Definition 9.2:
Let
be a ring and
a multiplicatively closed subset. Define
,
where the equivalence relation
is defined as
.
Equip this with addition
![{\displaystyle r/s+u/t:=(rt+us){\big /}st}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7f4e46958730beb6fe8a0650b33767526aa2e18)
and multiplication
.
The following two lemmata ensure that everything is correctly defined.
Lemma 9.3:
is an equivalence relation.
Proof:
For reflexivity and symmetry, nothing interesting happens. For transitivity, there is a little twist. Assume
and
.
Then there are
such that
and
.
But in this case, we have
;
note
because
is multiplicatively closed.
Lemma 9.4:
The addition and multiplication given above turn
into a ring.
Proof:
We only prove well-definedness; the other rules follow from the definition and direct computation.
Let thus
and
.
Thus, we have
and
for suitable
.
We want
![{\displaystyle (rp+as){\big /}sp=(uq+bt){\big /}tq}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a9d523ebe630d76fe2ba7d28cdfa3993f1236bb)
and
.
These translate to
![{\displaystyle x((rp+as)tq-(uq+bt)sp)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48cca10c75ccd811e25bbfc87df1488526fc56f4)
and
![{\displaystyle y(ratq-ubsp)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d26b3b52d4913ffa06fd4590a1b60cdcfe862e8a)
for suitable
. We get the desired result by picking
and observing
![{\displaystyle ij(ratq-ubsp)=ij(ratq-sauq+sauq-ubsp)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01ed6af3f8b4e7b3906e5424c06dc478d6f08b45)
and
.![{\displaystyle \Box }](https://wikimedia.org/api/rest_v1/media/math/render/svg/029b77f09ebeaf7528fc831fe57848be51f2240b)
Note that we were heavily using commutativity here.
Theorem 9.5 (properties of augmentation):
Let
a ring and
multiplicatively closed. Set
,
the projection morphism. Then:
is a unit.
for some
.
- Every element of
has the form
for suitable
,
.
- Let
be ideals. Then
, where
.
- Let
an ideal. If
, then
.
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
, then the rules for multiplication for
indicate that
is an inverse for
.
2.:
Assume
. Then there exists
such that
.
3.:
Let
be an arbitrary element of
. Then
.
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a85ae443778163b65fd8cda7b052b82ceae87bbd)
5.
Let
, that is,
. Then
, where
is a unit in
. Further,
is an ideal within
since
is a morphism. Thus,
.
Proof:
We first prove uniqueness. Assume there exists another such morphism
. Then we would have
.
Then we prove existence; we claim that
![{\displaystyle g(r/s):=f(r)(f(s))^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f41699b7adc72d8b2935642081fafe85e4e5638f)
defines the desired morphism.
First, we show well-definedness.
Firstly,
exists for
.
Secondly, let
, that is,
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57dc6c2aac4dd940374fd79349fe74276e948d23)
The multiplicativity of this morphism is visually obvious (use that
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f2b7044e6163d51c2fa61b53f356fd494b84616)
It is obvious that the unit is mapped to the unit.
Theorem 9.7:
Category theory context
Note that applying this construction to a ring
that is canonically an
-module over itself, we obtain nothing else but
canonically seen as an
-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.
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e1125c3f79be2eda8ff880cca0d2049ab9cce8f)
note that to get from the third row back to the second, we used that submodules are closed under multiplication by an element of
to equalize denominators and thus get a suitable
(
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd88a7d0646188465f78fdad9bed3839302e2c96)
to get from the second to the first row, we note
for a suitable
, and in particular for example
,
where
.
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49d9f358bdfd78260a9f0a3fa228b149367c4fff)
and prove that this is an isomorphism.
First we prove well-definedness. Indeed, if
, then
, hence
and thus
.
Then we prove surjectivity. Let
be given. Then obviously
is mapped to that element.
Then we prove injectivity. Assume
. Then
, where
and
, that is
for a suitable
. Then
and therefore
.
Theorem 9.10:
functor relating tensor product and fractions
Proof:
- Exercise 9.2.1: Let
be
-modules and
an ideal. Prove that
is a submodule of
and that
(this exercise serves the purpose of practising the proof technique employed for theorem 9.11).
Proof:
Let
and
. Then for all
,
. Hence the theorem by lemma 5.3.
Definition 9.14:
An
-module
is called faithful iff
.
Theorem 9.15:
Let
be a ring. Then
regarded as an
module over itself is faithful.
Proof: Let
such that
. Then in particular
.
Proof:
From the definition it is clear that
, since annihilating all elements of
is a stronger condition than only those of
.
Let now
and
, where
and
. Then
.
Definition 9.17:
Let
be an
-module (where
is a ring) and let
be a prime ideal. Then the localisation of
with respect to
, denoted by
,
is defined to be
with
; note that
is multiplicatively closed because
is a prime ideal.
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.
1. suffices.
Assume that
is a nonzero module, that is, we have
such that
. By theorem 9.11,
is an ideal of
. Therefore, it is contained within some maximal ideal of
, call
(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
we have
and therefore
in
.
The following theorems do not really describe local-global properties, but are certainly similar and perhaps related to those.
Theorem 9.20:
If
is a morphism, then the following are equivalent:
surjective.
surjective for all
multiplicatively closed.
surjective for all
prime.
surjective for all
maximal.
Proof: