Characterization of polynomial injection from Q to Q? I want to know if we can find (or characterize) all the polynomials $f(x) \in \mathbb{Q}[x]$ that induces an injection $f : \mathbb{Q} \rightarrow \mathbb{Q}$ by evaluation.
Some examples are $x, x^{3}$, and I would like to know how to find all of them. However, please don't hesitate to let me know any useful list of examples that narrow the possibilities of $f(x)$ or references that give ways to attack this problem. Thank you.
 A: When investigating $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_0$ with $a_n\ne0$ for injectivity on the rationals, you may scale $f$ vertically to ensure $a_n=1$, translate horizontally to ensure $a_{n-1}=0$, and translate vertically to ensure $a_0=0$.
Lastly, we can scale horizontally to ensure that $a_{n-2}$ is a square-free integer (or zero).
Thus wlog. $f(x)=x^n+a_{n-2}x^{n-2}+\cdots +a_1x$ with $a_{n-2}$ a square free integer or zero.
In degree $1$, we thus get $f(x)=x$, which is injective.
In degree $2$, we get $f(x)=x^2$, which is not injective.
In degree $3$, we get $f(x)=x^3+ax$. 
If $a\ge 0$, this is injective.
Otherwise note $f(x)-f(y)=(x-y)(x^2+xy+y^2+a)$, so $f$ is injective iff there is no $z=x-y\omega\in\mathbb Q(\omega)$ with $x\ne y$ and $N(z)=|z|^2=-a$, where $\omega=-\frac12+\frac{\sqrt 3}2i$ is a primitive third root of unity. 
The condition $x\ne y$, i.e. $z\notin(1-\omega)\mathbb Z$ can be ignored because if $N(z)=-a$ then also $N(\omega z)=-a$.
Consider the prime decomposition of $-a=|a|$ in the principle ideal domain $\mathbb Z[\omega]$.
If $p\in\mathbb Z$ is an inert prime in our extension (that is $p\equiv 2\pmod 3$), then $p|-a$ implies $p|z$, hence $p^2|-a$, contradiction.
If on the other hand all prime factors $p$ are not inert (i.e. split, $p\equiv 1\pmod 3$; or ramify, $p=3$), then taking one of the primes above each $p$ and multiplying these gives us $z$ with $N(z)=|a|$. (This includes the case $a=-1$ that is trivial anyway).
Thus $f(x)=x^3+ax$ with $a$ square-free is injective iff $a\ge 0$ or $a$ is divisible by at least one prime $\in\{2,5,11,17,23,29,41,47,\ldots\}$.
In degree $4$, we get $f(x)=x^4+ax^2+bx$ with $a$ squarefree.
This is even, hence not injective, if $b=0$.
We have
$$\frac{f(x)-f(y)}{x-y}=(x^3+x^2y+xy^2+y^3)+a(x+y)+b=(x+y)(x^2+y^2-a)+b$$
It is again an interesting problem whether this expression can be made $=0$ or not. For example, if $6|a$, $2\not |b$ and $3\Vert b$, we note that $x+y$ or $x^2+y^2-a$ must be divisible by $3$. If the latter, then $3|x$ and $3|y$, hence $f(x)\ne f(y)$ because $9\not | b$. And if the former, then $x^2+y^2-a=2x^2-a$ is even, but $b$ is not. We conclude that $f$ is injective. This looks more like a trick than a theory, but I'm sure something nice can be done - just less easy than in the quadratic case, not to mention degrees $\ge 5$.
A: For real polynomials, it must pass the horizontal line test. A real polynomial is monotonic iff it is always increasing or always decreasing. Thus, they are the integrals of polynomials that are non-negative or non-positive. So the question is whether a real polynomial is always monotone if its restriction to $\Bbb{Q}$ is monotone under $\Bbb{Q}$'s order.
If a real polynomial intersects a horizontal line at least twice, there is a local minimum or maximum between two of those points, and the height of such a point together with the height of the line give an interval of values which are visited at least twice by the function.
The rationals are a dense subset of the reals, meaning every neighborhood of a real number has at least one rational. Therefore, that interval of values that are repeated contains at least one rational. Therefore if a real polynomial is not monotonic, then the rational polynomial is not monotonic. Therefore if a rational polynomial is monotonic, then the real polynomial is monotonic.
Clearly every monotonic real polynomial is monotonic as a rational polynomial. Therefore a polynomial is monotonic over the rationals iff it is monotonic over the reals iff it passes the horizontal line test.
