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Math Help - Confused about Algebraic Geometry Theorem

  1. #1
    Senior Member slevvio's Avatar
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    Confused about Algebraic Geometry Theorem

    Let \phi: X \rightarrow Y be a regular map between algebraic sets.
    This induces a k-algebra homomorphism \phi^*: k[Y] \rightarrow k[X] given by f \mapsto f\phi

    I have a theorem here that is really confusing me, and help would be appreciated.

    It says " \phi^* is injective \iff \phi(X) \subseteq Y is dense "

    While proving this, the author takes density to mean \phi(X) \subseteq V_Y(I) \implies V_Y(I) = Y. What has this got to do with density in the usual sense?

    (where V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \} )
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Confused about Algebraic Geometry Theorem

    Quote Originally Posted by slevvio View Post
    Let \phi: X \rightarrow Y be a regular map between algebraic sets.
    This induces a k-algebra homomorphism \phi^*: k[Y] \rightarrow k[X] given by f \mapsto f\phi

    I have a theorem here that is really confusing me, and help would be appreciated.

    It says " \phi^* is injective \iff \phi(X) \subseteq Y is dense "

    While proving this, the author takes density to mean \phi(X) \subseteq V_Y(I) \implies V_Y(I) = Y. What has this got to do with density in the usual sense?

    (where V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \} )
    Because V_Y(I) is closed in the Zariski topology, and so \phi(X) being dense means the only closed set which contains it is the full space.
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  3. #3
    Senior Member slevvio's Avatar
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    Re: Confused about Algebraic Geometry Theorem

    thanks drexel.
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