This question is closely related to [What is the geometric object corresponding to a subalgebra in a polynomial ring](https://mathoverflow.net/questions/101176/what-is-the-geometric-object-corresponding-to-a-subalgebra-in-a-polynomial-ring).
There, it is asked, given a subalgebra of an algebra $S \subset R$ over a field $k$ what is the corresponding geometric object, in the sense of algebraic geometry.
The answer is that there are varieties $X,Y$ associated to $R,S$, respectively, along with a dense morphism $X \to Y$.
Here the term "variety" is a bit looser than often used.


My question is the same, but for a specific type of subalgebra. Given a $k$-algebra $R$ and an ideal $I \subset R$, let $S = k + I \subset R$ be the $k$-subalgebra generated by the ideal $I$. What is the geometric interpretation of the variety $\mathrm{Spec}(S)$ and the morphism $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$?


I am happy to assume that $R$ is a finite type $k$-algebra, and even that it is reduced. I am most concerned about the case when $R = k[x\_1, ..., x\_n]$ is just a polynomial ring.



Geometric object corresponding to subalgebra generated by an ideal?
The image of the composition $S \to R \to R/I$ is equal to the constants $k \subset R/I$. This indicates that the morphism of the zero locus $V(I) = Z \to \mathrm{Spec}(R) \to \mathrm{Spec}(S)$ factors through the structure morphism $Z \to \mathrm{Spec}(k)$, and so the image must be a point. Therefore the morphism corresponding to the inclusion of subalgebras must be one in which the distinguished closed subvariety $Z$ is collapsed to a point.


Because $S \to R$ is injective, the morphism $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ will be dense. That the morphism is dense is actually equivalent to the kernel being contained in the nilradical. So I believe one should be able to show without much difficulty something like $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ is universal with respect to being a dense morphism to a reduced scheme, sending $Z$ to a point.


