Rings and locales Let $A$ be a ring. Given a monoid (=multiplicative system)
$S \subseteq A$ one can define the localization of $A$ at $S$
as the $A$-algebra
$$j_S \colon A \to A[S^{-1}]$$
which is universal among the ones in which the elements of $S$
are invertible.
I'm trying to understand the construction of the spectrum of $A$
going through what would be the locale associated to it,
but I'm not confident about those ideas.
The actual question is the following:

Given monoids $(S_i)_{i \in I}$ one has inf and sup given by
    $$
  \bigwedge S_i = \bigcap S_i, \qquad
  \bigvee S_i = \bigcup_{n \in \mathbb{N}}
    \{x_{i_1} \cdots x_{i_n} \mid x_{i_k} \in S_{i_k}\}.
  $$
Is is true that the (co)distributive law
    $$
  \tag{1}
  \left(\bigwedge S_i\right) \vee T
  = \bigwedge (S_i \vee T)
  $$
  holds?

The endgoal would be to show that the (full subcategory of) localizations
form a locale and points correspond to prime ideals.
Working with localizations is equivalent to work with monoids
and inclusions.

I'm unable to prove the equality (1).
Here is the step where my attempt fails:
if $z \in \bigwedge (S_i \vee T)$ then $z = x_i \, y_i$
for $x_i \in S_i$ and $y_i \in T$; it is not clear to me
how to show that $z = x y$ for some $x \in \bigwedge S_i$ and $y \in T$.

I have the suspect that this construction of the (co)locale is wrong,
so I would appreciate if you were to point to the correct one.
 A: This is false.  For instance, let $A=\mathbb{Z}[x]/(2x)$, let $S_n$ be generated by $x^n$ and let $T$ be generated by $2$.  Then it is easy to see that $\bigwedge S_n=\{1\}$ so $0\in S_n\vee T$ for all $n$ but $0\not\in (\bigwedge S_n)\vee T$.
Note that you should not expect this to be the correct construction: localizations of $A$ are not the same thing as submonoids, since different submonoids can give the same localization,.  Moreover, open subsets of $\operatorname{Spec} A$ are not the same as localizations of $A$: only localizations with respect to finitely generated (or equivalently, singly generated) submonoids give open sets, and a general open set is a union of open sets that come from localizations.  Another reason to be suspicious of your construction is that it uses only the multiplicative structure of $A$, not the additive structure! (And $\operatorname{Spec} A$ as a locale absolutely does depend on the additive structure; for instance if $k$ is a field then $k[x]$ and $k[x,y]$ are multiplicatively isomorphic but their Specs are different.)
From the perspective of algebraic geometry, there is a very simple and natural way to describe $\operatorname{Spec} A$ as a locale: just take the poset of radical ideals in $A$ (by the usual correspondence, radical ideals correspond to closed sets of $\operatorname{Spec} A$ by an inclusion-reversing bijection, and thus to open sets by an inclusion-perserving bijection).
Here's one way to see radical ideals arising naturally from the perspective of localizations.   It is enough to consider only localizations with respect to a single element, so instead of arbitrary submonoids of $A$ we will just consider elements of $A$, representing the distinguished open subsets of $\operatorname{Spec}A$.  We will represent general open subsets as the collection of all the distinguished open subsets they contain, so our locale $L$ will consist of certain subsets of $A$.  That is, we think of an element $S\in L$ as representing the open set $U=\bigcup_{a\in S} D(a)\subseteq\operatorname{Spec} A$, and conversely $S=\{a\in A:D(a)\subseteq U\}$).  What properties do these subsets need to satisfy?  Well, if $D(a)$ is covered by the sets $D(b)$ for $b\in S$, then $a$ should be in $S$.  In other words, if the elements of $S$ generate the unit ideal in the localization $A[a^{-1}]$, then $a$ should be in $S$.  Or equivalently, if some power of $a$ is in the ideal of $A$ generated by $S$, then $a\in S$.  Now this condition looks familiar: it just says that $S$ must be a radical ideal.
To be fair, this may feel like cheating in that it takes for granted that "$D(a)$ is covered by elements $D(b)$" should mean "the elements $b$ generate the unit ideal in $A[a^{-1}]$".  Here is another construction that is very slick and may feel more natural.  Take the free frame on $A$ modulo relations that say $1_A$ is $1$, $0_A$ is $0$, $a\wedge b=ab$ for $a,b\in A$, and $a\vee b\geq a+b$.  Thinking of $a\in A$ as representing the distinguished open set $D(a)$, this means we are just formally making a frame out of these distinguished open sets, imposing the relation that $D(1)$ is the whole space, $D(0)$ is empty, $D(ab)=D(a)\wedge D(b)$, and $D(a)\vee D(b)$ contains $D(a+b)$.  Remarkably, it turns out that this frame is naturally isomorphic to the frame of radical ideals in $A$, by mapping $a\in A$ to the radical ideal generated by $a$.
For more discussion of this construction and other properties of the spectrum from a locale-theoretic perspective, you may be interested in section V.3 of Peter Johnstone's book Stone spaces.
