I need to prove the following:

"Prove that $\displaystyle f(X)$ is irreducible in $\displaystyle \mathbb{Q}[X]$ if and only if $\displaystyle f(\alpha X + \beta)$ is irreducible in $\displaystyle \mathbb{Q}[X]$"

I want to prove "if"; that is, $\displaystyle f(\alpha X + \beta)$ irreducible $\displaystyle \Rightarrow f(X)$ irreducible.

So, I think we have that $\displaystyle f(X)= a(X)b(X)$, with one of $\displaystyle \deg(a), \deg(b) = \deg(f)$.

The hint states that we should argue by contradiction. So, in this instance, we shall assume that $\displaystyle f(\alpha X + \beta)$ is reducible, and hence

$\displaystyle f(\alpha X + \beta) = g(\alpha X + \beta)h(\alpha X + \beta)$

where $\displaystyle \deg(g), \deg(h) < \deg(f)$

Now,

$\displaystyle f(\alpha X + \beta) \equiv [f(\alpha X + \beta) - f(X)] + f(X)$

and so,

$\displaystyle [f(\alpha X + \beta) - f(X)] + f(X) = g(\alpha X + \beta)h(\alpha X + \beta)$

we say, $\displaystyle f(\alpha X + \beta) - f(X)= \tilde{f}(\alpha X + \beta)$, and so

$\displaystyle f(X) = g(\alpha X + \beta)h(\alpha X + \beta) - \tilde{f}(\alpha X + \beta)$

with $\displaystyle \deg(\tilde{f}) < \deg(g)$, by the division algorithm. But, this is a contradiction, and so $\displaystyle \tilde{f}(\alpha X + \beta) = 0$. Thus,

$\displaystyle f(X) = g(\alpha X + \beta)h(\alpha X + \beta) = a(X)b(X)$

which contradicts the fact that $\displaystyle \mathbb{Q}[X]$ is a unique factorisation domain.

I think this is wrong. I'd be really grateful if anyone could pick out the flaws in this proof, and if possible, provide a better (or, correct) proof.

Thanks in advance,

HTale.