Algebras Over a Ring
Let A be any ring; we would like to look at A-modules with a compatible ring structure.
An –algebra is an -module , together with a multiplication operator such that
- becomes a commutative ring (with 1);
- multiplication is compatible with scalar multiplication:
In the second condition, we have used · for module multiplication and × for ring multiplication. That condition gives us
Hence, for any and we have
An alternative way of describing A-algebras is as follows.
- If B is an A-algebra, the map , is a ring homomorphism.
- Conversely, for any ring homomorphism , B takes the structure of an A-algebra, where the A-module map is:
First claim: clearly is additive and takes 1 to 1. Also
Second claim: verify the axioms as follows.
This completes the proof. ♦
Prove that the constructions in the lemma are mutually inverse.
If B is an A-algebra, an A-subalgebra is a subset of B which is an A-submodule as well as a subring.
If the A-algebra B corresponds to , prove that an A-subalgebra of B is precisely a subring of B containing .
P. S. Please do the above exercises. Don’t be lazy. 🙂
Let B and B’ be A-algebras. A homomorphism of A-algebras from B to B’ is an A-linear map which is also a ring homomorphism.
If we consider the alternative way of defining A-algebras, we get:
Suppose B and B’ are A-algebras corresponding to ring homomorphisms and . A homomorphism of A-algebras is precisely a ring homomorphism such that .
(⇒) Suppose is a homomorphism of A-algebras. For ,
(⇐) Suppose f is a ring homomorphism such that , so for all , . Hence for , ,
The following result follows from the corresponding results for A-modules and rings.
First Isomorphism Theorem.
If is a homomorphism of A-algebras, the image of f is an A-subalgebra of B’ and we get an isomorphism of A-algebras.
For A-algebras B and C, let be the set of A-algebra homomorphisms . Unlike the case of modules, this set has no canonical additive structure, since if are homomorphisms of A-algebras, usually is not (e.g. it does not take 1 to 1).
However, is useful for classifying certain constructions from A-algebras.
As an example fix . Any A-algebra homomorphism corresponds to a pair of elements such that . Thus we have a bijection:
where U(C) is the set of all units of C. Furthermore, any A-algebra homomorphism induces a map
which translates to the map g restricted to .
Thus, we obtain a natural isomorphism . To express this rigorously, one needs the language of category theory.
For any A-algebra C, let
Find A-algebras B and B’ such that
Finally observe that are not just sets, but have group structures as well. To specify these group structures rigorously, one requires B to have a Hopf algebra structure.
Generating a Subalgebra
If B is an A-algebra and is a collection of subalgebras, their intersection is clearly also a subalgebra of B. This follows from the fact that intersection of submodules (resp. subring) is a submodule (resp. subring).
For subset , let be the collection of all A-subalgebras of B containing S. Their intersection is the A-subalgebra of B generated by S; it is denoted by
This generated subalgebra can be concretely described as follows. For a finite sequence , and polynomial in k variables, we take the element . Taking the set of all these elements over all k, all and all then gives us .
An A-algebra B is finitely generated (or of finite type) if for some finite subset .
If , it suffices to take only polynomials f in exactly n variables and consider all , in which case we have a surjective homomorphism of A-algebras
In future articles, if B is an A-algebra, we say B is a finite A-algebra if it is finitely generated as an A-module; we say it is of finite type if it is finitely generated as an A-algebra.
Direct Product of Algebras
Let be a collection of A-algebras. Let be the set-theoretic product; we saw earlier that B has a natural structure of an A-module. It also has a natural structure of a ring via component-wise multiplication: , which is compatible with its A-module structure. Thus:
Theorem (Universal Property of Products).
For each define the projection map
clearly a homomorphism of A-algebras. This collection satisfies the following.
- For any A-algebra C and collection where each is an A-algebra homomorphism, there is a unique A-algebra homomorphism such that for each .
Easy exercise. ♦
Thus is the A-algebra analogue of the direct product. Is there an A-algebra analogue of the direct sum?
The natural inclination is to take the direct sum as an A-module then define multiplication component-wise, but it would not work since does not contain . The correct answer is obtained via tensor products, which will be covered in a later article.
In the next article, we will revisit algebraic geometry with a more general framework.