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Thread: Confused about Algebraic Geometry Theorem

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

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

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

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

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

    (where $\displaystyle V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \}$ )
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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 $\displaystyle \phi: X \rightarrow Y$ be a regular map between algebraic sets.
    This induces a $\displaystyle k$-algebra homomorphism $\displaystyle \phi^*: k[Y] \rightarrow k[X]$ given by $\displaystyle f \mapsto f\phi$

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

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

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

    (where $\displaystyle V_Y(I) = \{ y \in Y : f(y) = 0 \forall f \in I \}$ )
    Because $\displaystyle V_Y(I)$ is closed in the Zariski topology, and so $\displaystyle \phi(X)$ 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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