Question on reducibility over rationals.

Question

If we have a polynomial with real coefficients $P(x)$ for which $P(x)=q$ has a rational solution for all rational numbers q, does that mean that the coefficients actually have to be rational? I think this may have something to do with Gauss's Lemma but any help would be much appreciated. Thanks so much.

Answer 1

Let $V$ be the ${\mathbb Q}$-space spanned by $1$ and the coefficients of $P$. We can find a basis $(v_1=1,v_2,v_3,\ldots,v_r)$ of $V$ starting with $1$ : $v_1=1$. Then $P$ can be decomposed as

$$ P=\sum_{k=1}^r v_kP_k=P_1+\sum_{k=2}^r v_kP_k \tag{1} $$

where all the $P_k$ have rational coefficients.

For each $q\in{\mathbb Q}$, there is an $x_q\in{\mathbb Q}$ such that $P(x_q)=q$. From (1) we deduce $P_1(x_q)=q$ and $P_j(x_q)=0$ for $j> 1$. Now, the $x_q$ are obviously all distinct ; so there are infinitely many of them. We see that for $j>1$, $P_j$ has infinitely many roots, so it must be zero. Finally $P=P_1\in{\mathbb Q}[X]$ as wished.

Answer 2

Hint. Take $q=0$.${}{}{}{}{}{}{}{}{}{}{}{}{}{}$