# Wednesday March 4th

For $R$ a ring and $S\subset R$ such that $S^2 \subset S$ and $1\in S$, there exists a ring $S\inv R$ and a ring morphism $\iota: R\to S\inv R$ such that

1. $\iota(S) \subset (S\inv R)\units$, and
2. $\iota$ is universal for such morphisms, i.e every $R \mapsvia{f} T$ with $f(S) \subset T\units$ lifts to $S\inv R\mapsvia{\tilde f} T$.

Last time we constructed it as $R\cross S/\sim$ where $(r_1, s_1) \sim (r_2, s_2)$ iff there exists an $s \in S$ such that $sr_1 s_2 = sr_2 s_1$ (needed to obtain transitivity).

We then have $\iota(r) = [(r, 1)]$; what is its kernel?
If $(r, 1) \sim (0, 1)$ then there exists $s\in S$ such that $s\cdot r\cdot 1 = s\cdot1\cdot0 = 0$, so $\ann(r) \intersect \ker \iota \neq \emptyset$.
Note that if $0\in S$ then $\ker \iota = R$ and thus $S\inv R = 0$.
Conversely, if $0\not\in S$, then $\ann(1) \intersect S = \emptyset$, so $S\inv R \neq 0$. 
Thus $S\inv R = 0$ iff $0\in S$.

Example
: For $f\in R$, $R_f \definedas S\inv_f R$ where $S_f = \theset{1, f, f^2, \cdots}$, then $R_f = 0 \iff f\in \nil(R)$.

Definition (Saturated Multiplicatively Closed Sets)
: A  multiplicatively closed set $S$ is *saturated* iff for $s\in S, f\in R$ with $f$ dividing $S$, then $f\in S$.
  Denote the $\bar S$ the saturation of $S$ obtained by adding all divisors, then $S\inv R = \bar{S}\inv R$.
  
> Recall link to early problem of characterizing rings between $\ZZ$ and $\QQ$.
> There are more localizations than such rings, since localizing at $n$ is as good as localizing at $kn$.
 
If $R$ is a domain, then for any $S$ with $0\not\in S$, there is a diagram

\begin{center}
\begin{tikzcd}
R \arrow[hook, r] & S\inv R\arrow[hook, d] \\
 & K 
\end{tikzcd}
\end{center}

where $K = ff(R)$.

In any ring, take $S$ to be the nonzero divisors, then there is a maximal injective localization.
\begin{center}
\begin{tikzcd}
\iota R \arrow[hook, r] & S\inv R\arrow[hook, d] \\
 & \text{Total fraction field} 
\end{tikzcd}
\end{center}

> Can generalize results from domains to arbitrary rings this way.

Exercise
: Take $R_1, R_2$ nonzero rings, $R = R_1\cross R_2$, and take $S = R_1 \cross \{1\}$.
  What is $S\inv R$?
  (First figure out the kernel of the localization.)
  
## Pushing and Pulling

> Note that we can push/pull for quotients and get back what we started with -- want something similar for localization.

Consider the map $\iota: R\to S\inv R$.

Lemma
: $I\normal R$ implies that $\iota_*(I) = \theset{ \frac x s \suchthat x\in I, s\in S}$.

Proof
: Easy.

Proposition (Push-Pull Equality for Ideals in Localizations)
:   For all $J \normal S\inv R$, 
    $$
    \iota_* \iota^* J = J
    .$$

Proof
: Note that we always have containment, just need to show reverse containment.

Lemma
: For $I\normal R$, 
$$
i_*(I) = S\inv R \iff I\intersect S \neq \emptyset
.$$

Proposition (Properties of Spec for Localization)
:   \hfill 
    
    a. For $\mfp \in \spec(R)$, TFAE:

    - $\iota_* \mfp \in \spec(S\inv R)$
    - $\iota_* \subsetneq S\inv R$
    - $\mfp \intersect S = \emptyset$
      
    b. If $\mfp \intersect S = \emptyset$, then $\iota^* \iota_* \mfp = \mfp$.
    
Corollary
: $i^*$ and $i_*$ are mutually inverse, order-preserving bijections 
$$
\spec(S\inv R) 
\overset{i^*}{\underset{i_*}\rightleftarrows}
\theset{\mfp \in \spec(R) \suchthat \mfp\intersect S = \emptyset}
.$$

Lemma
: For $I\normal R$, $S$ a multiplicatively closed set, let $f: R\surjects R/I$ be the quotient map and $\bar{S} \definedas q(S)$. 
  Then 
  $$
  {S\inv R \over IS\inv R} &\mapsvia{\cong} \bar{S}\inv (R/I) \\
  \frac a s + IS\inv R &\mapsto \frac{a + I}{s + I}
  .$$
  
> Thus localizing commutes with taking quotients.

Let $\mfp \in \spec(R)$, then $S_\mfp \definedas R\set minus \mfp$ is multiplicatively closed.
(Note that localizing at any non-prime ideal gives you the zero ring.)
Let $R_\mfp \definedas S_{\mfp}\inv R$.

Proposition (Complements of Prime Ideals are Local? Extremely Important!)
: $R_\mfp$ is a local ring with a unique maximal ideal $\mfp R_\mfp$,

Proof
: The poset $\spec(R_\mfp) = \theset{q\in \spec(R) \suchthat q \intersect (R\setminus \mfp) = \emptyset \iff q\leq \mfp}$.

> This gives us a way to construct a local ring from *any* maximal ideal. 
> Perhaps the most important construction thus far.

Exercise
: Let $(R, \mfm)$ be local and $S\subset R$ be multiplicatively closed.
  Show that $S\inv R$ need **not** be local.
  
## Localization for Modules

Let $M$ be an $R\dash$module and $S\subset R$ multiplicatively closed.
We want $S\inv M$ to satisfy:

- $S\inv M$ is an $R\dash$module
- There is a morphism $M \to S\inv M$ such that for all $s\in S$, the map $S\inv M\mapsvia{[s]} S\inv M$ is an isomorphism, i.e. $S \to \endo_R(S\inv M)$ with $i(S) \subset \endo_R(S\inv M)$
- This is universal wrt the above property.

There are two potential constructions.

Construction 1:
Adapt the $S\inv R$ construction, defining $S\inv M =  M\cross S/\sim$.

Construction 2:
Define $S\inv M \definedas S\inv R \tensor_R M$, where $\iota: M \to S\inv M$ where $m\mapsto 1\tensor m$.

It can be checked that these both satisfy the appropriate Universal mapping property.

Exercise
: If $M$ is an $R\dash$module, then $M$ has an $S\inv R\dash$module structure iff $S$ acts invertibly (so $[s]: M\to M$ is invertible), and if so the structure is unique.