# Does there exist some $p(x) \in \mathbb{Q}[x]$, deg$(p) > 1$, which maps $\mathbb{Q}$ onto itself surjectively?

Clearly this is impossible for $$p$$ of even degree, and I imagine that Cardano’s formula quickly reveals it to be impossible in the cubic case, although I have not checked in detail. My guess is that no such $$p$$ exists. Does one exist? If so, is there an explicit example?

Failing a general yes or no answer, are there sufficient conditions to identify a non-surjective polynomial function?

• Nope cause you won’t hit $1/p$ for $p$ large (lemme know if I’m being stupid!). Sep 18, 2021 at 5:26
• Perhaps $\ \mathbb Q\setminus f(\mathbb Q)\$ is dense in $\ \mathbb Q$. Sep 18, 2021 at 6:45
• (…because wlog by scaling $f\in \Z[x]$, and then when $p$ doesn’t divide the leading coefficient $c_0$ of $f(x) =: c_0 x^d + \cdots + c_d\in \Z[x]$ the $p$-adic valuation of ($(a,b) = 1$) $f(a/b) = b^{-d} (c_0 a^d + c_1 a^{d-1} b + \cdots + c_d b^d) = b^{-d} (c_0 a^d + (\in b\Z))$ is either $\geq 0$ (when $(p,b)=1$) or a multiple of $d$ (when $p\vert b$, whence $(p,a) = 1$), so it can’t be $-1$.) Sep 18, 2021 at 22:48
• Stronger question answered here: mathoverflow.net/questions/186768/images-of-polynomials Sep 19, 2021 at 15:25

No, this can't happen. One way to prove this is via Hilbert irreducibility: The polynomial $$p(x) - t$$ is irreducible over $$\mathbb Q[x,t]$$, so there are infinitely many specializations $$t = c$$ with $$c \in \mathbb Q$$ such that $$p(x) - c$$ is irreducible in $$\mathbb Q[x]$$. Since the degree of $$p(x)$$ is greater than 1, it follows that for each such $$c$$ the polynomial $$p(x) - c$$ has no rational roots.

• Fascinating. This is the first I’ve heard about the Hilbert irreducibility theorem. I will read more about it. Thank you!
– Bma
Sep 18, 2021 at 4:45
• This shows that a similar statement holds when we replace the field of rational numbers by a Hilbertian field (e.g., a number field or finitely generated field of characteristic zero). Sep 18, 2021 at 19:28

John's answer is good, but an alternative, more elementary, method is to use the theory of height functions. Thus for a fraction in lowest terms $$a/b$$, we define the height of $$a/b$$ to be $$H(a/b) = \max\bigl\{ |a|,|b|\bigr\}.$$ One can show that if $$p(x)\in\mathbb Q[x]$$ is a polynomial of degree $$d$$, then there is a constant $$C_1(p)>0$$ depending only on $$p$$ such that $$H\bigl(p(a/b)\bigr) \ge C_1(p) H(a/b)^d.$$ From this one can prove a counting result: $$\# \bigl\{ a/b\in\mathbb Q : H\bigl(p(a/b)\bigr) < X \bigr\} \le C_2(p)\cdot X^{2/d}.$$ On the other hand, it's not hard to see that here is a constant $$C_3>0$$ such that $$\#\bigl\{ a/b\in\mathbb Q: H(a/b) < X\bigr\} \ge C_3\cdot X^2.$$ So the heights of the values of $$p(a/b)$$ grow too rapidly to cover all of $$\mathbb Q$$. (Indeed, part of the proof of Hilbert irreducibility uses an argument of this sort.)

• Should $f$ be $p$?
– user44143
Sep 18, 2021 at 18:20
• Note that Joe's answer also makes it clearer that something stronger is true: Not only is the set of points in $\mathbb Q \setminus p(\mathbb Q)$ infinite, but the image $p(\mathbb Q)$ is actually quite sparse in $\mathbb Q$. Sep 18, 2021 at 18:42
• @MattF. Yes, thanks, I'll fix that. I'm really used to the letter $p$ being a prime! Sep 18, 2021 at 19:45
• @JohnDoyle True, although a more refined version of Hilbert irreducibility is that the exceptional set is a thin set, so has exactly this sparsity property relative to the usual height function. So your answer already implicitly includes the sparisty of $\mathbb Q\smallsetminus p(\mathbb Q)$. Sep 18, 2021 at 23:11

I claim that no polynomial $$q$$ of degree greater than $$1$$ and rational coefficients can be a surjective mapping from $$\mathbb{Q}$$ to $$\mathbb{Q}$$.

Suppose that a polynomial $$q$$ of degree greater than $$1$$ is surjective from $$\mathbb{Q}$$ to $$\mathbb{Q}$$. For simplicity, by replacing $$q(x)$$ with $$p(x)=\alpha(q(\beta x)-\gamma)$$ where $$\alpha,\beta,\gamma$$ are rational with $$\alpha,\beta\neq 0$$, we can assume that $$p(x)$$ is a surjective monic polynomial with constant term $$0$$ and integer coefficients. Suppose now that $$p(x)=x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x$$ where the coefficients $$a_{1},\dots,a_{n-1}$$ are integers.

If $$\alpha,\beta$$ are integers with $$p(x)=\frac{\alpha}{\beta}$$, then $$\beta x^{n}+\dots+\beta a_{1}x=\alpha$$, so by the rational root theorem, $$x$$ must be of the form $$\frac{r}{s}$$ where $$r$$ is a factor of $$\alpha$$ and $$s$$ is a factor of $$\beta$$. In particular, in the case where $$\beta=1$$, if $$p(x)=\alpha$$, then $$x$$ must be a factor of $$\alpha$$. Therefore, $$p$$ must restrict to a surjective function from $$\mathbb{Z}$$ to $$\mathbb{Z}$$. This is impossible.

• Thanks for a great answer. This is the most elementary one yet.
– Bma
Sep 19, 2021 at 7:13
• I don't see how replacing $q$ by $\alpha (q(\beta x)-\gamma )$ produces a monic polynomial with integer coefficients.
– abx
Oct 6, 2021 at 7:50
• Suppose that $q(x)=a_{n}x^{n}+\dots+a_{0}$. Let $q_{1}(x)=q(x)-a_{0}$. Then $q_{1}(x)=a_{n}x^{n}+\dots+a_{1}x$. Let $q_{2}(x)=q_{1}(a)/a_{n}$. Then $q_{2}(x)=x^{n}+b_{n-1}x^{n-1}+\dots+b_{1}x$. Now select a non-zero integer $b$ such that $b^{n-k}b_{k}$ is an integer for $1\leq k<n$. Then let $p(x)=x^{n}+b\cdot b_{n-1}x^{n-1}+\dots+b^{n-1}\cdot b_{1}x$. Then $p(x)=b^{n}q_{2}(x/b)$. Oct 6, 2021 at 10:42
• Oh, sure! Thank you.
– abx
Oct 7, 2021 at 7:42