# Commutative Algebra/Fractions, annihilator

## Fractions within rings

Definition 9.1:

Let ${\displaystyle R}$ be a commutative ring, and let ${\displaystyle S\subseteq R}$ be an arbitrary subset. ${\displaystyle S}$ is called multiplicatively closed iff the following two conditions hold:

1. ${\displaystyle 1\in S}$
2. ${\displaystyle a,b\in S\Rightarrow ab\in S}$

Definition 9.2:

Let ${\displaystyle R}$ be a ring and ${\displaystyle S\subseteq R}$ a multiplicatively closed subset. Define

${\displaystyle S^{-1}R:=\{r/s|r\in R,s\in S\}/\sim _{S}}$,

where the equivalence relation ${\displaystyle \sim _{S}}$ is defined as

${\displaystyle r/s\sim _{S}u/t:\Leftrightarrow \exists i\in S:i(rt-su)=0}$.

${\displaystyle r/s+u/t:=(rt+us){\big /}st}$

and multiplication

${\displaystyle r/s\cdot u/t:=(ru){\big /}st}$.

The following two lemmata ensure that everything is correctly defined.

Lemma 9.3:

${\displaystyle \sim _{S}}$ is an equivalence relation.

Proof:

For reflexivity and symmetry, nothing interesting happens. For transitivity, there is a little twist. Assume

${\displaystyle r/s\sim _{S}u/t}$ and ${\displaystyle u/t\sim _{S}v/w}$.

Then there are ${\displaystyle i,j\in S}$ such that

${\displaystyle i(rt-su)=0}$ and ${\displaystyle j(uw-tv)=0}$.

But in this case, we have

${\displaystyle ijt(rw-vs)=ij(rwt-vst)=ij(rwt-suw+suw-vst)=0}$;

note ${\displaystyle ijt\in S}$ because ${\displaystyle S}$ is multiplicatively closed.${\displaystyle \Box }$

Lemma 9.4:

The addition and multiplication given above turn ${\displaystyle S^{-1}R}$ into a ring.

Proof:

We only prove well-definedness; the other rules follow from the definition and direct computation.

Let thus ${\displaystyle r/s\sim _{S}u/t}$ and ${\displaystyle a/p\sim _{S}b/q}$.

Thus, we have ${\displaystyle i(rt-su)=0}$ and ${\displaystyle j(aq-bp)=0}$ for suitable ${\displaystyle i,j\in S}$.

We want

${\displaystyle (rp+as){\big /}sp=(uq+bt){\big /}tq}$

and

${\displaystyle ra{\big /}sp=ub{\big /}tq}$.

These translate to

${\displaystyle x((rp+as)tq-(uq+bt)sp)=0}$

and

${\displaystyle y(ratq-ubsp)=0}$

for suitable ${\displaystyle x,y\in S}$. We get the desired result by picking ${\displaystyle x=y=ij}$ and observing

${\displaystyle ij(ratq-ubsp)=ij(ratq-sauq+sauq-ubsp)=0}$

and

${\displaystyle ij((rp+as)tq-(uq+bt)sp)=ij(rptq+astq-uqsp-btsp)=0}$.${\displaystyle \Box }$

Note that we were heavily using commutativity here.

Theorem 9.5 (properties of augmentation):

Let ${\displaystyle R}$ a ring and ${\displaystyle S\subseteq R}$ multiplicatively closed. Set

${\displaystyle \pi _{S}:R\mapsto S^{-1}(R),\pi _{S}(r):=r/1}$,

the projection morphism. Then:

1. ${\displaystyle s\in S\Rightarrow \pi _{S}(s)}$ is a unit.
2. ${\displaystyle \pi _{S}(r)=0\Rightarrow rs=0}$ for some ${\displaystyle s\in S}$.
3. Every element of ${\displaystyle S^{-1}R}$ has the form ${\displaystyle \pi _{S}(r)\pi _{S}(s)^{-1}}$ for suitable ${\displaystyle r\in R}$, ${\displaystyle s\in S}$.
4. Let ${\displaystyle I,J\leq R}$ be ideals. Then ${\displaystyle S^{-1}(I\cdot J)=S^{-1}(I\cdot J)}$, where
${\displaystyle S^{-1}I:=\{i/s|i\in I,s\in S\}}$.
1. Let ${\displaystyle I\leq R}$ an ideal. If ${\displaystyle S\cap I\neq \emptyset }$, then ${\displaystyle \pi _{S}(I)=S^{-1}R}$.

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 ${\displaystyle s\in S}$, then the rules for multiplication for ${\displaystyle S^{-1}R}$ indicate that ${\displaystyle 1/s}$ is an inverse for ${\displaystyle \pi _{S}(s)=s/1}$.

2.:

Assume ${\displaystyle r/1=0=0/1}$. Then there exists ${\displaystyle s\in S}$ such that ${\displaystyle s(r-0)=sr=0}$.

3.:

Let ${\displaystyle r/s}$ be an arbitrary element of ${\displaystyle S^{-1}R}$. Then ${\displaystyle r/s=r/1\cdot 1/s=\pi _{S}(r)\pi _{S}(s)^{-1}}$.

4.

{\displaystyle {\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 ${\displaystyle I\cap S\neq \emptyset }$, that is, ${\displaystyle s\in S\cap I}$. Then ${\displaystyle \pi _{S}(s)\in \pi _{S}(I)}$, where ${\displaystyle \pi _{S}(s)}$ is a unit in ${\displaystyle S^{-1}R}$. Further, ${\displaystyle \pi _{S}(I)}$ is an ideal within ${\displaystyle S^{-1}R}$ since ${\displaystyle \pi _{S}}$ is a morphism. Thus, ${\displaystyle \pi _{S}(I)=S^{-1}R}$.${\displaystyle \Box }$

Theorem 9.6 (universal property):

Let ${\displaystyle R}$ be a ring, ${\displaystyle S\subseteq R}$ multiplicatively closed, let ${\displaystyle T}$ be another ring and let

${\displaystyle f:R\to T}$

be a morphism, such that for all ${\displaystyle s\in S}$, ${\displaystyle f(s)\in T^{\times }}$. Then there exists a unique morphism

${\displaystyle g:S^{-1}R\to T}$

such that

${\displaystyle f=g\circ \pi _{S}}$.

Proof:

We first prove uniqueness. Assume there exists another such morphism ${\displaystyle g'}$. Then we would have

${\displaystyle g'(r/s)=g'(r/1)g'(s/1)^{-1}=f(r)f^{-1}=g(r/1)g(s/1)^{-1}}$.

Then we prove existence; we claim that

${\displaystyle g(r/s):=f(r)(f(s))^{-1}}$

defines the desired morphism.

First, we show well-definedness.

Firstly, ${\displaystyle f(s)^{-1}}$ exists for ${\displaystyle s\in S}$.

Secondly, let ${\displaystyle r/s\sim _{S}u/t}$, that is, ${\displaystyle i(rt-su)=0}$. Then

{\displaystyle {\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 ${\displaystyle f\circ \pi _{S}}$ is a morphism and commutativity); additivity is proven as follows:

{\displaystyle {\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.${\displaystyle \Box }$

Theorem 9.7:

Category theory context

## Fractions within modules

Definition 9.8:

Let ${\displaystyle R}$ be a ring, ${\displaystyle S\subseteq R}$ a multiplicative subset of ${\displaystyle R}$ and ${\displaystyle M}$ an ${\displaystyle R}$-module. Set ${\displaystyle S^{-1}R}$ to be the ring ${\displaystyle R}$ augmented by inverses of ${\displaystyle S}$. We define the ${\displaystyle S^{-1}R}$-module ${\displaystyle S^{-1}M}$ as follows:

${\displaystyle S^{-1}M:=\left\{m/s{\big |}m\in M\right\}/\sim _{S}}$ (the formal fractions),

where again

${\displaystyle m/s\sim _{S}n/t:\Leftrightarrow \exists u\in S:u(tm-sn)=0}$,

${\displaystyle m/s+n/t:=(tm+sn)/st}$

and module operation

${\displaystyle r/s~~m/t:=rm/st}$.

Note that applying this construction to a ring ${\displaystyle R}$ that is canonically an ${\displaystyle R}$-module over itself, we obtain nothing else but ${\displaystyle S^{-1}R}$ canonically seen as an ${\displaystyle S^{-1}R}$-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 ${\displaystyle M}$ be an ${\displaystyle R}$-module, let ${\displaystyle S\subseteq M}$ be a multiplicatively closed subset of ${\displaystyle R}$, and let ${\displaystyle N,K\leq M}$ be submodules. Then

1. ${\displaystyle S^{-1}(N+K)=S^{-1}N+S^{-1}K}$,
2. ${\displaystyle S^{-1}(N\cap K)=S^{-1}N\cap S^{-1}K}$, and
3. ${\displaystyle S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N)}$;

in the first two equations, all modules are seen as submodules of ${\displaystyle S^{-1}M}$ (as above with ${\displaystyle S^{-1}I}$), and in the third isomorphy relation, the modules are seen as independent ${\displaystyle S^{-1}R}$-modules.

Proof:

1.

{\displaystyle {\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 ${\displaystyle R}$ to equalize denominators and thus get a suitable ${\displaystyle t\in S}$ (${\displaystyle S}$ is closed under multiplication).

2.

{\displaystyle {\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 ${\displaystyle n/u=k/v\Leftrightarrow w(vn-uk)=0}$ for a suitable ${\displaystyle w\in S}$, and in particular for example

${\displaystyle m/s=wvn/wvu}$,

where ${\displaystyle wvn=wuk\in N\cap K}$.

3.

We set

${\displaystyle \varphi :S^{-1}(M/N)\to (S^{-1}M)/(S^{-1}N),\varphi ((m+N)/s):=m/s+S^{-1}N}$

and prove that this is an isomorphism.

First we prove well-definedness. Indeed, if ${\displaystyle m+N=m'+N}$, then ${\displaystyle m-m'\in N}$, hence ${\displaystyle (m-m')/s\in S^{-1}N}$ and thus ${\displaystyle m/s+S^{-1}N=m'/s+S^{-1}N}$.

Then we prove surjectivity. Let ${\displaystyle m/s+S^{-1}N}$ be given. Then obviously ${\displaystyle (m+N)/s}$ is mapped to that element.

Then we prove injectivity. Assume ${\displaystyle m/s\in S^{-1}N}$. Then ${\displaystyle m/s=n/t}$, where ${\displaystyle n\in N}$ and ${\displaystyle t\in S}$, that is ${\displaystyle u(tm-sn)=0}$ for a suitable ${\displaystyle u\in S}$. Then ${\displaystyle utm\in N}$ and therefore ${\displaystyle (m+N)/s=(utm+N)/uts=0}$.${\displaystyle \Box }$

Theorem 9.10:

functor relating tensor product and fractions

Theorem 9.11:

Let ${\displaystyle M,N}$ be ${\displaystyle R}$-modules and ${\displaystyle S\subseteq R}$ multiplicatively closed. Then

${\displaystyle S^{-1}M\otimes _{S^{-1}R}S^{-1}N\cong S^{-1}(M\otimes _{R}N)}$.

Proof:

### Exercises

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

## The annihilator, faithfulness

Definition 9.12:

Let ${\displaystyle R}$ be a ring, ${\displaystyle M}$ a module over ${\displaystyle R}$ and ${\displaystyle S\subseteq M}$ an arbitrary subset. Then the annihilator of ${\displaystyle S}$ with respect to ${\displaystyle M}$ is defined to be the set

${\displaystyle {\text{Ann}}_{R}(S):=\{r\in R|\forall s\in S:rs=0\}}$.

Theorem 9.13:

Let ${\displaystyle R}$ be a ring, ${\displaystyle M}$ a module over ${\displaystyle R}$ and ${\displaystyle S\subseteq M}$ an arbitrary subset. Then ${\displaystyle {\text{Ann}}_{R}(S)}$ is an ideal of ${\displaystyle R}$.

Proof:

Let ${\displaystyle a,b\in {\text{Ann}}_{R}(S)}$ and ${\displaystyle r\in R}$. Then for all ${\displaystyle s\in S}$, ${\displaystyle (ra-b)s=r(as)-bs=0}$. Hence the theorem by lemma 5.3.${\displaystyle \Box }$

Definition 9.14:

An ${\displaystyle R}$-module ${\displaystyle M}$ is called faithful iff ${\displaystyle {\text{Ann}}_{R}M=\{0\}}$.

Theorem 9.15:

Let ${\displaystyle R}$ be a ring. Then ${\displaystyle R}$ regarded as an ${\displaystyle R}$ module over itself is faithful.

Proof: Let ${\displaystyle s\in R}$ such that ${\displaystyle \forall r\in R:rs=0}$. Then in particular ${\displaystyle s1=0}$.${\displaystyle \Box }$

Theorem 9.16:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module and ${\displaystyle S\subseteq M}$ an arbitrary subset. Let ${\displaystyle \langle S\rangle =:N\leq M}$ be the submodule of ${\displaystyle M}$ generated by ${\displaystyle S}$. Then ${\displaystyle {\text{Ann}}_{R}S={\text{Ann}}_{R}N}$.

Proof:

From the definition it is clear that ${\displaystyle {\text{Ann}}_{R}N\subseteq {\text{Ann}}_{R}S}$, since annihilating all elements of ${\displaystyle N}$ is a stronger condition than only those of ${\displaystyle S}$.

Let now ${\displaystyle t\in {\text{Ann}}_{R}S}$ and ${\displaystyle x_{1}s_{1}+\cdots x_{n}s_{n}\in N}$, where ${\displaystyle x_{j}\in R}$ and ${\displaystyle s_{j}\in S}$. Then ${\displaystyle t(x_{1}s_{1}+\cdots x_{n}s_{n})=x_{1}ts_{1}+\cdots x_{n}ts_{n}=0+0+\cdots +0=0}$.${\displaystyle \Box }$

## Local properties

Definition 9.17:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module (where ${\displaystyle R}$ is a ring) and let ${\displaystyle p\leq R}$ be a prime ideal. Then the localisation of ${\displaystyle M}$ with respect to ${\displaystyle p}$, denoted by

${\displaystyle M_{p}}$,

is defined to be ${\displaystyle S^{-1}M}$ with ${\displaystyle S:=R\setminus p}$; note that ${\displaystyle S}$ is multiplicatively closed because ${\displaystyle p}$ 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:

1. ${\displaystyle M}$ has property (*).
2. ${\displaystyle S^{-1}M}$ has property (*) for all multiplicatively closed ${\displaystyle S\subseteq R}$.
3. ${\displaystyle M_{p}}$ has property (*) for all prime ideals ${\displaystyle p\leq R}$.
4. ${\displaystyle M_{m}}$ has property (*) for all maximal ideals ${\displaystyle m\leq R}$.

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. ${\displaystyle \Rightarrow }$ 1. suffices.

Assume that ${\displaystyle N}$ is a nonzero module, that is, we have ${\displaystyle n\in N}$ such that ${\displaystyle n\neq 0}$. By theorem 9.11, ${\displaystyle \operatorname {Ann} _{R}(n):=\operatorname {Ann} _{R}(\{n\})}$ is an ideal of ${\displaystyle R}$. Therefore, it is contained within some maximal ideal of ${\displaystyle R}$, call ${\displaystyle m}$ (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 ${\displaystyle s\in R\setminus m}$ we have ${\displaystyle sn\neq 0}$ and therefore ${\displaystyle n/1\neq 0/1}$ in ${\displaystyle M_{m}}$.${\displaystyle \Box }$

The following theorems do not really describe local-global properties, but are certainly similar and perhaps related to those.

Theorem 9.20:

If ${\displaystyle f:M\to N}$ is a morphism, then the following are equivalent:

1. ${\displaystyle f:M\to N}$ surjective.
2. ${\displaystyle f_{S}:S^{-1}M\to S^{-1}N}$ surjective for all ${\displaystyle S\subseteq R}$ multiplicatively closed.
3. ${\displaystyle f_{p}:M_{p}\to N_{p}}$ surjective for all ${\displaystyle p\leq R}$ prime.
4. ${\displaystyle f_{m}:M_{m}\to N_{m}}$ surjective for all ${\displaystyle m\leq R}$ maximal.

Proof: