Direct Image Sheaf at Generic Point of an Irreducible Component

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In summary, f#(OX) contains information about the regular functions on f-1(U) for any open set U in Y. If X is irreducible and projective and f is constant, then f#(OX) is a skyscraper sheaf with stalk k supported on the image point of f in Y, so it contains very little information about X. In general, if f:X-->Y is a projective morphism with every fiber connected, and Y is any normal variety, then f#(OX) = OY, so it essentially contains no information about X. In the case of a non-constant map of smooth curves, the stalk of f#(OX) is the direct sum of n copies of the
  • #1
LorenzoMath
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I do not know if this counts as number theory, but I came to this question while studying number theory, so I post this here.

Suppose X and Y are reduced schemes of finite type over a field k and f:X->Y is a morphism. Ox denotes the structural sheaf of X.

Question
What is a stalk of the direct image sheaf f[tex]_{*}[/tex]Ox at y in Y?

Since this question is too general, here is a specific question I encountered.

In addition to the assumptions above, let f be flat and y be a generic point of some irreducible component of Y. In this case, is f[tex]_{*}[/tex]Ox[tex]_{y}[/tex] isomorphic to the direct sum of Ox[tex]_{x}[/tex], where x runs through the preimage of y?

Thanks
 
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  • #2
CORRECTION

All the superscripts should be subscripts.
 
  • #3
LorenzoMath said:
All the superscripts should be subscripts.
To get the LaTeX to work, you should put entire expressions/formulae inside the [ itex ] ... [ /itex ] tags. When you put individual symbols in the tags, you get the formatting problems you've observed.
 
  • #4
LorenzoMath said:
Question
What is a stalk of the direct image sheaf f[tex]_{*}[/tex]Ox at y in Y?
Well, we can make a direct computation:

[tex](f_*\mathcal{O}_{X})_y =
\mathop{\mathrm{colim}}_{U \ni y} (f_*\mathcal{O}_{X})(U)
= \mathop{\mathrm{colim}}_{U \ni y} \mathcal{O}_{X}(f^{-1}(U))
= \mathop{\mathrm{colim}}_{f^{-1}(y) \subseteq V} \mathcal{O}_{X}(f^{-1}(U))
= \mathcal{O}_{X, f^{-1}(y)}[/tex]
 
  • #5
I guess I wasn't clear about my question. What I wanted to ask was if we can express [itex]\mathcal{O}_X,_{f^{-1}(y)}[/itex] in terms of [itex]\mathcal{O}_X,_x[/itex]. Here [itex]x[/itex] is a scheme-theoretic point(s). This shouldn't be possible in general, but when X and Y are reduced schemes of finite type, the morphism f is flat, and y is a generic point, I feel the following equation holds:
[itex]f_*\mathcal{O}_X,_y(=\mathcal{O}_X_{,f^{-1}(y)})=\oplus{\mathcal{O}_X_{,x}}[/itex], where the sum is taken over [itex]x\in{f^{-1}(y)}[/itex].
The main reason why I believe this is that f is an open morphism under these conditions.
 
  • #6
I knew what what I posted didn't answer your question, but I had hoped it would point you in a useful direction. However, I notice what I wrote is only sometimes true. That'll teach me to do algebraic geometry early in the morning. :frown: (But at least it's true in the cases I was imagining!)


Anyways, my first thoughts are to wonder what happens when the relative dimensions are different!

For example, if [itex]Y = \mathrm{spec}\, k[/itex], then [itex](f_* \mathcal{O_X})_y = \mathcal{O_X}(X)[/itex]. What happens when X is the affine line? Or the projective line? Your direct sum appears clearly wrong.

And the other direction; what if Y is a line, and X a closed point on the line, and y is any other point? Then [itex](f_* \mathcal{O_X})[/itex] is a skyscraper sheaf concentrated at X, and its stalk at y is the trivial module. I suppose this agrees with your hypothesis, though.



In the case where X and Y are the same dimension, then [itex]f^{-1}(y)[/itex] consists of the generic points of the irreducible components of X... so we can safely reduce the problem to the case where X is connected.

Can your question be completely reduced to the case where:
[tex]Y = \mathrm{spec}\, R[/tex]
[tex]X = \mathrm{spec}\, R[\alpha][/tex]
? (where [itex]\alpha[/itex] is algebraic over R. And flatness is imposed, if you like) Can you prove your conjecture in this case?

(And is the case of same dimension the one you are actually interested in?)
 
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  • #7
here are some old notes from a course i taught a few years ago.

Q: Exactly what information is contained in f# (lower star) (OX)?
Look at the definition. For any U in Y open, we have f#(OX)(U) = OX(f-1(U)) = regular functions on f-1(U). So the information in f#(OX) is related to knowing what types of sets in X have the form f-1(U).

Cases where f#(OX) contains as little information as possible.
If X is irreducible and projective and f is constant for example, then the only non empty set of form f-1(U) in X is X itself. In this case f#(OX) is a skyscraper sheaf with stalk k supported on the image point of f in Y. There is very little information here about X, but perhaps we do see that f is constant and that X is connected.

More generally, if Z is a projective variety, Y is any variety, and X = ZxY, and f:ZxY-->Y is the projection, then f-1(U) = ZxU, so an element of f#(OX)(U), i.e. a regular function on f-1(U), is determined by its restriction to {p}xU for any p in X, i.e. a regular function on U in Y. Thus in this case we have f#(OX) = OY. Consequently in this case f#(OX) recovers Y, but contains no information at all about X. [this projection seems to be the primordial example of a flat map.]

In general, if f:X-->Y is a projective morphism with every fiber connected, and Y is any normal variety, then f#(OX) = OY, so again f#(OX) contains essentially no information about X. Recall that if X is a projective variety then every morphism out of X is a projective morphism, and more generally a projective morphism X-->Y is one that factors via an isomorphism of X with a closed subvariety of P^nxY, followed by the projection P^nxY-->Y.

Suppose that f:X-->Y is any projective morphism. Then the fibers f-1(y) over points y in Y are all finite unions of projective varieties. Therefore for any open set U in Y containing the point y, the only regular functions in OX(f-1(U)) = f#(OX)(U) are constant on every connected component of the fiber f-1(y). Thus f#(OX) can contain little information about X and f other than at most the connected components of the fibers. We shall see below that it contains exactly this information.

I hope this is correct and helpful.

by the way, maybe you are thinking of a non constant map of smooth curves. then the stalk of lower star of the upstairs structure sheaf seems to be the direct sum of n copies of the stalk downstairs, where n = the degree of the map. this is not the actual number of preimages, but the algebraic number of preimages, where some preimages may be counted more than once.
 
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  • #8
see kempf, lemma 7.5.1.b, p. 92, or shafarevich, thm. 3, page 169.
 
  • #9
oh i didnt read your specific question where you took y a generic point on a component.

then maybe you want x a generic point of a component of the preimage of y?

im not very strong with generic points, i like geometric points better.

well i see now hurkyl is way ahead of me in considering these possibilities.
 
  • #10
mathwonk said:
well i see now hurkyl is way ahead of me in considering these possibilities.
But behind in others; my brain started in 'algebraic extension' land (e.g. curves over a field equipped with a projection down to the line, or the spectrum of a subring of a number field with its projection down to the spectrum of the integers), and is still sort of stuck there.
 
  • #11
I really appreciate your detailed suggestions and comments.

What I originally wanted to do was to prove the projection formula of intersection theory, namely [itex]f_*(a.f^*b)=f_*a.b.[/itex] for a flat [itex]f[/itex]. I explicitly wrote down the both sides using Serre's Tor formula. After a little formal manipulation of Tor, it turned out that I only need to compute the generic fiber of [itex]f_*\mathcal{O}_X.[/itex] I will post the proof here soon so that someone may kindly check where I went wrong.
 
  • #12
yes, i wanted x to be the generic points of irreducible components of f^(-1)(Z).
 

Related to Direct Image Sheaf at Generic Point of an Irreducible Component

1. What is a Direct Image Sheaf?

A Direct Image Sheaf is a mathematical concept used in algebraic geometry. It is a type of sheaf that describes how a function or morphism transforms the local data of one space into the local data of another space.

2. How is a Direct Image Sheaf defined?

A Direct Image Sheaf is defined as the sheaf of germs of sections of a sheaf on a topological space that is induced by a continuous map between two spaces.

3. What is the purpose of a Direct Image Sheaf?

The purpose of a Direct Image Sheaf is to provide a way to study the relationship between different spaces and their local data. It allows for the comparison and analysis of functions or morphisms between spaces.

4. How is a Direct Image Sheaf computed?

A Direct Image Sheaf is computed using the direct image functor, which takes a sheaf on one space and maps it to a sheaf on another space. This functor preserves the local structure and allows for the calculation of the sheaf on the target space.

5. What are the applications of a Direct Image Sheaf?

Direct Image Sheaves have various applications in mathematics, particularly in algebraic geometry and topology. They are used to study the properties of functions or mappings between spaces, and can also be applied to problems in differential equations and complex analysis.

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