The following are some observations with a very broad question, so maybe this is a reference-request.
Let $f \colon R \to S$ be morphism of commutative rings.
It's clear that $f(R^\times) \subset S^\times$, i.e. $f$ maps units to units.
This inclusion is usually proper, and if $f(r)$ is a unit it does not imply that $r$ is a unit (example for both statements: $R = \mathbb{Z}$, $S = \mathbb{Q}$).
Now, the set $f^\times := f^{-1}(S^\times)$ is multiplicatively closed and contains $1$, so we can localize $R$ at $f^\times$.
What can we say about the localization $(f^\times)^{-1} R$?
I think the following should be valid examples:
-
$(id_R^\times)^{-1} R = R$
-
if $R$ is an integral domain, $S$ its field of fractions, and $f$ the inclusion, then $(f^\times)^{-1} R = S$
More generally, I think the following could be true (but I did not check):
If $D \subset R$ is multiplicatively closed and $f \colon R \to D^{-1} R$ is the localization map, then $(f^\times)^{-1} R = D^{-1} R$.
Note that in all of these examples, $S$ was secretely a localization with localization map $f$.
I was also wondering if maybe $S = (f^\times)^{-1} R$ could be true in general?
Edit 1: $S = (f^\times)^{-1} R$ is probably not true in general, since the localization is the smallest ring such that the elements of $f^{-1}(S^\times)$ become units, and I would think that $S$ is often a bigger ring.
Edit 2:
Note also that we have adjoint functors
$$
S \otimes_R – \dashv f^* \colon R\text{-mod} \to S\text{-mod}
$$
and
$$
(f^\times)^{-1} R \otimes_R – \dashv \tilde{f}^*
\colon R\text{-mod} \to (f^\times)^{-1} R\text{-mod}
$$
where $\tilde{f} \colon R \to (f^\times)^{-1} R$ is the localization map.
I don't know what to do with those, though.
Best Answer
Here is a geometric point of view.
Let $R$ be a commutative ring and $\operatorname{Spec}(R)$ be its spectrum.
A localization of $R$ is an $R$-algebra of the form $U^{-1}R$ for some multiplicative subset $U \subseteq R$.
A germ of $\operatorname{Spec}(R)$ is a set of the form $\{F \subseteq \operatorname{Spec}(R): F \supseteq E, F\textrm{ open}\}$ for some subset $E \subseteq \operatorname{Spec}(R)$.
It turns out that there is a dictionary between localizations of $R$ and germs of $\operatorname{Spec}(R)$, which I will explain below.
Let $U$ be a multiplicative subset of $R$. Its localization is defined as $U^{-1}R$.
We define $\overline U = \{a \in R: \exists b \in R, ab \in U\}$, which is again a multiplicative subset.
One can show that $\overline U$ is the maximal multiplicative subset which has the same localization as $U$, in the following sense: consider all multiplicative subsets $V \subseteq R$ such that $V \supseteq U$ and the canonical morphism $U^{-1}R \simeq V^{-1}R$ is an isomorphism, then $\overline U$ is such a subset, and every such subset is contained in $\overline U$.
It follows that $\overline{\overline U} = \overline U$.
Let $E$ be a subset of $\operatorname{Spec}(R)$. Its germ is defined as $\{F \subseteq \operatorname{Spec}(R): F \supseteq E, F\textrm{ open}\}$.
We define $\overline E = \bigcap_{F \supseteq E} F$, where $F$ ranges over all open subsets of $\operatorname{Spec}(R)$ containing $E$.
One can show that $\overline E$ is the maximal subset which has the same germ as $E$.
It follows that $\overline{\overline E} = \overline E$.
For a multiplicative subset $U \subseteq R$, we define $E(U) = \{\mathfrak p \in \operatorname{Spec}(R): \mathfrak p \cap U = \emptyset\}$. Equivalently, $E(U)$ is the set-theoretic image of the canonical map $\operatorname{Spec}(U^{-1}R) \rightarrow \operatorname{Spec}(R)$.
For a subset $E \subseteq\operatorname{Spec}(R)$, we define $U(E) = \{u \in R: u \not\in \mathfrak p, \forall \mathfrak p \in E\}$, which is a multiplicative subset of $R$.
We then have:
The proof is mostly tautological. The only nontrivial step is the following: if $U$ is a multiplicative subset of $R$ and $a \in R$ satisfies $aR \cap U = \emptyset$, then there exists a prime ideal $\mathfrak p\in \operatorname{Spec}(R)$ containing $a$ such that $\mathfrak p \cap U =\emptyset$.
To prove this, use Zorn's Lemma on the set of ideals $I \subseteq R$ such that $a \in I$ and $I \cap U = \emptyset$, and show that any maximal element of this set is a prime ideal.
The above results establish a dictionary between localizations of $R$ and germs of $\operatorname{Spec}(R)$.
In geometric terms: for $U$ and $E$ such that $U = U(E)$ and $E = E(U)$, the localization $U^{-1}R$ is the ring of "germs" of functions that are defined on an open neighborhood of $E$, and the germ of $E$ consists of all open sets on which every function in $U$ doesn't vanish.
Back to the original question, we have a homomorphism $f\colon R\rightarrow S$ which induces a map $f^*: \operatorname{Spec}(S) \rightarrow \operatorname{Spec}(R)$.
The multiplicative set $f^{\times}$ is defined as $\{a \in R: f(a) \in S^\times\}$.
It is easy to show that $U(im(f^*)) = f^{\times}$, where $im(f^*)$ denotes the set-theoretic image of $f^*$.
Thus the localization of $f^\times$ corresponds, via the dictionary, to the germ of the image of $f^*$.
To see how this can be used, let us solve the two questions in the original post.
We show that $(f^\times)^{-1}R\simeq D^{-1}R$ in the case that $f$ is the canonical morphism $R \rightarrow D^{-1}R$.
Here we have $im(f^*) = E(D)$ by definition, and hence $f^\times = U(im(f^*)) = U(E(D)) = \overline D$. It follows that $(f^\times)^{-1}R = \overline D^{-1}R \simeq D^{-1}R$.
We show that $(f^\times)^{-1}R$ is in general not isomorphic to $S$. This is clear, as $(f^\times)^{-1}R$ only depends on the image of $f^*$, which means that it cannot distinguish between two morphisms that have the same image.
For example, the induced morphisms of the two homomorphisms $f_1: \Bbb Z \rightarrow \Bbb Z/p\Bbb Z$ and $f_2:\Bbb Z \rightarrow \Bbb Z/p^2\Bbb Z$ have the same image.