Symmetric algebras

Motivation

The tensor algebra construction created, from each $R$ -module $M$ , a "minimal" $R$ -algebra $T (M)$ . In other words, beginning from the additive operation in $M$ and the $R$ -scaling on $M$ , it created a structure that also had an internal multiplication (compatible with those structures). The universal property formally encoded the "minimality" of the construction, in that all $R$ -module morphisms from $M$ to $R$ -algebras $A$ "lifted" to $R$ -algebra morphisms from $T (M) .$ There were no additional properties imposed on $T (M)$ other than those required to have an $R$ -algebra. In particular, the $R$ -algebra $T (M)$ was not guaranteed to be commutative (and rarely ever is).

Can we modify our construction so that the $R$ -algebra we obtain is also commutative?

The desired universal property

As with the tensor algebra functor, there is a functor from the category of $R$ -modules to the category of commutative $R$ -algebras. It is analogous to any free construction and the tensor algebra construction, in that it is left adjoint to a forgetful functor:

(Primary) Universal property of the symmetric algebra

Let $U : R - CAlg \to R - Mod$ be the forgetful functor from the category of commutative $R$ -algebras to the category of $R$ -modules. Then there is a functor $S : R - Mod \to R - CAlg$ together with a natural bijection

{Hom}_{R - CAlg} (S (M), A) \tilde{\to} {Hom}_{R - Mod} (M, U (A)) .

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

As with any object satisfying a universal property, we can now deduce many properties of $S (M)$ :

It is a commutative $R$ -algebra 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 $S f : S (M) \to S (N)$ ;
$R$ -algebra morphisms to commutative $R$ -algebras $S (M) \to A$ are in natural bijection with $R$ -module morphisms $M \to U (A)$ . As a special case:
- The identity $R$ -algebra morphism $S (M) \to S (M)$ corresponds to an $R$ -module morphism $M \to U (S (M))$ .
Since $S$ is a left adjoint it commutes with all colimits; in particular, it commutes with coproducts.

The construction

We already have a construction that takes an $R$ -module $M$ and creates an $R$ -algebra $T (M)$ with most of the properties we want. To obtain a commutative $R$ -algebra, then, it's reasonable to consider a quotient of $T (M)$ that forces a commutativity relation in the quotient ring.

Definition of symmetric algebra

Suppose $R$ is a commutative ring (with unity) and $M$ is an $R$ -module. Let $C (M) \subseteq T (M)$ be the ideal generated by elements of the form $m_{1} \otimes m_{2} - m_{2} \otimes m_{1}$ for $m_{1}, m_{2} \in M$ . The symmetric algebra of $M$ is quotient

S (M) = T (M) / C (M) .

Some notes are in order. First, the ideal $C (M)$ is generated by homogeneous elements of degree 2, which implies $C (M)$ is a graded ideal (with degree $k$ the submodule denoted $C^{k} (M)$ ) and the quotient ring $S (M)$ is a graded ring. The homogeneous component of degree $k$ is

S^{k} (M) = T^{k} (M) / C^{k} (M) .

This $R$ -module is called the $k^{th}$ symmetric power of $M$ . One can show that the submodule $C^{k} (M)$ is generated by all elements of the form

(m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}) - (m_{σ (1)} \otimes m_{σ (2)} \otimes \dots \otimes m_{σ (k)},

where $m_{i} \in M$ and $σ \in S_{k}$ .

Note that since $C (M)$ is generated (as an ideal) by degree 2 homogenous elements, we have $C^{0} (M) = C^{1} (M) = 0$ and hence have $S^{0} (M) = R$ and $S^{1} (M) = M$ .

Optional notation

It is common to drop the tensor symbol between elements when working in $S (M)$ ; e.g., to write simply $m_{1} m_{2}$ rather than $m_{1} \otimes m_{2} + C (M)$ for the image in $S (M)$ of the element $m_{1} \otimes m_{2} \in T (M)$ .

Examples

Let $V$ be an $n$ -dimensional vector space over a field $F$ . Then $S (V)$ is isomorphic as a graded $F$ -algebra to $F [x_{1}, \dots, x_{n}]$ . If $B = {v_{1}, \dots, v_{n}}$ is a basis for $V$ as an $F$ -vector space, then a basis for $S^{k} (V)$ is ${v_{1}^{a_{1}} v_{2}^{a_{2}} \dots v_{n}^{a_{n}} ∣ a_{i} \geq 0, a_{1} + a_{2} + \dots + a_{n} = k} .$ In particular, the dimension of the $F$ -vector space $S^{k} (V)$ is $(\binom{n + k - 1}{n - 1})$ .