we could use?

No, $f(x)=x$ is irreducible and $f(x^2)=x^2$ is not.
• Similarly, $f(x)=x-a^2$ is irreducible and $f(x^2)=x^2-a^2=(x-a)(x+a)$ is not. Sep 10, 2017 at 21:43
But now you might wonder if it's still true that $f(x^2)$ is irreducible if we are given any irreducible $f(x) \in \mathbb Q[x]$ with $\deg(f) > 1$. Now, if $\alpha \in \mathbb Q$ is a root of $f(x^2)$, then $\alpha^2 \in \mathbb Q$ is a root of $f(x)$ and hence $(x - \alpha^2)$ is a factor of $f(x)$. This means if $f(x)$ is not linear then the degree of the factors of $f(x^2)$ must be at least $2$. So I looked for examples with $\deg(f) = 2$ with $f(x^2)$ having two quadratic factors. Remembering a factoring trick, I found the following example.
Take $f(x) = x^2 + 4 \in \mathbb Q[x]$ which is irreducible since it has no roots in $\mathbb Q$. Then we have, $$f(x^2) = x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 - 2x + 2)(x^2 + 2x + 2)$$ and so $f(x^2)$ is not irreducible.
If $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb Z[x]$$ is an Eisenstein polynomial, i.e, there exists a prime number $$p$$ such that $$p$$ divides each $$a_i$$ for $$0\leq i, $$p$$ does not divide $$a_n$$ and $$p^2$$ does not divide $$a_0$$, then $$f(x^n)$$ is also an Einsenstein polynomial and thus irreducible over $$\mathbb Q$$.