# Algebras Over a Ring

Let *A* be any ring; we would like to look at *A*-modules with a compatible ring structure.

Definition.An –

algebrais 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-

subalgebrais 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

homomorphismof 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(orof 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!