$\widetilde{S}=R\backslash \cup m$, where the union is taken over all $m\in m-\text{Spec}(R)\cap V(I)$

Question

For $I$ an $R$-ideal consider the multiplicatively closed set $S=1+I$. Show that

1. $\widetilde{S}=R\backslash \cup m$, where the union is taken over all $m\in m-\text{Spec}(R)\cap V(I)$
2. $m-\text{Spec}(S^{-1}R)$ and $m-\text{Spec}(R/I)$ are homeomorphic.

Proof of (1) : By definition, $\widetilde{S}=R\backslash \bigcup\limits_{\substack{\text{prime} \\ p\cap S=\emptyset}} p$. We need to show that the set of $p$ where $p$ is prime and $p\cap S=\emptyset$ is equivalent to the set of $m-\text{Spec}(R)\cap V(I)$. So, \begin{equation*} \begin{aligned} m-\text{Spec}(R)\cap V(I) & = \{p\in \text{Spec}(R) | p\supset I \text{ and $p$-maximal ideal} \} \\ & = \{p | p\supset I \text{ and $p$-maximal ideal and prime} \} \\ & = \{p | p\cap I=\emptyset \text{ and $p$ is prime} \} \\ & = \{p | p\cap (1+I) =\emptyset \text{ and $p$ is prime} \} \\ & = \{p | p\cap S =\emptyset \text{ and $p$ is prime} \} \end{aligned} \end{equation*} This is true since $S=1+I\subset 1+p$, then $S\subset p$ because $1\in S$ and $S$ is multiplicatively closed.

Proof of (b): We need to show that $\varphi : m-\text{Spec}(S^{-1}R)\to m-\text{Spec}(R/I)$ is a continuous bijection. Note that since $I$ is an $R$-ideal, $$ S^{-1}R =RS=R(1+I)=R+IR =R+I=R/I$$ Hence $m-\text{Spec}(S^{-1}R)=m-\text{Spec}(R/I)$. Furthermore $\varphi$ is a continuous bijection.

Are the proofs above correct?

Answer 1

A couple notes about a)

(1) Since maximal ideals are prime, it's redundant to say "$p$ is maximal and prime."

(2) $\{p | p\supset I \text{ and $p$ maximal} \} = \{p | p\cap I=\emptyset \text{ and $p$ is prime} \} $ makes no sense at all. If $p \supset I$ then clearly $p \cap I = I$. The remainder of the proof also does not make any sense.

The sets of primes you are looking at will usually not coincide. Consider for example any integral domain that is not a field and any proper ideal $I$. You have that $(0) \cap (I+1) = \emptyset$ and $(0)$ is prime but not maximal.

What you can do is approach the set equality directly :

Suppose $x \in p$ with $p \cap (I+1) = \emptyset$. Either $(x,I) = R$ or $(x,I)$ is contained in some maximal ideal $m$. In the first case we have $ax + i = 1$ for some $a \in R, i \in I$. I claim this is a contradiction (show this!). Thus $x$ is contained in some maximal ideal which contains $I$. The reverse inclusion is even simpler.

Your proof of b) also does not make any sense to me. What do those equalities mean???

Update Remember that a localization at a multiplicative set is equivalent to the localization at the saturation of that multiplicative set. So we can replace $S^{-1}R$ by $\tilde{S}^{-1}R$. We have a homeomorphism $\phi$ between the maximal ideals of $\tilde{S}^{-1}R$ and the maximal ideals of $R$ disjoint from $\tilde{S}$. We also have a homeomorphism from the maximal ideals of $R$ containing $I$ and the maximal ideals of $R/I$. I claim that a maximal ideal of $R$ contains $I$ iff it is disjoint from $\tilde{S}$.

One direction is obvious. For the other direction, suppose that $m$ is a maximal ideal disjoint from $\tilde{S}$. By the previous part, $m$ is contained in the union of maximal ideals which contains $I$. It follows that $m$ contains $I$ and we are done (proof provided in spoiler below, but a good exercise to work out).

Suppose that $m$ is a maximal ideal contained in the union of all maximal ideals that contain $I$. Clearly $m + I$ is also contained in the union of all maximal ideals that contain $I$. Hence $(m + I)/I$ is contained in all the maximal ideals of $R/I$, and it follows $(m+I)/I$ is contained in some maximal ideal of $R/I$. Therefore $m+I$ is contained in some maximal ideal containing $I$, and $m$ is too. Since $m$ is maximal, $m$ contains $I$.