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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?

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3 Answers 3

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No, $f(x)=x$ is irreducible and $f(x^2)=x^2$ is not.

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    $\begingroup$ Similarly, $f(x)=x-a^2$ is irreducible and $f(x^2)=x^2-a^2=(x-a)(x+a)$ is not. $\endgroup$
    – Mike F
    Sep 10, 2017 at 21:43
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As Tsemo already answered, this is false.

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.

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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 $0\leq i<n$, $p$ 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 $\mathbb Q$.

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