Commutative Algebra/Fractions, annihilator

Fractions within rings

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Definition 9.1:

Let   be a commutative ring, and let   be an arbitrary subset.   is called multiplicatively closed iff the following two conditions hold:

  1.  
  2.  

Definition 9.2:

Let   be a ring and   a multiplicatively closed subset. Define

 ,

where the equivalence relation   is defined as

 .

Equip this with addition

 

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

 

and

 .

These translate to

 

and

 

for suitable  . We get the desired result by picking   and observing

 

and

 . 

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:

  1.   is a unit.
  2.   for some  .
  3. Every element of   has the form   for suitable  ,  .
  4. Let   be ideals. Then  , where
 .
  1. 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.

 

5.

Let  , that is,  . Then  , where   is a unit in  . Further,   is an ideal within   since   is a morphism. Thus,  . 

Theorem 9.6 (universal property):

Let   be a ring,   multiplicatively closed, let   be another ring and let

 

be a morphism, such that for all  ,  . Then there exists a unique morphism

 

such that

 .

Proof:

We first prove uniqueness. Assume there exists another such morphism  . Then we would have

 .

Then we prove existence; we claim that

 

defines the desired morphism.

First, we show well-definedness.

Firstly,   exists for  .

Secondly, let  , that is,  . Then

 

The multiplicativity of this morphism is visually obvious (use that   is a morphism and commutativity); additivity is proven as follows:

 

It is obvious that the unit is mapped to the unit. 

Theorem 9.7:

Category theory context

Fractions within modules

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Definition 9.8:

Let   be a ring,   a multiplicative subset of   and   an  -module. Set   to be the ring   augmented by inverses of  . We define the  -module   as follows:

  (the formal fractions),

where again

 ,

with addition

 

and module operation

 .

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.

Theorem 9.9 (properties of the augmented module):

Let   be an  -module, let   be a multiplicatively closed subset of  , and let   be submodules. Then

  1.  ,
  2.  , and
  3.  ;

in the first two equations, all modules are seen as submodules of   (as above with  ), and in the third isomorphy relation, the modules are seen as independent  -modules.

Proof:

1.

 

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.

 

to get from the second to the first row, we note   for a suitable  , and in particular for example

 ,

where  .

3.

We set

 

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

Theorem 9.11:

Let   be  -modules and   multiplicatively closed. Then

 .

Proof:

Exercises

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  • 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).

The annihilator, faithfulness

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Definition 9.12:

Let   be a ring,   a module over   and   an arbitrary subset. Then the annihilator of   with respect to   is defined to be the set

 .

Theorem 9.13:

Let   be a ring,   a module over   and   an arbitrary subset. Then   is an ideal of  .

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  . 

Theorem 9.16:

Let   be an  -module and   an arbitrary subset. Let   be the submodule of   generated by  . Then  .

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  . 

Local properties

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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.

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.   has property (*).
  2.   has property (*) for all multiplicatively closed  .
  3.   has property (*) for all prime ideals  .
  4.   has property (*) for all maximal ideals  .

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:

  1.   surjective.
  2.   surjective for all   multiplicatively closed.
  3.   surjective for all   prime.
  4.   surjective for all   maximal.

Proof: