Is there a nice algebraic structure that can be imposed on the set of polynomials passing through a collection of points in $\mathbb{R}^2$ without duplicated x-coordinates?
Consider the polynomials $\mathbb{R}[x]$ and a collection of $n+1$ points $(u_1, v_1) , (u_2, v_2) , \dots , (u_n, v_n), (u_{n+1}, v_{n+1})$ . Additionally,
$$\forall \;1 \le i \lt n + 1 \mathop{.} u_i \lt u_{i+1}$$
There's a unique polynomial of minimal degree whose graph passes through all those points. The degree of that polynomial is at most $n$. However, if we look at polynomials of any degree, then there are uncountably many that pass through those points. I'm curious what kind of structure they have and if it's possible to use that structure to "pick out" the unique polynomial of minimal degree.
If $v_1 = v_1 = \dots = v_{n} = v_{n+1} = 0$ , then our choice of structure seems fairly straightforward.
The set of polynomials whose set of roots contains $u_1, \dots, u_{n+1}$ forms a vector space over $\mathbb{R}$ , since they're closed under scalar multiplication and pointwise addition. $\lambda x \mathop{.}0$ is our polynomial of least degree, and also the additive identity, and also the only polynomial fixed by scalar multiplication.
In the more general case where $v_i \ne 0$ for at least one $v_i$ , is there a structure we can impose on the set of polynomials passing through those points?
I haven't made much progress trying to figure this out, but there are at least two operations that can, as a bare minimum, produce new polynomials in $F$ given inputs.
If we let $F \subsetneq \mathbb{R}[x]$ be the name of the collection of our polynomials and $f_1, f_2, f_3$ be arbitrary elements of $F$ then
$$ f_1 + f_2 – f_3 \in F $$
$$ f_1 + \alpha \, f_1 – \alpha \, f_2 \in F \;\;\;\;\;\;\text{where $\alpha \in \mathbb{R}$} $$
But this is just leveraging the fact that the roots of a difference of polynomials in $F$ like $f_4 – f_5$ must include $\vec{u}$ and doesn't seem like a promising direction.
Best Answer
Let $f$ and $g$ be two polynomials passing through these $n+1$ points. Then the difference $f-g$ has a zero at each of the $u_i$, and hence it is a multiple of $\prod_{i=1}^{n+1}(x-u_i)$. So the set of polynomials passing through these points is of the form $$\left\{f+h\prod_{i=1}^{n+1}(x-u_i):\ h\in\Bbb{R}[x]\right\}=f+\left(\prod_{i=1}^{n+1}(x-u_i)\right),$$ for some $f\in\Bbb{R}[x]$, which is a coset of an ideal in $\Bbb{R}[x]$.
If $v_i=0$ for all $i$ then $f\in\left(\prod_{i=1}^{n+1}(x-u_i)\right)$, and so the set is an $\Bbb{R}[x]$-ideal. This makes the set a real vector space in particular, and a subspace of $\Bbb{R}[x]$ of codimension $n+1$.
If $v_i\neq0$ for some $i$ then the set is a nontrivial coset of an $\Bbb{R}[x]$-ideal, and not an abelian group. It is a real affine space, however; it is a translation of a linear subspace of $\Bbb{R}[x]$ of codimension $n+1$.
Note that such an $f$ can be constructed explicitly, for example the Lagrange polynomial.