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Algebraic Geometry

  1. Mar 4, 2007 #1
    I'm having trouble understanding the importance of dominance with regards rational maps.

    I have the definition that a rational map phi (from some affine variety V to an affine variety W) is dominant if the image of V is not contained in a proper subvariety of W.

    In my lecture notes there is some remark about how if we compose phi with another such map psi going in the reverse direction W to V (to get a map V to V) then it may not make sense unless phi is dominant.

    Whilst reading around the subject I have noticed that the notion of dominance crops up in a lot of theorem hypotheses so I really feel I ought to get a grip on it.

    My problem is this: Rational maps are not generally functions anyway since they may have undefined points. OK. Now, in the remark when it talks about the composition 'not making sense', one would usually assume by the phrase 'not making sense' that it has undefined points - but why are we suddenly bothered about this now when we weren't before?

    Any help would be greatly appreciated.
  2. jcsd
  3. Mar 4, 2007 #2


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    A rational map must be defined on an dense subset of the domain, right? Well, what if the image of f lies in the complement of the set where g is defined?

    Explicit examples are fun: let's use R^2.

    Let f be the map (x, y) --> (1/x, 0)
    Let g be the map (x, y) --> (0, 1/y)

    What is their composite? (Either way)

    Also, I suspect that dominant maps are epimorphisms (f is epimorphic iff g o f = h o f implies g=h), and so play a similar role that surjective maps would in topology.
    Last edited: Mar 4, 2007
  4. Mar 4, 2007 #3

    matt grime

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    You should think of dominance as being 'essentially surjective'. It is the definition that makes rational maps composable. As you say there might be a set of points on which they are not defined - that is a subvariety of dimension 0. Each composition of dominant maps, adds at most finitely many more points that are bad. And a finite sum of finitely many things is still finite, so a dimension 0 subvariety. If you dropped the dominant assumption you can get a situation as in Hall's post above.
    Last edited: Mar 4, 2007
  5. Mar 4, 2007 #4
    I didn't know this but it certainly makes sense now you say it. Topology wasn't a prerequisite for this course so the lecturer has avoided topological notions. (I have done toplogy though.) I know about the Zariski topology so it seems that dense here means a subset A of our variety for which there is no closed set (i.e. a subvariety) which both contains A and is contained in V. So nicely enough NOT being dominant means that our image of V is not a dense subset of W. It's starting to make sense now.

    Of course the image of either composition is the empty set.
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