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Math Help - Spec(A), ring homomorphism

  1. #1
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    Spec(A), ring homomorphism

    Let \phi: A \rightarrow B be a ring homomorphism. Let X=\text{Spec}(A) and Y= \text{Spec}(B). If q \in Y, then \phi^{-1} is a prime ideal of A, i.e., a point of X. Hence \phi induces a map \phi^*:Y \rightarrow X. Show that:

    If \phi is surjective, then \phi^* is a homeomorphism of Y onto the closed subset V(\text{ker}(\phi)) of .

    [ V(E) denotes the set of all prime ideals of A which contain E].

    Here is one way to do it:

    By the "Ideal Correspondence Theorem," we know that
    {\color{red} \phi^*(Y)=V(\text{ker}(\phi))} and {\color{red}\phi^*} induces a bijective map from {\color{red}Y} to {\color{red}V(\text{ker}(\phi))}. Then I need to show \phi^{*-1} is continuous and \phi^*is continuous to show it is a homeomorphism.

    I don't understand the sentence in red. How do we know \phi^* induces a bijective map from Y to V(\text{ker}(\phi))?
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  2. #2
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    this is by the ideal correspondence theorem, there is a bijective correspondence according to this theorem.
    Last edited by GaloisTheory1; February 5th 2009 at 09:55 PM.
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  3. #3
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    Quote Originally Posted by poincare4223 View Post
    Let \phi: A \rightarrow B be a ring homomorphism. Let X=\text{Spec}(A) and Y= \text{Spec}(B). If q \in Y, then \phi^{-1} is a prime ideal of A, i.e., a point of X. Hence \phi induces a map \phi^*:Y \rightarrow X. Show that:

    If \phi is surjective, then \phi^* is a homeomorphism of Y onto the closed subset V(\text{ker}(\phi)) of .

    [ V(E) denotes the set of all prime ideals of A which contain E].

    Here is one way to do it:

    By the "Ideal Correspondence Theorem," we know that {\color{red} \phi^*(Y)=V(\text{ker}(\phi))} and {\color{red}\phi^*} induces a bijective map from {\color{red}Y} to {\color{red}V(\text{ker}(\phi))}. Then I need to show \phi^{*-1} is continuous and \phi^*is continuous to show it is a homeomorphism.

    I don't understand the sentence in red. How do we know \phi^* induces a bijective map from Y to V(\text{ker}(\phi))?
    My attempt is as follows:

    Lemma 1. If \phi: A \rightarrow B be a ring homomorphism mapping 1_{A} to 1_{B} and P a prime ideal in B. This induces a map \phi^*:Spec(B) \rightarrow Spec(A) such that \phi^*(P)=\phi^{-1}(P).

    By lemma 1 and an ideal correspondence theorem, (\phi^*)^{-1}(V(ker (\phi)) is V(\phi(ker (\phi))B) . The latter corresponds the all prime ideals in B, which is spec(B).

    Since the inverse image of a closed subset in Spec(A) is a closed subset in Spec(B), \phi^* is a continuous map in the Zariski topology.
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