Tensor algebras

Motivation

Given a commutative ring $R$ and two $R$ -modules $M$ and $N$ , the $R$ -module $M \otimes_{R} N$ 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 $M \otimes_{R} M$ 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 $U : R -Alg \to R -Mod$ be the forgetful functor from the category of $R$ -algebras to the category of $R$ -modules. Then there is functor $T : R -Mod \to R -Alg$ together with a natural bijection

τ_{M, A} : {Hom}_{R -Alg} (T (M), A) \tilde{\to} {Hom}_{R -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 $T (M)$ :

It is an $R$ -algebra that we can associate to the $R$ -module $M$ ;
The construction is functorial, so that if $f : M \to N$ is an $R$ -module morphism then there is a corresponding $R$ -algebra morphism $T (f) : T (M) \to T (N)$ ;
$R$ -algebra morphisms $T (M) \to A$ are in natural bijection with $R$ -module morphisms $M \to U (A)$ .
The identity $R$ -algebra morphism $T (M) \to T (M)$ corresponds to an $R$ -module morphism $M \to U (T (M))$ . Classically, this is viewed as an inclusion $M ↪ T (M)$ onto the degree 1 summand. (This is really a component of the unit of the adjunction.)
Since $T$ is a left adjoint it commutes with all colimits; in particular, it commutes with coproducts (which in these categories are called direct sums). In other words $T (⨁_{x \in X} M_{x}) ≃ ⨁_{x \in X} T (M_{x})$ .

The construction

Since $M \otimes_{R} M$ 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 $T^{0} (M) = R$ and for each positive integer $i$ define the $i^{th}$ tensor power of $M$ to be the $R$ -module

T^{i} (M) = \underset{i times}{\underset{⏟}{M \otimes_{R} M \otimes_{R} \dots \otimes_{R} M}} .

The elements of $T^{i} (M)$ are called $i$ -tensors.

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

T (M) = ⨁_{i = 0}^{\infty} T^{i} (M) .

Every element of $T (M)$ is a finite formal linear combination of $i$ -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:

(m_{1} \otimes \dots \otimes m_{i}) \cdot (m_{1}^{'} \otimes \dots \otimes m_{j}^{'}) := m_{1} \otimes \dots \otimes m_{i} \otimes m_{1}^{'} \otimes \dots \otimes m_{j}^{'} .

The multiplication on sums of tensors is defined via the distributive laws. Note that this multiplication satisfies $T^{i} (M) T^{j} (M) \subseteq T^{i + j} (M)$ . 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 $T^{0} (M) = R$ and $T^{1} (M) = M$ .

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

Examples

Let $R = Z$ and $M = Q / Z$ . One can verify that $Q / Z \otimes_{Z} Q / Z ≃ 0$ , and so the tensor algebra is simply
$T (Q / Z) ≃ Z \oplus (Q / Z) .$
We can also unwind the multiplication rule. Let's look at a specific example. Two elements in $Z \oplus (Q / Z)$ are $2 \oplus [\frac{1}{3}]$ and $3 \oplus [\frac{4}{5}]$ . When we add these two elements we get the element
$(2 + 3) \oplus [\frac{1}{3} + \frac{4}{5}] = 5 \oplus [\frac{17}{15}] = 5 \oplus [\frac{2}{15}] .$
Notice that this is not what we would get conventionally by adding the rational numbers $(2 + \frac{1}{3}) + (3 + \frac{4}{5}) = 6 + \frac{2}{15}$ .

When we multiply these two elements we get the element
$\begin{aligned} (2 \oplus [\frac{1}{3}]) \otimes (3 \oplus [\frac{4}{5}]) & = (2 \cdot 3) \oplus (2 \cdot [\frac{4}{5}] + 3 \cdot [\frac{1}{3}]) \oplus ([\frac{1}{3}] \otimes [\frac{4}{5}]) \\ = 6 \oplus [\frac{13}{5}] \oplus 0 \\ = 6 \oplus [\frac{3}{5}] . \end{aligned}$
Let $R = Z$ and $M = Z_{n}$ . One can verify that $Z_{n} \otimes_{Z} Z_{n} ≃ Z_{n}$ , and so the tensor algebra is
$T (Z_{n}) ≃ Z \oplus Z_{n} \oplus Z_{n} \oplus \dots ≃ Z [x] / (n x) .$
For that final isomorphism, the correspondence is that each finite formal sum $c_{0} \oplus [c_{1}] \oplus [c_{2}] \oplus \dots$ maps to the coset represented by the polynomial $c_{0} + c_{1} x + c_{2} x^{2} + \dots$ .
Suppose $R = k$ is a field and $V$ is a finite-dimensional $k$ -vector space. Let $B = {v_{1}, \dots, v_{n}}$ be a basis for $V$ as an $k$ -vector space. Then a basis for the $F$ -vector space $T^{i} (V)$ is
${v_{j_{1}} \otimes v_{j_{2}} \otimes \dots \otimes v_{j_{i}} ∣ v_{j_{l}} \in B} .$
(When $i = 0$ , the basis is simply ${1}$ .) In particular, $T^{i} (V)$ is a $k$ -vector space of dimension $n^{i}$ .

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