Applying a polynomial to the roots of a polynomial with rational coefficients Let $p \in \mathbb{Q}[t]$ be an irreducible polynomial. Write $p(t) = (t - \alpha_1)(t - \alpha_2) \dotsb (t - \alpha_n)$. Does $q(t) = (t - r(\alpha_1))(t - r(\alpha_2)) \dotsb (t - r(\alpha_n))$ belong to $\mathbb{Q}[t]$, where $r \in \mathbb{Q}[t]$? I think this should be true, but the proof has eluded me. 
 A: Yes.  For instance, considering the $\alpha_i$ as formal variables, the coefficients of $q(t)$ are symmetric in the $\alpha_i$.  This means they can be written as polynomials (with coefficients in $\mathbb{Q}$) in the elementary symmetric functions of the $\alpha_i$.  But the elementary symmetric functions of the $\alpha_i$ are (up to sign) just the coefficients of $p(t)$, and so are rational.  Thus the coefficients of $q(t)$ are rational.
A: Yes, this is true. Aside from the proof in Eric Wofsey's answer here are two other proofs. 
Galois theory
The roots of $p$ generate a Galois extension of $\mathbb{Q}$ with some finite Galois group $G$ acting on the roots whose fixed field is $\mathbb{Q}$. $G$ has the property that $g(r(\alpha_i)) = r(g \alpha_i)$ for any polynomial $r$ with rational coefficients, so it also permutes the $r(\alpha_i)$. This implies that the coefficients of $\prod (t - r(\alpha_i))$ are fixed by the action of $G$ and hence are rational.
(Eric Wofsey's answer can also be thought of in these terms: polynomials are a Galois extension of symmetric polynomials with full Galois group $S_n$.)
Explicit construction
$p$ has a companion matrix $M$, which is among other things a square matrix with rational entries whose characteristic polynomial is $p$ and hence whose eigenvalues are the roots of $p$. Now, $r(M)$ is again a square matrix  with rational entries. Its eigenvalues are the $r(\alpha_i)$ (with the same eigenvectors as $M$), and so its characteristic polynomial, which must also have rational entries, is $\prod (t - r(\alpha_i))$. 
This proof gives the easiest way I know to actually compute the coefficients of $\prod (t - r(\alpha_i))$. 
