6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 Jan 14 '18 at 21:00
  • 3
    $\begingroup$ 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 Jan 14 '18 at 21:03
  • 1
    $\begingroup$ That's definitely a pushout and not a coequalizer. $\endgroup$ – Qiaochu Yuan Jan 14 '18 at 21:27
  • $\begingroup$ Of course it is a coequalizer: just take the fiber product of the quotient morphism with itself. $\endgroup$ – Jason Starr Jan 14 '18 at 21:44
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.