# Spec(A), ring homomorphism

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• Feb 5th 2009, 05:19 PM
poincare4223
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 $V(\text{ker}(\phi))$?
• Feb 5th 2009, 06:03 PM
GaloisTheory1
this is by the ideal correspondence theorem, there is a bijective correspondence according to this theorem.
• Feb 5th 2009, 06:58 PM
aliceinwonderland
Quote:

Originally Posted by poincare4223
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.