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This question is closely related to 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.

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    $\begingroup$ Even if $R$ is a finitely generated $k$-algebra, typically the algebra $S$ is not finitely generated. For instance, let $R$ be the polynomial ring in two variables, $R=k[t,u]$. This is a graded semigroup $k$-algebra $S=\bigoplus_{m,n\geq 0} R_{m,n}$ with $R_{m,n}$ equal to the $k$-span of $t^mu^n.$ Let $I$ be the monomial ideal $\langle t \rangle$ in $R$. Then the algebra $S$ is a monomial $k$-subalgebra of $R$ with nonzero weight space $S_{m,n}$ if and only if $m\geq 1$. The semigroup $\{(m,n) \in \mathbb{Z}_{\geq 0}^2| m\geq 1\}$ is not finitely generated. $\endgroup$
    – Jason Starr
    Commented Jan 14, 2018 at 21:00
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    $\begingroup$ In those cases where $S$ is a finitely generated $k$-algebra, then $\text{Spec}(S)$ is the cofiber coproduct (also called a "pushout", "coequalizer", "dilatation", etc.) for the pair of $k$-morphisms consisting of the closed immersion $i:\text{Spec}(R/I)\to \text{Spec}(R)$ and the constant $k$-morphism $j:\text{Spec}(R/I)\to \text{Spec}(k)$. $\endgroup$
    – Jason Starr
    Commented Jan 14, 2018 at 21:03
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    $\begingroup$ That's definitely a pushout and not a coequalizer. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2018 at 21:27
  • $\begingroup$ Of course it is a coequalizer: just take the fiber product of the quotient morphism with itself. $\endgroup$
    – Jason Starr
    Commented Jan 14, 2018 at 21:44

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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.

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