Looking at evaluating polynomials at a point via schemes

Question

Let $R$ be some commutative ring with identity and fix some $a \in R$. Then, consider the evaluation map: $$\epsilon_a: R[x] \to R: f(x) \mapsto f(a)$$ We know that this is a surjective homomorphism (in fact, it gives the isomorphism $R[x]/(x-a) \cong R$), so it gives us a closed subscheme $\text{Spec} R \to \mathbb{A}^1_R$. My question is then: is there a geometric way to imagine this closed embedding of scheme, particularly in regards to the process of evaluating of polynomial $f(x)$ at $a$ ?

Answer 1

Let $S:=\operatorname {Spec}R$ and let $\pi:\mathbb A^1_R\to S$ be the canonical projection, dual to the embedding of rings $R\to R[x]$.
The fiber of $\mathfrak p\in S$ under $\pi$ is the affine line $\pi^{-1}(\mathfrak p)=\mathbb A^1_{\kappa (\mathfrak p)}$ over the residue field $\kappa (\mathfrak p)=\operatorname {Frac}(R/\mathfrak p)$.
So the geometric picture is clear: the scheme $\mathbb A^1_R$ is obtained by gluing together the affine lines over $\kappa (\mathfrak p)$ where $\mathfrak p$ runs through $S$.
And what is the morphism $j_a:S\to \mathbb A^1_R$ dual to your evaluation $\epsilon_a:R[x]\to R$ ?
It is the map $j_a:S\to \mathbb A^1_R$ associating to the prime ideal $\mathfrak p\in S$ the prime ideal $j_a(\mathfrak p) \subset R[x]$ consisting of the set of polynomials $f\in R[x]$ such that $f(a)\in \frak p$.
The point $j_a(\mathfrak p)$ can also be seen as the rational point $\langle x-\bar a\rangle \in \mathbb A^1_{\kappa (\mathfrak p)}=\pi^{-1}(\mathfrak p)$,where $\bar a \in \kappa (\mathfrak p)$ is the image of $a$ under the natural map $R\to \kappa (\mathfrak p)=\operatorname {Frac}(R/\mathfrak p)$.