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Commutative Algebra/Fractions, annihilator

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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:

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: