Commutative Algebra/Fractions, annihilator
Fractions within rings
editDefinition 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:
- 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.
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
editDefinition 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
- ,
- , and
- ;
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
edit- 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
editDefinition 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
editDefinition 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:
- has property (*).
- has property (*) for all multiplicatively closed .
- has property (*) for all prime ideals .
- 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:
- surjective.
- surjective for all multiplicatively closed.
- surjective for all prime.
- surjective for all maximal.
Proof: