Injectivity preservation for polynomial maps under field extension

Question

Question

Question. Which subfields $\def\Q{\Bbb{Q}}\def\K{\Bbb{K}}\def\R{\Bbb{R}}\Q\subset\K\subset\R$ have the following "injectivity preservation property" (IPP): any polynomial $P\in\K[X]$ such that (the induced map) $P:\K\to\K$ is injective remains injective when viewed $P:\R\to\R$.

Observations.

1. $\Q$ doesn't have the IPP
2. The field $\def\A{\Bbb{A}}\A$ of real algebraic numbers has the IPP
3. If $\def\L{\Bbb{L}}\Q\subset\K\subset\L\subset\R$ are subfields then
   • If $\K$ has the IPP then so does $\L$

Context

The question came up in a comment discussion on a youtube video. The initial question was: "do rational polynomials which induce an injection over $\Q$ induce one over $\R$"; the counter-example below was given there (though with a mistake in the computation.) The next question was then: which fields have this "injectivity preservation property"? The observation that the field $\A$ of real algebraic number has IPP was given there, too.

Proofs

$\Q$ doesn't have the IPP

Proof. We first notice that the following transformations performed on polynomials preserve injectivity (both over $\K$ and $\R$) $$P(X) \rightsquigarrow aP(bX + c)$$ with $(a, b, c) \in \K^* \times \K^* \times \K$. We may thus work with unitary polynomials with zero constant term. Under such transformation all (rational) degree 2 polynomials are equivalent to $X^2$ and thus not injective. All (rational) degree 3 polynomials are equivalent to $X^3 - pX$ for some $p\in \K$.

We are going to produce a rational degree 3 polynomial of this form which is injective over $\Q$ but not injective over $\R$. We will therefore look for such polynomials with $p>0$ (these are automatically non injective over $\R$.)

Finding collisions in $P$ means finding distinct rational numbers $x\neq y$ with $P(x) = P(y)$ i.e. $x^2 + xy + y^2 = p$ i.e., setting $\sigma = \frac{x + y}{2}$ and $\delta =\frac{x - y}{2}$: $$3\sigma^2 + \delta^2 = p$$

Now we set $p=2$. We must show that there are no rational solutions to $$3\sigma^2 + \delta^2 = 2$$ i.e. there are no integer solutions to $$3u^2 + v^2 = 2w^2$$ (besides the trivial $(0,0,0)$.) Considering the above modulo 8 we see that any solution to the above must have $u,v,w$ all even. This proves that there are no nontrivial solutions and that $X^3 - 2X$ is injective $\Q \to \Q$

The field $\def\A{\Bbb{A}}\A$ of real algebraic numbers has the IPP

Consider $P\in\A[X]$. If $P$ has even degree then it won't be injective $\A\to\A$ since for large enough $\def\N{\Bbb{N}}N\in\N$ the polynomial $P-N$ will have 2 real roots. These are real algebraic numbers and hence $P$ can't be injective over $\A$. We thus assume that $P$ has odd degree. We shall prove that if it is non injective over $\R$ then it is non injective over $\A$.

Injectivity of (unitary) real polynomials $Q$ over $\R$ is equivalent to $Q' \geq 0$ on $\R$. Thus if $P$ is non injective one may find a nontrivial maximal (for $\subset$) real interval $[a,b]\subset\R$ such that $P' < 0$ on $]a,b[$ and $P' \geq 0$ on $]-\infty, a]$. Then $a<b$ are real roots of $P'$ hence real algebraic and so $c=\frac{a+b}2$ is real algebraic. The polynomial $P - P(c) \in \A[X]$ has at least 2 real roots (one $<a$, the other $c$), these are algebraic over $\Q$ and so $P$ is noninjective over $\A$.