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 $S = \mathbb{Z}\backslash\{0\}$?

  • 1
    $\begingroup$ 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 $0\in S$. $\endgroup$
    – KReiser
    Oct 21, 2019 at 22:29

1 Answer 1


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.

  • $\begingroup$ 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 $S\not\subset \mathbb{Z}-\cup_{i=1}^{n}(p_{i})$. $\endgroup$
    – user707512
    Oct 20, 2019 at 19:38

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