Let $A$ be a commutative ring. $I$ be an ideal of $A$, $S$ be a multiplicative closed subset. We know that :
- there is 1-1 correspondence between the prime ideals $\mathfrak{p}\in Spec A$ containing $I$ and prime ideals of $A/I$.
- there is 1-1 correspondence between prime ideals $\mathfrak{p}\in Spec A$ that does not meet $S$ and prime ideals of $S^{-1}A$.
My questions are:
-
How can I show that $Spec(A/I)$ and the set of prime ideals $\mathfrak{p}\in Spec A$ containing $I$ are homeomorphic?
-
The same question for $Spec(S^{-1}A)$ and the set of prime ideals $\mathfrak{p}\in Spec A$ that does not meet $S$ ?
I can show that if $f : A\longrightarrow B$ is a ring homomorphism then $\bar{f}: Spec B\rightarrow Spec A$ is continuous. Can I apply it in this case ?
An exercise from the book of Matsumura even claim that $Spec(S^{-1}A)$ is neither open nor closed subset of $SpecA$. I can not find the example for this. Could anyone help me? Thanks.
Best Answer
The homomorphisms $A \to A/I$ and $A \to S^{-1}A$ induce continuous maps of spectra in the other direction. These are injective and you know the images. You want them to be homeomorphisms onto their images.
There's probably a better way to think about this, but given an ideal of $A/I$ defining a closed set there, what do you think the associated ideal of $A$ should be?
For an example, remember that the most common multiplicative sets are $S = \{1, f, f^2, \dots\}$ and $S = A - \mathfrak{p}$. The first kind always produces an open subset, but the second need not. Try looking in $\operatorname{Spec} \mathbb{Z}$.