Quotients by finite equivalence relations (Q2918462)
From MaRDI portal
| This is the item page for this Wikibase entity, intended for internal use and editing purposes. Please use this page instead for the normal view: Quotients by finite equivalence relations |
scientific article; zbMATH DE number 6092081
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Quotients by finite equivalence relations |
scientific article; zbMATH DE number 6092081 |
Statements
6 October 2012
0 references
geometric quotient
0 references
categorical quotient
0 references
equivalence relation
0 references
math.AG
0 references
Quotients by finite equivalence relations (English)
0 references
This article is concerned with the existence of a geometric quotient \(X/R\) of a scheme \(X\) by a finite equivalence relation \(R\rightrightarrows X\). A typical example is the relation given by the normalisation morphism \(f:X\to Y\) of a non-normal scheme \(Y\).NEWLINENEWLINEIn general, even if \(X\) is Noetherian and a categorical quotient \((X/R)^{\mathrm{cat}}\) exists, it may be non-Noetherian and the geometric quotient may either not exist or be different from \((X/R)^{cat}\). The author gives many examples in section 2 of this type of phenomenon. As one example, let \(f:X\to Y\) be a finite surjective morphism and \(R:=(X\times_YX)^{\mathrm{red}}\subset X\times X\) the corresponding set-theoretic equivalence relation (note that \(X\times_Y X\) may fail to be reduced). Then \(X/R\) exists and \(X/R\to(X/R)^{\mathrm{cat}}\) is a finite and universal homeomorphism, but need not be an isomorphism. In particular, if \(X\) is the normalisation of \(Y\), then \(X/R\) is the weak normalisation of \(Y\).NEWLINENEWLINESection 3 contains some elementary results on the existence of geometric quotients. Section 4 consists of an inductive plan for constructing \(X/R\), which works well in finite characteristic; in characteristic \(0\), on the other hand, a certain inductive assumption proves quite restrictive. In section 5, the author proves that, if \(X\) is an algebraic space which is essentially of finite type over a Noetherian \({\mathbb F}_p\)-scheme \(S\) and \(R\rightrightarrows X\) is a finite set-theoretic equivalence relation, then \(X/R\) exists (Theorem 6). This result remains valid for algebraic spaces for which the Frobenius map \(F^q:X\to X^{(q)}\) is finite. Section 6 contains a discussion of gluing.NEWLINENEWLINEIn the appendix, Claudiu Raicu constructs some interesting examples. The following question remains open in characteristic \(0\): let \(R\subset X\times X\) be a scheme-theoretic equivalence relation such that the projections \(R\rightrightarrows X\) are finite; does \(X/R\) exist?NEWLINENEWLINEFor the entire collection see [Zbl 1242.14002].
0 references