The tensor algebra construction created, from each $R$-module $M$, a "minimal" $R$-algebra $\mathcal{T}(M)$. 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 $\mathcal{T}(M).$ There were no additional properties imposed on $\mathcal{T}(M)$ other than those required to have an $R$-algebra. In particular, the $R$-algebra $\mathcal{T}(M)$ 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 $U:R-\mathbf{\text{CAlg}}\to R-\mathbf{\text{Mod}}$ be the forgetful functor from the category of commutative $R$-algebras to the category of $R$-modules. Then there is a functor $\mathcal{S}:R-\mathbf{\text{Mod}}\to R-\mathbf{\text{CAlg}}$ together with a natural bijection

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

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

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 $\mathcal{S}f:\mathcal{S}(M)\to \mathcal{S}(N)$;

$R$-algebra morphisms to commutative $R$-algebras $\mathcal{S}(M)\to A$ are in natural bijection with $R$-module morphisms $M\to U(A)$. As a special case:

The identity $R$-algebra morphism $\mathcal{S}(M)\to \mathcal{S}(M)$ corresponds to an $R$-module morphism $M\to U(\mathcal{S}(M))$.

Since $\mathcal{S}$ 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 $\mathcal{T}(M)$ with most of the properties we want. To obtain a commutative $R$-algebra, then, it's reasonable to consider a quotient of $\mathcal{T}(M)$ 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 $\mathcal{C}(M)\subseteq \mathcal{T}(M)$ 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

$$\mathcal{S}(M)=\mathcal{T}(M)/\mathcal{C}(M).$$

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

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

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

Optional notation

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

Examples

Let $V$ be an $n$-dimensional vector space over a field $F$. Then $\mathcal{S}(V)$ is isomorphic as a graded $F$-algebra to $F[{x}_{1},\dots ,{x}_{n}]$. If $\mathcal{B}=\{{v}_{1},\dots ,{v}_{n}\}$ is a basis for $V$ as an $F$-vector space, then a basis for ${\mathcal{S}}^{k}(V)$ is$$\{{v}_{1}^{{a}_{1}}{v}_{2}^{{a}_{2}}\cdots {v}_{n}^{{a}_{n}}\mid {a}_{i}\ge 0,{\textstyle \phantom{\rule{0.278em}{0ex}}}{a}_{1}+{a}_{2}+\cdots +{a}_{n}=k\}.$$In particular, the dimension of the $F$-vector space ${\mathcal{S}}^{k}(V)$ is $(\genfrac{}{}{0ex}{}{n+k-1}{n-1})$.