# Commutative Algebra/Fractions, annihilator
## Fractions within rings

Definition 9.1:
Let $R$ be a commutative ring, and let $S\subseteq R$ be an arbitrary subset. $S$ is called multiplicatively closed iff the following two conditions hold:
* $1\in S$
* $a,b\in S\Rightarrow ab\in S$
Definition 9.2:
Let $R$ be a ring and $S\subseteq R$ a multiplicatively closed subset. Define
    $S^{-1}R:=\{r/s|r\in R,s\in S\}/\sim _{S}$,
where the equivalence relation $\sim _{S}$ is defined as
    $r/s\sim _{S}u/t:\Leftrightarrow \exists i\in S:i(rt-su)=0$.
Equip this with addition
    $r/s+u/t:=(rt+us){\big /}st$
and multiplication
    $r/s\cdot u/t:=(ru){\big /}st$.
The following two lemmata ensure that everything is correctly defined.
Lemma 9.3:
$\sim _{S}$ is an equivalence relation.
Proof:
For reflexivity and symmetry, nothing interesting happens. For transitivity, there is a little twist. Assume
    $r/s\sim _{S}u/t$ and $u/t\sim _{S}v/w$.
Then there are $i,j\in S$ such that
    $i(rt-su)=0$ and $j(uw-tv)=0$.
But in this case, we have
    $ijt(rw-vs)=ij(rwt-vst)=ij(rwt-suw+suw-vst)=0$;
note $ijt\in S$ because $S$ is multiplicatively closed.$\Box$
Lemma 9.4:
The addition and multiplication given above turn $S^{-1}R$ into a ring.
Proof:
We only prove well-definedness; the other rules follow from the definition and direct computation.
Let thus $r/s\sim _{S}u/t$ and $a/p\sim _{S}b/q$.
Thus, we have $i(rt-su)=0$ and $j(aq-bp)=0$ for suitable $i,j\in S$.
We want
    $(rp+as){\big /}sp=(uq+bt){\big /}tq$
and
    $ra{\big /}sp=ub{\big /}tq$.
These translate to
    $x((rp+as)tq-(uq+bt)sp)=0$
and
    $y(ratq-ubsp)=0$
for suitable $x,y\in S$. We get the desired result by picking $x=y=ij$ and observing
    $ij(ratq-ubsp)=ij(ratq-sauq+sauq-ubsp)=0$
and
    $ij((rp+as)tq-(uq+bt)sp)=ij(rptq+astq-uqsp-btsp)=0$.$\Box$
Note that we were heavily using commutativity here.
Theorem 9.5 (properties of augmentation):
Let $R$ a ring and $S\subseteq R$ multiplicatively closed. Set
    $\pi _{S}:R\mapsto S^{-1}(R),\pi _{S}(r):=r/1$,
the projection morphism. Then:
* $s\in S\Rightarrow \pi _{S}(s)$ is a unit.
* $\pi _{S}(r)=0\Rightarrow rs=0$ for some $s\in S$.
* Every element of $S^{-1}R$ has the form $\pi _{S}(r)\pi _{S}(s)^{-1}$ for suitable $r\in R$, $s\in S$.
* Let $I,J\leq R$ be ideals. Then $S^{-1}(I\cdot J)=S^{-1}(I\cdot J)$, where
       $S^{-1}I:=\{i/s|i\in I,s\in S\}$.
* Let $I\leq R$ an ideal. If $S\cap I\neq \emptyset$, then $\pi _{S}(I)=S^{-1}R$.
We will see further properties like 4. when we go to modules, but we can't phrase it in full generality because in modules, we may not have a product of two module elements.
Proof:
1.:
If $s\in S$, then the rules for multiplication for $S^{-1}R$ indicate that $1/s$ is an inverse for $\pi _{S}(s)=s/1$.
2.:
Assume $r/1=0=0/1$. Then there exists $s\in S$ such that $s(r-0)=sr=0$.
3.:
Let $r/s$ be an arbitrary element of $S^{-1}R$. Then $r/s=r/1\cdot 1/s=\pi _{S}(r)\pi _{S}(s)^{-1}$.
4.
    ${\begin{aligned}r/s\in S^{-1}(I\cdot J)&\Leftrightarrow r/s=ij/t,i\in I,j\in J,t\in S\\&\Leftrightarrow r/s=(i/t)(j/1),i\in I,j\in J,t\in S\\&\Leftrightarrow r/s\in S^{-1}I\cdot S^{-1}J\end{aligned}}$
5.
Let $I\cap S\neq \emptyset$, that is, $s\in S\cap I$. Then $\pi _{S}(s)\in \pi _{S}(I)$, where $\pi _{S}(s)$ is a unit in $S^{-1}R$. Further, $\pi _{S}(I)$ is an ideal within $S^{-1}R$ since $\pi _{S}$ is a morphism. Thus, $\pi _{S}(I)=S^{-1}R$.$\Box$
Theorem 9.6 (universal property):
Let $R$ be a ring, $S\subseteq R$ multiplicatively closed, let $T$ be another ring and let
    $f:R\to T$
be a morphism, such that for all $s\in S$, $f(s)\in T^{\times }$. Then there exists a unique morphism
    $g:S^{-1}R\to T$
such that
    $f=g\circ \pi _{S}$.
Proof:
We first prove uniqueness. Assume there exists another such morphism $g'$. Then we would have
    $g'(r/s)=g'(r/1)g'(s/1)^{-1}=f(r)f^{-1}=g(r/1)g(s/1)^{-1}$.
Then we prove existence; we claim that
    $g(r/s):=f(r)(f(s))^{-1}$
defines the desired morphism.
First, we show well-definedness.
Firstly, $f(s)^{-1}$ exists for $s\in S$.
Secondly, let $r/s\sim _{S}u/t$, that is, $i(rt-su)=0$. Then
    ${\begin{aligned}g(r/s)&=g(itr/its)\\&=f(itr)(f(its))^{-1}\\&=f(isu)(f(its))^{-1}\\&=g(isu/its)=g(u/t).\end{aligned}}$
The multiplicativity of this morphism is visually obvious (use that $f\circ \pi _{S}$ is a morphism and commutativity); additivity is proven as follows:
    ${\begin{aligned}g(r/s+u/t)&=g\left((rt+su){\big /}st\right)\\&=f(rt+su)(f(st))^{-1}\\&=f(rt)(f(st))^{-1}+f(su)(f(st))^{-1}\\&=g(r/s)+g(u/t).\end{aligned}}$
It is obvious that the unit is mapped to the unit.$\Box$
Theorem 9.7:
Category theory context

## Fractions within modules

Definition 9.8:
Let $R$ be a ring, $S\subseteq R$ a multiplicative subset of $R$ and $M$ an $R$-module. Set $S^{-1}R$ to be the ring $R$ augmented by inverses of $S$. We define the $S^{-1}R$-module $S^{-1}M$ as follows:
    $S^{-1}M:=\left\{m/s{\big |}m\in M\right\}/\sim _{S}$ (the formal fractions),
where again
    $m/s\sim _{S}n/t:\Leftrightarrow \exists u\in S:u(tm-sn)=0$,
with addition
    $m/s+n/t:=(tm+sn)/st$
and module operation
    $r/s~~m/t:=rm/st$.
Note that applying this construction to a ring $R$ that is canonically an $R$-module over itself, we obtain nothing else but $S^{-1}R$ canonically seen as an $S^{-1}R$-module over itself, since multiplication and addition coincide. Thus, we have a generalisation here!
That everything is well-defined is seen exactly as in the last section; the proofs carry over verbatim.
Theorem 9.9 (properties of the augmented module):
Let $M$ be an $R$-module, let $S\subseteq M$ be a multiplicatively closed subset of $R$, and let $N,K\leq M$ be submodules. Then
* $S^{-1}(N+K)=S^{-1}N+S^{-1}K$,
* $S^{-1}(N\cap K)=S^{-1}N\cap S^{-1}K$, and
* $S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N)$;
in the first two equations, all modules are seen as submodules of $S^{-1}M$ (as above with $S^{-1}I$), and in the third isomorphy relation, the modules are seen as independent $S^{-1}R$-modules.
Proof:
1.
    ${\begin{aligned}m/s\in S^{-1}(N+K)&\Leftrightarrow m/s=(n+k)/t,t\in S,n\in N,k\in K\\&\Leftrightarrow m/s=n/t+k/t,t\in S,n\in N,k\in K\\&\Leftrightarrow m/s\in S^{-1}N+S^{-1}K;\end{aligned}}$
note that to get from the third row back to the second, we used that submodules are closed under multiplication by an element of $R$ to equalize denominators and thus get a suitable $t\in S$ ($S$ is closed under multiplication).
2.
    ${\begin{aligned}m/s\in S^{-1}(N\cap K)&\Leftrightarrow m/s=l/t,t\in S,l\in N\cap K\\&\Leftrightarrow m/s=n/u=k/v,u,v\in S,n\in N,k\in K\\&\Leftrightarrow m/s\in S^{-1}N\cap S^{-1}K;\end{aligned}}$
to get from the second to the first row, we note $n/u=k/v\Leftrightarrow w(vn-uk)=0$ for a suitable $w\in S$, and in particular for example
    $m/s=wvn/wvu$,
where $wvn=wuk\in N\cap K$.
3.
We set
    $\varphi :S^{-1}(M/N)\to (S^{-1}M)/(S^{-1}N),\varphi ((m+N)/s):=m/s+S^{-1}N$
and prove that this is an isomorphism.
First we prove well-definedness. Indeed, if $m+N=m'+N$, then $m-m'\in N$, hence $(m-m')/s\in S^{-1}N$ and thus $m/s+S^{-1}N=m'/s+S^{-1}N$.
Then we prove surjectivity. Let $m/s+S^{-1}N$ be given. Then obviously $(m+N)/s$ is mapped to that element.
Then we prove injectivity. Assume $m/s\in S^{-1}N$. Then $m/s=n/t$, where $n\in N$ and $t\in S$, that is $u(tm-sn)=0$ for a suitable $u\in S$. Then $utm\in N$ and therefore $(m+N)/s=(utm+N)/uts=0$.$\Box$
Theorem 9.10:
functor relating tensor product and fractions
Theorem 9.11:
Let $M,N$ be $R$-modules and $S\subseteq R$ multiplicatively closed. Then
    $S^{-1}M\otimes _{S^{-1}R}S^{-1}N\cong S^{-1}(M\otimes _{R}N)$.
Proof:

### Exercises

* Exercise 9.2.1: Let $M,N$ be $R$-modules and $I\leq R$ an ideal. Prove that $IM:=\{im|i\in I,m\in M\}$ is a submodule of $M$ and that $(M\otimes _{R}N)/I(M\otimes _{R}N)\cong (M/I)\otimes _{R/I}(N/I)$ (this exercise serves the purpose of practising the proof technique employed for theorem 9.11).

## The annihilator, faithfulness

Definition 9.12:
Let $R$ be a ring, $M$ a module over $R$ and $S\subseteq M$ an arbitrary subset. Then the annihilator of $S$ with respect to $M$ is defined to be the set
    ${\text{Ann}}_{R}(S):=\{r\in R|\forall s\in S:rs=0\}$.
Theorem 9.13:
Let $R$ be a ring, $M$ a module over $R$ and $S\subseteq M$ an arbitrary subset. Then ${\text{Ann}}_{R}(S)$ is an ideal of $R$.
Proof:
Let $a,b\in {\text{Ann}}_{R}(S)$ and $r\in R$. Then for all $s\in S$, $(ra-b)s=r(as)-bs=0$. Hence the theorem by lemma 5.3.$\Box$
Definition 9.14:
An $R$-module $M$ is called faithful iff ${\text{Ann}}_{R}M=\{0\}$.
Theorem 9.15:
Let $R$ be a ring. Then $R$ regarded as an $R$ module over itself is faithful.
Proof: Let $s\in R$ such that $\forall r\in R:rs=0$. Then in particular $s1=0$.$\Box$
Theorem 9.16:
Let $M$ be an $R$-module and $S\subseteq M$ an arbitrary subset. Let $\langle S\rangle =:N\leq M$ be the submodule of $M$ generated by $S$. Then ${\text{Ann}}_{R}S={\text{Ann}}_{R}N$.
Proof:
From the definition it is clear that ${\text{Ann}}_{R}N\subseteq {\text{Ann}}_{R}S$, since annihilating all elements of $N$ is a stronger condition than only those of $S$.
Let now $t\in {\text{Ann}}_{R}S$ and $x_{1}s_{1}+\cdots x_{n}s_{n}\in N$, where $x_{j}\in R$ and $s_{j}\in S$. Then $t(x_{1}s_{1}+\cdots x_{n}s_{n})=x_{1}ts_{1}+\cdots x_{n}ts_{n}=0+0+\cdots +0=0$.$\Box$

## Local properties

Definition 9.17:
Let $M$ be an $R$-module (where $R$ is a ring) and let $p\leq R$ be a prime ideal. Then the localisation of $M$ with respect to $p$, denoted by
    $M_{p}$,
is defined to be $S^{-1}M$ with $S:=R\setminus p$; note that $S$ is multiplicatively closed because $p$ is a prime ideal.
Definition 9.18:
A property which modules can have (such as being equal to zero) is called a local-global property iff the following are equivalent:
* $M$ has property (*).
* $S^{-1}M$ has property (*) for all multiplicatively closed $S\subseteq R$.
* $M_{p}$ has property (*) for all prime ideals $p\leq R$.
* $M_{m}$ has property (*) for all maximal ideals $m\leq R$.
Theorem 9.19:
Being equal to zero is a local-global property.
Proof:
We check the equivalence of 1. - 4. from definition 9.12. Clearly, 4. $\Rightarrow$ 1. suffices.
Assume that $N$ is a nonzero module, that is, we have $n\in N$ such that $n\neq 0$. By theorem 9.11, $\operatorname {Ann} _{R}(n):=\operatorname {Ann} _{R}(\{n\})$ is an ideal of $R$. Therefore, it is contained within some maximal ideal of $R$, call $m$ (unfortunately, we have to refer to a later chapter, since we wanted to separate treatments of different algebraic objects. The required theorem is theorem 12.2). Then for $s\in R\setminus m$ we have $sn\neq 0$ and therefore $n/1\neq 0/1$ in $M_{m}$.$\Box$
The following theorems do not really describe local-global properties, but are certainly similar and perhaps related to those.
Theorem 9.20:
If $f:M\to N$ is a morphism, then the following are equivalent:
* $f:M\to N$ surjective.
* $f_{S}:S^{-1}M\to S^{-1}N$ surjective for all $S\subseteq R$ multiplicatively closed.
* $f_{p}:M_{p}\to N_{p}$ surjective for all $p\leq R$ prime.
* $f_{m}:M_{m}\to N_{m}$ surjective for all $m\leq R$ maximal.
Proof: