When is the set of prime ideals that don't intersect a monoid zariski open?

Question

The $R$ be a commutative ring and $S$ be a multiplicative subset. When is the set $ \left\{ \mathfrak p\in \operatorname{Spec}R:\mathfrak p\cap S=\emptyset \right\}$ Zariski open in $\operatorname{Spec}R$?

Answer 1

You're looking at the set of prime ideals $P$ such that every $s \in S$ is nonzero as an element of $R/P$, or more geometrically such that every $s \in S$ does not vanish at $P$. This condition can be checked on a set of generators of $S$, and this set of prime ideals is the intersection of the corresponding sets for each generator of $S$; these are open by definition.

So the answer is yes if $S$ is finitely generated and otherwise there's no reason to expect the answer to be yes. For example, take $S$ generated by the primes congruent to $1 \bmod 4$ in $\mathbb{Z}$.

Answer 2

Here is an "elementary" proof that $S$ must be finitely generated*, supposing that $U_S = \{\mathfrak{p} \mid \mathfrak{p}\cap S = \emptyset\}$ is open.

For each $\mathfrak{q}\in U_S$, the openness of $U$, together with the fact that the topology on $\operatorname{Spec} R$ is generated by sets of the form $D(f) :=\{\mathfrak{p} \mid f\notin\mathfrak{p}\}$, allows us to choose some $f_\mathfrak{q}$ such that $D(f_\mathfrak{q}) \subset U_S$.

Since $U_S$ is quasi-compact (it is the image of the natural continuous map $\operatorname{Spec} R_S\to\operatorname{Spec} R$), we can find some finite set $\{f_1,\ldots , f_n\}$ such that $\{D(f_i)\}$ covers $U_S$. The set $\{f_i\}$ generates the unit ideal in $R_S$. Indeed, any prime ideal containing each $f_i$ cannot lie in $U_S$. Write $1 = \sum_i c_i f_i$ in $R_S$.

If $s\in S$, then $s$ is invertible in each $R_{f_i}$, so there exist $t_i \in R$, $K\in\mathbb{N}$ such that $t_i s = f_i^K$. By taking powers of $1 = \sum_i c_i f_i$, this shows that $s\cdot a = 1$ for some $a\in R_S$ whose denominator is a multiplicative combination of the denominators $c_1, \ldots , c_n$. In other words, $1/s = 1/s'$ (in $R_S$) for some $s'$ generated by the denominators of $c_1, \ldots , c_n$, i.e. $ss'' = s' s''$ for some $s''\in S$.

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* This is not quite accurate; we have only showed that $S$ is equivalent to a finitely generated set. This is an important restriction when $S$ is not assumed to be saturated.

For example, the set $S=\{1\} \cup \{X^i Y^j \mid i,j\geq 1 \}$ is a multiplicatively closed set in $k[X,Y]$ that is not finitely generated, but $U_S$, the set of prime ideals not containing $X$ or $Y$, is open in $\operatorname{Spec} k[X,Y]$. This happens because the saturation of $S$ is the multiplicative set $\{X^i Y^j \mid i,j\geq 0\}$, which is finitely generated.