Algebras Over a Ring
Let A be any ring; we would like to look at A-modules with a compatible ring structure.
Definition.
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.
Lemma 1.
- 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:
Proof
First claim: clearly is additive and takes 1 to 1. Also
Second claim: verify the axioms as follows.
This completes the proof. ♦
Exercise A
Prove that the constructions in the lemma are mutually inverse.
Definition.
If B is an A-algebra, an A-subalgebra is a subset of B which is an A-submodule as well as a subring.
Exercise B
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. 🙂
Homomorphisms
Definition.
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:
Proposition 1.
Suppose B and B’ are A-algebras corresponding to ring homomorphisms
and
. A homomorphism of A-algebras is precisely a ring homomorphism
such that
.
Proof
(⇒) 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.
Representability
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.
Exercise C
For any A-algebra C, let
Find A-algebras B and B’ such that
.
Note
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).
Definition.
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
.
Definition.
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
.
Proof
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.
In the definition of algebra , should the ” compatible multiplication x: B x B -> B ” be ” A x B -> B” ?
Sorry, my mistake.
I wonder why
has no canonical additive structure (under the section representability). Could you please clarify?
Because if
are homomorphisms of A-algebras, then
is not. For example, the map does not take 1 to 1.
Added a remark in the main text. Thanks!