Algebras

We've seen how the notion of a ring acting on an abelian group leads to the structure of a module. Can a ring act on another ring? Or, to phrase the question a bit differently, if a ring R acts on a module M, is it possible for M to have a second operation that's both compatible with the given R-action and makes M into a ring? These two questions lead to the following two equivalent definitions of a structure known as an algebra.

Rings have unity

Recall our standing assumption on rings, namely that all rings have a multiplicative identity ("unity") and that ring morphisms send unities to unities. In other words, our rings are always drawn from the category Ring, of rings with unity together with unity-preserving ring morphisms.

Definition of an algebra (via modules)

Let R be a commutative ring. An R-algebra is an R-module A that is also equipped with a multiplication that makes A into a ring, with the following compatibility property between the R-action and the multiplication in A:

r(a1a2)=(ra1)a2=a1(ra2)

for all rR and a1,a2A.

Definition of an algebra (via rings)

Let R be a commutative ring. An R-algebra is a ring A together with a ring morphism f:RA whose image is contained in the center of A.

Let's quickly verify these two definitions are actually equivalent. First suppose A is an R-algebra in the first sense. For the sake of this analysis, let's use a to denote the action of an element rR on an element aA, and reserve a (or no notation at all) for a product of elements in A. Then A is a ring (with unity) and we can consider the map f:RA defined by rr1A. We claim this is a ring morphism whose image is in the center of A. First note we certainly have f(1R)=1R1A=1A, since part of the assumption of the R-action on the module A is that the identity element 1R acts as the identity on
A. Next note that properties of the R-action on the module A guarantee that

f(r+r)=(r+r)1A=r1A+r1A=f(r)+f(r).

Finally, observe that

f(rr)=(rr)1A=r(r1A)(by the properties of the R-action on the module A)=rf(r)=r(1Af(r))=(r1A)f(r)(by compatibility of the R-action with the product in A)=f(r)f(r).

So, our map f:RA really is a ring morphism. Moreover, for every rR the compatibility condition guarantees that for every aA we have

f(r)a=(r1A)a=r(1Aa)=r(a1A)=a(r1A)=af(r).

Thus, f(r) is in the center of A.

Conversely, suppose f:RA is a ring morphism whose image is contained in the center of A. Then A is an abelian group (under its additive operation) and we can define a set map :R×AA by ra=f(r)a. We claim this defines a left action of R on A. First note that

(r+r)a=f(r+r)a=(f(r)+f(r))a=f(r)a+f(r)a=ra+ra,

and

(rr)a=f(rr)a=f(r)f(r)a=rf(r)a=r(ra).

We also have

r(a1+a2)=f(r)(a1+a2)=f(r)a1+f(r)a2=ra1+ra2

and

1Ra=f(1R)a=1Aa=a.

So, we have indeed defined a left action of R on A, giving A the structure of an R-module. We also have

r(a1a2)=f(r)(a1a2)=(f(r)a1)a2=(ra1)a2,

and also (since the image of f is in the center of A)

r(a1a2)=f(r)(a1a2)=(f(r)a1)a2=(r1f(r))a2=r1(f(r)a2)=r1(ra2).
Associative? Unital?

We assume rings have unity, which means we're assuming every algebra also has unity. There is an alternative definition without that assumption, which one would call a non-unital algebra.

There is also an alternative definition that results in a very similar structure to an algebra, with the notable exception that the multiplication in A is not (assumed to be) associative. When the multiplication is not associative such a structure is called a non-associative algebra.

We will not worry about these slightly more general structures.

Examples

  1. Every ring is a Z-algebra. For each ring A, there is a unique ring morphism ZA and the image of that ring morphism is always contained in the center of A.
  2. If A is a commutative ring, then A is itself an A-algebra. More generally, if A is a ring and RA is a subring of the center of A, then A is an R-algebra.
  3. The ring Mn(R) of n×n matrices with entries in a commutative ring R is an R-algebra. The ring morphism f:RMn(R) sends each ring element r to rIn, the diagonal matrix with r on the diagonal.
  4. More generally, if M is an R-module then the endomorphism ring EndR(M) is an R-algebra.
  5. The field of complex numbers C is a commutative R-algebra, via the inclusion morphism RC.
  6. The quaternions H is an R-algebra but not a C-algebra, as the complex numbers are not in the center of the quaternions.
  7. The polynomial ring R[x1,,xn] is the free commutative R-algebra on the set {x1,,xn}.
  8. The tensor algebra T(M) of an R-module M is an R-algebra.
  9. The symmetric algebra S(M) and exterior algebra (M) of an R-module M are both R-algebras.

Note that in the above examples we often said things like "Every ring A is a Z-algebra ... ", which is somewhat incorrect. We should really be careful to specify the categories in which we are working, saying things like "There is a functor F:RingZ-Alg with object function sending each ring A to the Z-algebra defined by the unique ring morphism ZA." Can you figure out how to make each example above formally correct?

Morphisms of algebras

As might be expected, morphisms of R-algebras should be maps that respect "the algebraic structures." From the ring-centric definition, that would mean:

Definition of an algebra morphism (via rings)

Suppose A and B are two R-algebras. An R-algebra morphism from A to B is a ring morphism ϕ:AB such that ϕ(ra)=rϕ(a) for every rR and aA.

In other words, if f:RA and g:RB are the ring morphisms giving A and B their R-actions, then we should have a commutative diagram

Definition of algebra morphism (via modules)

Suppose A and B are two R-algebras. An R-algebra morphism from A to B is an R-module morphism ϕ:AB that is also a ring morphism of A and B as rings. In other words, it is a set map satisfying the following properties:

  1. ϕ(1A)=1B
  2. ϕ(a1+a2)=ϕ(a1)+ϕ(a2) for all a1,a2A
  3. ϕ(a1a2)=ϕ(a1)ϕ(a2) for all a1,a2A
  4. ϕ(ra)=rϕ(a) for all rR and aA.

We can now form the category R-Alg whose objects are R-algebras and morphisms are morphisms of R-algebras. Note that, by construction, we have forgetful functors both to the category Ring of rings and to the category R-Mod of R-modules:

The forgetful functor to rings is not full (i.e., surjective on hom sets): for two R-algebras A and B, there are ring morphisms ψ:U2(A)U2(B) that are not compatible with the given R-actions (and hence are not of the form U2(ϕ) for some R-algebra morphism ϕ:AB).

The same is true for the forgetful functor to R-modules, for analogous reasons.

Category theoretic interpretation

Much like a ring is a monoid object in the category of abelian groups, an R-algebra is a monoid object in the category of R-modules. Since I haven't (yet) typed up any notes on monoidal categories and monoid objects, I won't say any more here.


Suggested next notes

Tensor algebras
Symmetric algebras
Exterior algebras