In this post, I will introduce yet another “universal” construction over an $R$-module for $R$ a commutative ring. Recall that the tensor algebra can be seen as the free algebra over a given module. It is therefore natural to expect important constructions to be given by quotients of this algebra. Here is a first example : the symmetric algebra.

Definition: The symmetric algebra of an $R$-module $M$ is the quotient of the tensor algebra $T(M)$ by the ideal $C(M)$ generated by the elements of the form $m_1 \otimes m_2 - m_2 \otimes m_1,$ for all $m_1, m_2 \in M.$ It is denoted by $S(M).$

Let’s see what we got here. Denoting by $C^k(M)$ the set of all finite linear combinations of elements in $T^k(M)$ of the form

$m_1 \otimes \cdots \otimes m_{i-1} \otimes (m_i \otimes m_{i+1} - m_{i+1} \otimes m_i) \otimes m_{i+2} \otimes \cdots \otimes m_k,$

we have $C(M) = \bigoplus_{k=2}^{\infty}C^k(M).$ Notice that by expanding this expression, we see that $C^k(M)$ is generated by the elements of $T^k(M)$ which are the difference of two k-tensors that differ only by a transposition of two adjacent entries (the i’th and (i+1)’th entries are swapped). Therefore, quotienting $T(M)$ by $C(M)$ result in giving us the right to swap two adjacent entries, ie:

$m_1 \otimes \cdots m_i \otimes m_{i+1} \otimes \cdots \otimes m_k \equiv m_1 \otimes \cdots \otimes m_{i+1} \otimes m_i \otimes \cdots \otimes m_k$

(modulo $C(M)).$ But since for any $k,$ adjacent transpositions in $S_k$ generate all of the permutation group $S_k,$ tensors in $T(M)$ which can be obtained from each others by permuting their entries are all identified in $S(M).$ In other words, the k’th symmetric power $T^k(M)/C^k(M),$ denoted $S^k(M),$ is equal to $T^k(M)$ modulo the submodule generated by all elements of the form

$m_1 \otimes m_2 \otimes \cdots \otimes m_k - m_{\sigma(1)} \otimes m_{\sigma(2)} \otimes \cdots \otimes m_{\sigma(k)}$

for all $m_i \in M$ and all $\sigma \in S_k.$

In particular, $S(M)$ is a commutative algebra. In fact, it satisfies the universal property with respect to maps from a module to a commutative algebra. It is also universal with respect to symmetric multilinear maps. Recall that a k-multilinear map $\varphi : M^k \to N$ is symmetric if $\varphi(m_1, \ldots, m_k) = \varphi(m_{\sigma(1)}, \ldots \varphi(m_{\sigma(k)}.$

Theorem:

1. (Universal Property for Symmetric Multilinear Maps) Let $\varphi : M^k \to N$ be a symmetric k-multilinear map with respect to $R,$ then there is a unique $R$-module homomorphism $\phi : S^k(M) \to N$ such that $\varphi = \phi \circ \iota$ where $\iota$ is the canonical inclusion map  from $M^k$ to $T^k(M)$ modulo $C(M).$
2. (Universal Property for maps to commutative $R$-algebras) Let $A$ be a commutative $R$-algebra and $\varphi : M \to A$ an $R$-module homomorphism, then there is a unique $R$-algebra homomorphism $\phi : S(M) \to A$ such that $\phi|_M = \varphi.$

Proof: The proof of both assertions is very similar to the proof of the universal property of tensor algebras. For 1, the map is multilinear so it induces a unique homomorphism $\tilde{\phi}: T^k(M) \to N$ such that

$\tilde{\phi}(m_1 \otimes \cdots \otimes m_k) = \varphi(m_1, \ldots, m_k).$

But then $\tilde{\phi}$ is symmetric thus $C^k(M)$ is contained in the kernel of $\tilde{\phi}.$ Hence, the $R$-module homomorphism $\phi : S^k(M) \to N$ defined by

$\phi (m_1 \otimes \cdots \otimes m_k \quad \text{mod }C(M)) = \tilde{\phi}(m_1 \otimes \cdots \otimes m_k)$

is well defined and has the desired properties.

For 2, we proceed as in the proof linked above. For $\varphi : M \to A,$ we define $\phi : S(M) \to A$ by

$\phi(m_1 \otimes \cdots \otimes m_k \quad \text{mod }C(M)) = \varphi(m_1) \cdots \varphi(m_k)$

which is well-defined since $A$ is commutative.

QED

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.