5
$\begingroup$

Let $k$ be an algebraically closed field, $A$ a finitely generated $k$-algebra. Let $x,y$ be two distinct closed points of $\mathrm{Spec}(A)$. Is there an affine $k$-scheme of finite type obtained from $\mathrm{Spec}(A)$ by "gluing $x$ and $y$"? More precisely: let $\mathfrak{m}_x$ and $\mathfrak{m}_y$ be the maximal ideals of $A$ corresponding to $x$ and $y$, and $I=\mathfrak{m}_x\cap \mathfrak{m}_y$. Define $B$ to be the $k$-subalgebra of $A$ generated by $k$ and the elements of $I$. Is $B$ a finitely generated algebra over $k$?

In Serre's book "Algebraic groups and class fields" he carries out the construction for smooth curves (i.e. $A$ normal and of dimension 1) but I don't see how to generalize this to higher dimension.

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ For more general varieties, I guess this follows from the affine case if any two points lie in an affine subset (e.g., quasi-projective varieties). However otherwise it might fail; possibly some people here know counterexamples (or even it might hold that if two points belong to a common affine open subset then gluing the points does not yield a variety?). $\endgroup$
    – YCor
    Commented Oct 18, 2019 at 20:45
  • 3
    $\begingroup$ A fairly general procedure of gluing schemes along subschemes is studied in a paper of Karl Schwede: math.utah.edu/~schwede/Papers/SchemeWithoutPoints.pdf $\endgroup$
    – Sándor Kovács
    Commented Oct 18, 2019 at 21:12
  • 1
    $\begingroup$ Can you explain why Spec$B$ is obtained by gluing $x$ and $y$ on Spec$A$? I don't understand... $\endgroup$
    – user141691
    Commented Nov 13, 2019 at 14:04

1 Answer 1

Reset to default
5
$\begingroup$

$B$ is finitely generated. First notice that $B\subset A$ is an integral extension, since if $f\in A$, then $(f-f(x))(f-f(y))\in I$, giving you an integral equation. Then, it is easy to find a finitely generated algebra $C\subset B$ such that $C\subset A$ is integral. Then, $A$ is a finite $C$-module and thus so is $B$. Then, $B$ is finitely generated.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Awesome! Thank you $\endgroup$
    – Chris
    Commented Oct 18, 2019 at 20:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .