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Direct Image Sheaf

  1. Mar 7, 2008 #1
    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
     
  2. jcsd
  3. Mar 7, 2008 #2
    CORRECTION

    All the superscripts should be subscripts.
     
  4. Mar 7, 2008 #3

    Hurkyl

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    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.
     
  5. Mar 7, 2008 #4

    Hurkyl

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    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]
     
  6. Mar 7, 2008 #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.
     
  7. Mar 7, 2008 #6

    Hurkyl

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    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?)
     
    Last edited: Mar 7, 2008
  8. Mar 7, 2008 #7

    mathwonk

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    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.
     
    Last edited: Mar 7, 2008
  9. Mar 7, 2008 #8

    mathwonk

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    see kempf, lemma 7.5.1.b, p. 92, or shafarevich, thm. 3, page 169.
     
  10. Mar 8, 2008 #9

    mathwonk

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    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.
     
  11. Mar 8, 2008 #10

    Hurkyl

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    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.
     
  12. Mar 11, 2008 #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.
     
  13. Mar 11, 2008 #12
    yes, i wanted x to be the generic points of irreducible components of f^(-1)(Z).
     
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