Is the symmetric algebra direct sum of $k$-th symmetric powers?

Question

Let $R$ be a commutative ring and $M$ be an $R$-module.

Let $S(M)$ be the symmetric algebra of $M$ and $S^n(M)$ be $n$-th symmetric power of $M$.

Then is $S(M)$ the internal direct sum of $S^n(M)$? Why? If not, what is an example?

Answer 1

The direct sum $\bigoplus S^n(M)$ is a commutative algebra over $R$ by concatenation of monomials and the fact that $S^0(M) = R$. Given any commutative algebra $T$ over $R$ equipped with an $R$-module map $f:M \rightarrow T$, we can define a map $\bigoplus S^n(M) \rightarrow T$ by just taking $m_1 \otimes \ldots \otimes m_k \mapsto f(m_1)\cdots f(m_k)$. The fact that $T$ is commutative and $f$ is an $R$-module map means this is well-defined, and it's a map of algebras. And any other algebra map $g:\bigoplus S^n(M) \rightarrow T$ that agrees with $f$ as an $R$-module map has to do the same thing on $S^1(M) = M$, which is enough to see that $g = f$.

This means that $\bigoplus S^n(M)$ is a perfectly good model of the free functor $$free:R\text{-mod} \rightarrow \text{comm-}R\text{-alg}$$

and so your favorite definition of $S(M)$ has a canonical isomorphism with the model I just gave, and so inherits an internal grading from it.

(Please check everything I've said more carefully than I have and let me know if I overlooked something...)