Fractions within rings
Let be a commutative ring, and let be an arbitrary subset. is called multiplicatively closed iff the following two conditions hold:
Let be a ring and a multiplicatively closed subset. Define
where the equivalence relation is defined as
Equip this with addition
The following two lemmata ensure that everything is correctly defined.
is an equivalence relation.
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.
The addition and multiplication given above turn into a ring.
We only prove well-definedness; the other rules follow from the definition and direct computation.
Let thus and .
Thus, we have and for suitable .
These translate to
for suitable . We get the desired result by picking and observing
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.
If , then the rules for multiplication for indicate that is an inverse for .
Assume . Then there exists such that .
Let be an arbitrary element of . Then .
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
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.
Category theory context
Fractions within modules
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),
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.
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).
to get from the second to the first row, we note for a suitable , and in particular for example
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 .
functor relating tensor product and fractions
Let be -modules and multiplicatively closed. Then
- 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
Let be a ring, a module over and an arbitrary subset. Then the annihilator of with respect to is defined to be the set
Let be a ring, a module over and an arbitrary subset. Then is an ideal of .
Let and . Then for all , . Hence the theorem by lemma 5.3.
An -module is called faithful iff .
Let be a ring. Then regarded as an module over itself is faithful.
Proof: Let such that . Then in particular .
Let be an -module and an arbitrary subset. Let be the submodule of generated by . Then .
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 .
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.
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 .
Being equal to zero is a local-global property.
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.
If is a morphism, then the following are equivalent:
- surjective for all multiplicatively closed.
- surjective for all prime.
- surjective for all maximal.