Structure on $\mathbb{R}$ such that homomorphisms $\mathbb{R} \to \mathbb{R}$ are exactly polynomials?

Question

If we consider $\mathbb{R}$ as a vector space over the field $\mathbb{R}$, then the maps $\mathbb{R} \to \mathbb{R}$ preserving this structure are exactly the linear maps $x \mapsto ax$. In contrast, if we think of $\mathbb{R}$ as an affine space instead, the maps preserving this structure are exactly the affine maps $x \mapsto ax + b$.

Is it possible to "forget" even more structure, and have a structure on $\mathbb{R}$ such that the maps $\mathbb{R} \to \mathbb{R}$ preserving this structure are exactly the polynomials?

I'll leave it up to you what ring to take coefficients of polynomials from. If anyone could do it for $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{C}$ instead, I'll be equally happy.

Answer 1

The structure is that of an affine variety or affine algebraic set. In general, as a set, it is the zero-locus in $\mathbb{A}^n(k)=k^n$ of some finite family of polynomials (that generate a prime ideal for a variety). For $\mathbb{A}^n(k)$ itself we take the zero polynomial. This is usually considered over $k=\mathbb{C}$, or some other algebraically closed field, not $\mathbb{R}$. However, one can restrict to considering real points in it. In this case we will take the affine line $\mathbb{A}^1(k)$. Morphisms of affine varieties are functions that are polynomials in each coordinate, in this case, a single coordinate. What they preserve is the coordinate ring, in this case the ring of all polynomials, compositions of polynomials with polynomials are again polynomials. This makes it a type of ringed space.

Answer 2

Conifold's answer is the right one: The structure of $\mathbb{R}$ as the affine line over $\mathbb{R}$ is the canonical one in which the endomorphisms are the polynomial maps in $\mathbb{R}[x]$.

I just want to follow up on Rob Arthan's comment, since you use the model theory tag. When we view $\mathbb{R}$ as an affine variety, we are not viewing it as a structure in the sense of model theory. In fact, there is no way of equipping $\mathbb{R}$ with constant, function, and relation symbols in such a way that the homomorphisms $\mathbb{R}\to \mathbb{R}$ are exactly the polynomials.

Let's suppose that we're viewing $\mathbb{R}$ as an $L$-structure in such a way that every polynomial map $\mathbb{R}\to \mathbb{R}$ is a homomorphism. Suppose $P$ is an $n$-ary relation symbol in the language, and suppose $\mathbb{R}\models P(a_1,\dots,a_n)$ for some distinct $a_1,\dots,a_n\in \mathbb{R}$. Now for any distinct $b_1,\dots,b_n\in \mathbb{R}$, there is some polynomial $f\in \mathbb{R}[x]$ of degree $n-1$ so that $f(a_i) = b_i$ for all $1\leq i \leq n$. Since $f$ is a homomorphism, $\mathbb{R}\models P(b_1,\dots,b_n)$. Thus if $P$ holds of any $n$-tuple of distinct elements, it holds of every $n$-tuple of distinct elements.

A similar argument shows that for each relation symbol in the language, if the relation symbol holds of some tuple (possibly with repeats), then it holds of every tuple that has the same pattern of repeats. For example, if $\mathbb{R}\models P(a,b,a)$ with $b\neq a$, then $\mathbb{R}\models P(c,d,c)$ for all $c\neq d$. The same argument also shows that there can be no constant symbols in the language, and any $n$-ary function symbol in the language must be interpreted to always return one its $n$ inputs, and which input it returns only depends on the pattern of repeats in the tuple of inputs (apply the argument to the graph relation $g(x_1,\dots,x_{n-1}) = x_n$).

Now you can check that in a trivial language like this, any function $\mathbb{R}\to \mathbb{R}$ is a homomorphism. In particular, there are many endomorphisms which are not polynomials.