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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?

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  • $\begingroup$ Nope cause you won’t hit $1/p$ for $p$ large (lemme know if I’m being stupid!). $\endgroup$
    – alpoge
    Sep 18, 2021 at 5:26
  • $\begingroup$ Perhaps $\ \mathbb Q\setminus f(\mathbb Q)\ $ is dense in $\ \mathbb Q$. $\endgroup$
    – Wlod AA
    Sep 18, 2021 at 6:45
  • $\begingroup$ (…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 $-1$.) $\endgroup$
    – alpoge
    Sep 18, 2021 at 22:48
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    $\begingroup$ Stronger question answered here: mathoverflow.net/questions/186768/images-of-polynomials $\endgroup$ Sep 19, 2021 at 15:25

3 Answers 3

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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.

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    $\begingroup$ Fascinating. This is the first I’ve heard about the Hilbert irreducibility theorem. I will read more about it. Thank you! $\endgroup$
    – Bma
    Sep 18, 2021 at 4:45
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    $\begingroup$ 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). $\endgroup$ Sep 18, 2021 at 19:28
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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.)

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  • $\begingroup$ Should $f$ be $p$? $\endgroup$
    – Matt F.
    Sep 18, 2021 at 18:20
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    $\begingroup$ 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 $\mathbb Q$. $\endgroup$
    – John Doyle
    Sep 18, 2021 at 18:42
  • $\begingroup$ @MattF. Yes, thanks, I'll fix that. I'm really used to the letter $p$ being a prime! $\endgroup$ Sep 18, 2021 at 19:45
  • $\begingroup$ @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 $\mathbb Q\smallsetminus p(\mathbb Q)$. $\endgroup$ Sep 18, 2021 at 23:11
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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.

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  • $\begingroup$ Thanks for a great answer. This is the most elementary one yet. $\endgroup$
    – Bma
    Sep 19, 2021 at 7:13
  • $\begingroup$ I don't see how replacing $q$ by $\alpha (q(\beta x)-\gamma )$ produces a monic polynomial with integer coefficients. $\endgroup$
    – abx
    Oct 6, 2021 at 7:50
  • $\begingroup$ 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 $p(x)=b^{n}q_{2}(x/b)$. $\endgroup$ Oct 6, 2021 at 10:42
  • $\begingroup$ Oh, sure! Thank you. $\endgroup$
    – abx
    Oct 7, 2021 at 7:42

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