1
$\begingroup$

I am reading a book titled "Lectures on An Introduction to Grothendieck's Theory of the Fundamental Group" by J.P. Murre. I am in the chapter 4 titled "Fundamental groups". Here he fixes a base locally noetherian scheme $S$ and defines a category $\mathscr{C}$ taking the objects to be locally noetherian schemes with finite etale maps to $S$. Morphisms with in the objects are finite etale maps between them.

At the beginning of this chapter he is stating the basic properties of this category, like existence of fibre product and disjoint union. I am reading the proof of the existence of the quotient by a finite subgroup $\mathfrak{g}$ of the automorphisms of an object $X$. (I am unable to write proper notation for $\mathfrak{g}$, because I do not know what it is. What is the notation?) He states that the problem is enough to be solved locally and goes on to define a finite subgroup of automorphisms of $\operatorname{Spec}(A)$ corresponding to $\mathfrak{g}$. I am unable to understand what this means. It is not even clear to me why the problem is enough to be solved for affine open subschemes. I have attached the snapshot below. Snapshot

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ The notation is \mathfrak{} (Fraktur "g"). I hope you don't mind me fixing it for you. $\endgroup$
    – M.G.
    Commented Aug 3, 2017 at 18:10
  • $\begingroup$ I always forget \mathfrak{} $\endgroup$
    – Grobber
    Commented Aug 3, 2017 at 18:14
  • $\begingroup$ Evidently $G$ is an isomorphic copy of $\mathfrak g$ acting on $A$. More or less by definition the category of affine $\operatorname{Spec} B$-schemes is opposite to that of $B$-algebras, so naming a subgroup $\mathfrak g$ of the automorphism group of $X=\operatorname{Spec} A$ over $\operatorname{Spec} B$ is the same as naming a(n isomorphic) subgroup $G$ of the automorphism group of the $B$-algebra $A$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 3, 2017 at 18:51
  • $\begingroup$ But why does $\mathfrak{g}$ act on $\operatorname{Spec} A$? $\endgroup$
    – Grobber
    Commented Aug 3, 2017 at 18:58
  • $\begingroup$ By definition, $\mathfrak g$ is a finite subgroup of the automorphism group of $X$, so it acts by automorphisms on $X$, and $X$ is in the case considered $\operatorname{Spec}A$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 3, 2017 at 22:34

1 Answer 1

Reset to default
4
$\begingroup$

I think you have described the morphisms in the category wrong.

Morphisms should be finite etale maps forming a commutative triangle with the structure maps to $S$.

If this is so, then we can take $\operatorname{Spec} B$ to be an affine open in $S$, and then $\operatorname{Spec} A$ to be the fiber product of $X$ with $\operatorname{Spec} B$ over $\operatorname{Spec} A$. The existence of a map $\operatorname{Spec} A \to \operatorname{Spec} A$ induced by a map $X \to X$ commuting with the projection to $S$ is a simple diagram chase with fiber products. This is the map Milne is discussing.

Improve this answer
$\endgroup$

You must log in to answer this question.

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