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Thread: Spec(A), ring homomorphism

  1. #1
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    Spec(A), ring homomorphism

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

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

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

    Here is one way to do it:

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

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

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

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

    Here is one way to do it:

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

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

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

    By lemma 1 and an ideal correspondence theorem, $\displaystyle (\phi^*)^{-1}(V(ker (\phi)) $ is $\displaystyle 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), $\displaystyle \phi^*$ is a continuous map in the Zariski topology.
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