# Multiplicative closed subsets $S\subset \mathbb{Z}$ such that $\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$ is open in $\mathrm{Spec}(\mathbb Z)$.

I want to find all multiplicatively closed subsets $$S\subset \mathbb{Z}$$ such that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ is open in $$\mathrm{Spec}(\mathbb{Z})$$, where $$\varphi(\mathfrak{q}) = f^{-1}(\mathfrak{q})$$, where $$f$$ is the natural homomorphism between $$\mathbb{Z}$$ and $$S^{-1}\mathbb{Z}$$.

My attempt: Notice that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \{\mathfrak{p}\in \mathrm{Spec}(\mathbb{Z})\mid \mathfrak{p}\cap S = \emptyset\}$$. The case when $$0\in S$$ is clear, since then $$S^{-1}\mathbb{Z} = \{0\}$$ and thus $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \varphi(\emptyset) = \emptyset$$ which is clearly open.

Now suppose that $$0\notin S$$, notice that since $$\mathbb{Z}$$ is a PID every closed set is of the form $$V((n))$$ for some $$n\in \mathbb{Z}$$. Thus for $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ to be open we need that $$X_{S} := \mathrm{Spec}(\mathbb{Z})\setminus\varphi(\mathrm{Spec}(S^{-1}(\mathbb{Z})) = V((n))$$ for some $$n\in\mathbb{Z}$$.

If $$n=0$$, then we have $$V((0)) = \mathrm{Spec}(\mathbb{Z})$$. And this is the case if and only if $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \emptyset$$. Thus we need $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$. Now my claim is that the only possibility is that $$S = \mathbb{Z}\setminus\{0\}$$, but I have problems proving this.

If $$n\neq 0$$, then $$n$$ has finitely many prime divisors, say $$p_{1},...,p_{r}$$. Then we have $$V((n)) = \bigcup_{i=1}^{r}\{(p_{i})\}$$. Then we have $$X_{S} = \bigcup_{i=1}^{r}\{(p_{i})\}$$ if and only if $$(p_{i})\cap S \neq \emptyset$$ and $$\forall \pi$$ prime in $$\mathbb{Z}$$ with $$\pi\neq p_{i}$$ $$\forall i$$ we have $$(\pi)\cap S = \emptyset$$. From here I also have problems completing this case.

New attempt: Using the cofiniteness of the topology on $$\mathrm{Spec}(\mathbb{Z})$$ we have, assuming that $$\varphi(\mathrm{Spec}(\mathbb{Z}))$$ is open that:

$$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \emptyset$$, which implies that $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$.

$$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \mathrm{Spec}(\mathbb{Z})$$, which implies that $$S=\{1\}$$, or $$S=\{-1,1\}$$.

Or we have a finite set of primes $$\{p_{1},...,p_{r}\}$$ such that $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z})) = \{\mathfrak{p}\in\mathrm{Spec}(\mathbb{Z})\rvert p_{1},...,p_{r}\notin \mathfrak{p}\}$$. Rewriting this last equality gives $$\{\mathfrak{p}\in\mathrm{Spec}(\mathbb{Z})\rvert \mathfrak{p}\cap S=\emptyset\} = \{(\pi)\rvert \pi\in\mathbb{Z} \ \text{prime such that} \ \pi\neq p_{i} \ \forall i\in\{1,...,r\}\}\cup\{(0)\}$$. This implies that $$S\subset \cup_{i=1}^{r}\{(p_{i})\}\cup \{-1,1\}$$ and $$S\cap (\pi)=\emptyset = S\cap(0)$$ for every $$\pi\in\mathbb{Z}$$ prime with $$\pi\neq p_{i}$$ for all $$i$$.

In all the cases, assuming that you have a multiplicatively closed $$S$$ satisfying one of the conditions above it is easy to show that in those cases $$\varphi(\mathrm{Spec}(S^{-1}\mathbb{Z}))$$ is indeed open.

Only question left: Does the condition $$\mathrm{Spec}(S^{-1}\mathbb{Z}) = \emptyset$$ imply something stronger, like $$S = \{0\}$$, or $$S = \mathbb{Z}\backslash\{0\}$$?

• For your "only question left": $\operatorname{Spec} R =\emptyset$ is equivalent to $R=0$. It is easy to check that if we localize at a multiplicative subset not containing a zero divisor then the original ring embeds in to this localization and if we localize at a multiplicative set containing a zero divisor, then we collapse our ring to zero. Thus $\operatorname{Spec} S^{-1}\Bbb Z=\emptyset$ is equivalent to $0\in S$. Oct 21, 2019 at 22:29

Note that the topology on $$Spec(\mathbb{Z})$$ is cofinite.
Let $$S$$ be a multiplicative closed sub set of $$\mathbb{Z}$$. Then $$Spec(S^{-1}\mathbb{Z})$$ is an open subset of $$Spec(\mathbb{Z})$$ if and only if there exists a subset $$A=\{p_1,...,p_n\}$$ of prime number ( $$A$$ may be empty) such that $$S\subseteq \mathbb{Z}-\cup_i\langle p_i\rangle$$ and for any other subset $$\{q_j\}$$ of prime numbers we have $$S\not\subseteq \mathbb{Z}-\cup_j\langle q_j\rangle$$.
Proof: If $$Spec(S^{-1}\mathbb{Z})$$ is an open subset of $$Spec(\mathbb{Z})$$, then either $$Spec(S^{-1}\mathbb{Z})=\{\}$$ or there are prime number $$p_1,...,p_n$$ such that $$Spec(S^{-1}\mathbb{Z})=\{P\in Spec(\mathbb{Z})\mid S\cap P=\{\}\}=\{P\in Spec(\mathbb{Z})\mid p_1,...,p_n\not\in P\}$$. Thus either $$A$$ is empty or $$A=\{p_1,...,p_n\}$$ such that $$S\subseteq \mathbb{Z}-\cup_i\langle p_i\rangle$$ and for any other subset $$\{q_j\}$$ of prime numbers we have $$S\not\subseteq \mathbb{Z}-\cup_j\langle q_j\rangle$$. The converse is trivial.
• If you have a $P\in\mathrm{Spec}(\mathbb{Z})$ with $S\cap P=\emptyset$ and $p_{1},...,p_{n}\notin P$ one could have that $p_{i}\in S$ for some $i$ right? So this means that we might have $S\not\subset \mathbb{Z}-\cup_{i=1}^{n}(p_{i})$.