# Monday January 27th

## Localization

Consider rings $T$ such that $\ZZ \subseteq T \subseteq \QQ$, and let $P$ be a set of prime numbers.
We've shown that if $P, Q$ are two sets of prime numbers, then $\ZZ_p = \ZZ_q \iff \ZZ_p \cong \ZZ_q \iff P = Q$.

Let $R$ be a domain with fraction field $K$.
Let $P$ be a set of mutually nonassociate prime elements.
Note that $p\in R$ is a prime element iff $(p)$ is a prime ideal.
We say $x, y$ are associates iff there exists a $u\in R\units$ such that $y=ux$.
Since we're in a domain, (exercise) this is equivalent to $(x) = (y)$.

**Fact:**
We can then consider $$R_P definedas R\left[\theset{\frac 1 p \suchthat p\in P}\right],$$ and the fact is that the previous statement still holds.

But if $R = \ZZ$, we also have (exercise) if $Z\subset T \subset \QQ$ then $T = \ZZ_p$ for a unique $P$.

**Exercise:** 
How do we find such a $P$? 
This comes down to looking at $\frac a b \in T$ with $\gcd(a, b) = 1$, then $\frac 1 b \in T$.

> Hint: In a PID, $\gcd(a, b)$ exists and is a $\ZZ\dash$linear combination of $a$ and $b$.
> The solution should work for an arbitrary PID.

Let $R$ be a domain and $S$ multiplicatively closed (so $(S, \cdot) \leq (R, \cdot)$ is a submonoid).
Then $S$ is *primal* if $S$ is generated as a monoid by its prime elements.
Suppose that $S$ is *saturated*, i.e. if $s\in S$ and $r\in R$ with $r\divides s$, then $r\in S$.

> Can always add in all divisors.

We can then define the localization of $R$ at $S$, 

\begin{align*}
R_s \definedas \theset{ \frac a s \suchthat a\in R, s\in S }
.\end{align*}

This satisfies $R \subset R_S \subset K$, and is a multiplicative partial group completion.
If we took nonzero elements, this would yield exactly the fraction field.

Theorem (Negata)
: Let $R$ be a Noetherian domain with $S\subset R$ primal as above.
  If $R_S$ is a UFD, then $R$ is a UFD.

Exercise
: Show that the converse holds.

> Fraction fields are always UFDs?
> Localizing makes it easier for irreducibles to be prime.
> This helps prove that some interesting rings are UFDs.

## Modules

If $M$ is an $R\dash$module, then an $R\dash$submodule $N \leq M$ is a subgroup of $(M, +)$ such that $R\actson N \subset N$.

Every ring $R$ is an $R\dash$module over itself, and the $R\dash$submodules of $R$ are precisely the ideals of $R$.

> Can express certain concepts about rings/commutative algebra in the language of modules.

A morphism of $R\dash$modules $f: M\to N$ is a homomorphism $(M, +) \to (N, +)$ such that $f(r\actson m) = r\actson f(m)$.

**Exercise:**
Any module morphism that is a bijection is an isomorphism. 
(Usually true in algebraic settings.)

We can form quotient modules $\frac M N$ which is an $R\dash$module with $r\actson (m + N) = (r\actson m) + N$, and $M \to \frac M N$ is a surjective morphism.

If $I\normal R$ is an $R\dash$submodule of $R$, then $R/I$ is an $R\dash$module.
We have $\ann(R/I) = I$.

**Fact:**
Every ideal in $R$ is the annihilator of some $R\dash$module.

**Fact:**
Suppose $R$ is a ring such that every every nonzero $R\dash$module is faithful, then $R$ is a field.
The converse also holds.

> General idea: we study rings by looking at modules over them.

For an $R\dash$module $M$ and $S\subset M$, then we can consider $\generators{S}$ the $R\dash$submodule generated by $S$.
We can write this as 
$$
\intersect_{N\st S\subset N\subseteq RM} N = \theset{\sum_{i=1^n} r_i s_i \suchthat r_i\in R, s_i\in S}
.$$

We say $R$ is finitely generated iff there exists a finite generating set $S \subset M$.
We say $M$ is cyclic iff it is generated by a single element, i.e. $M = \generators{s}$.

Let $\theset{M_i}_{i\in I}$ be a family of $R\dash$modules.
Let $\prod_{i\in I} M_i$ be the cartesian product with a coordinate-wise $R\dash$action be the direct product.
Let 
$$
\bigoplus_{i\in I}M_i = \theset{(x_i) \in \prod M_i \suchthat x_i \neq 0 \text{ for finitely many } i}
,$$ 
which is a submodule of $\prod M_i$.
When $I$ is finite, these are equal.

Recall:
If $R$ is a PID and $M$ is a finitely generated $R\dash$module, then there exist finitely many cyclic $R\dash$ modules $\theset{C_1, \cdots, C_n}$, then $M \cong \bigoplus C_i$.

**Exercise:**
Let $R$ be a ring and $C$ a cyclic $R\dash$module, then show that $C \cong R /\ann(C)$ as $R\dash$modules.

> We'll later see that the class of rings $R$ such that every $R\dash$module is free are exactly fields.

*Remark:*
Let $I\normal R$, then $I$ is cyclic as an $R\dash$module iff $I$ is principal.

**Exercise:**

a. Let $I\normal R$ for $R$ a domain, then $I$ is indecomposable, i.e. $I\neq M_1 \oplus M_2$ for any nonzero $M_1, M_2$ $R\dash$modules.

b. If $R$ is additionally Noetherian and not a PID, then there exists an $I\normal R$ where $I$ is finitely generated, not principal, and so $I$ is not a cyclic $R\dash$module.

> Converse to structure theorem! Mild assumptions negate cyclic direct sum decomposition.