Tensor algebras

Motivation


Given a commutative ring R and two R-modules M and N, the R-module MRN satisfies a universal property that captures the idea of being able to multiply elements in M with elements in N. In the special case M=N, the module MRM gives us a way to compute products of elements in M. It doesn't quite give us a ring structure on M, however, since it doesn't allow us to multiply together more than two elements in M.

Question

Can we amplify our tensor product construction to obtain a module with a full ring structure, capturing the desire of layering a multiplicative operation on the R-module M?

Yes. Let's see how.

A desired universal property


With the above motivation in mind, we are looking for a functor from the category of R-modules to the category of R-algebras that is "free" in the same sense as the free R-module construction on a set S; i.e., that is left adjoint to the corresponding forgetful functor.

A universal property of the tensor algebra

Let U:R-AlgR-Mod be the forgetful functor from the category of R-algebras to the category of R-modules. Then there is functor T:R-ModR-Alg together with a natural bijection

τM,A:HomR-Alg(T(M),A)HomR-Mod(M,U(A)).

In other words, the functor T is a left adjoint of the forgetful functor U.

Even without having seen the construction yet, such a property gives us a way to think about T(M):

The construction


Since MRM captures the idea of multiplying two elements in M, the construction/definition below probably comes as no surprise.

Definition of tensor algebra

Suppose R is a commutative ring (with unity) and M is an R-module. Set T0(M)=R and for each positive integer k define the kth tensor power of M to be the R-module

Tk(M)=MRMRRMk times.

The elements of Tk(M) are called k-tensors.

Then define the tensor algebra of M to be the R-module

T(M)=k=0Tk(M).

Every element of T(M) is a finite formal linear combination of k-tensors.

As the name implies, the R-module T(M) has a (natural) structure of an R-algebra. The multiplication on simple tensors is defined by concatenation of tensors:

(m1mi)(m1mj):=m1mimimj.

The multiplication on sums of tensors is defined via the distributive laws. Note that this multiplication satisfies Ti(M)Tj(M)Ti+j(M). In other words, the tensor algebra T(M) has the structure of a graded ring.

It's worth noting that, as a ring, the tensor algebra T(M) is generated by the elements of T0(M)=R and T1(M)=M.

At some point we should verify that this construction satisfies the claimed universal property. For now, however, let's look at some examples.

Examples


  1. Let R=Z and M=Q/Z. One can verify that Q/ZZQ/Z0, and so the tensor algebra is simply

    T(Q/Z)Z(Q/Z).

    We can also unwind the multiplication rule. Let's look at a specific example. Two elements in Z(Q/Z) are 2[13] and 3[45]. When we add these two elements we get the element

    (2+3)[13+45]=5[1715]=5[215].

    Notice that this is not what we would get conventionally by adding the rational numbers (2+13)+(3+45)=6+215.

    When we multiply these two elements we get the element

    (2[13])(3[45])=(23)(2[45]+3[13])([13][45])=6[135]0=6[35].
  2. Let R=Z and M=Zn. One can verify that ZnZZnZn, and so the tensor algebra is

    T(Zn)ZZnZnZ[x]/(nx).

    For that final isomorphism, the correspondence is that each finite formal sum c0[c1][c2] maps to the coset represented by the polynomial c0+c1x+c2x2+.

  3. Suppose R=F is a field and V is a finite-dimensional F-vector space. Let B={v1,,vn} be a basis for V as an F-vector space. Then a basis for the F-vector space Tk(V) is

    {vi1vi2vikvijB}.

    (When k=0, the basis is simply {1}.) In particular, Tk(V) is an F-vector space of dimension nk.

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