Lemma 15.9.13. Let $A$ be a ring. Let $M$ be an $A$-module. Then $C = \text{Sym}_ A^*(M)$ is smooth over $A$ if and only if $M$ is a finite projective $A$-module.

Proof. Let $\sigma : C \to A$ be the projection onto the degree $0$ part of $C$. Then $J = \mathop{\mathrm{Ker}}(\sigma )$ is the part of degree $> 0$ and we see that $J/J^2 = M$ as an $A$-module. Hence if $A \to C$ is smooth then $M$ is a finite projective $A$-module by Algebra, Lemma 10.139.4.

Conversely, assume that $M$ is finite projective and choose a surjection $A^{\oplus n} \to M$ with kernel $K$. Of course the sequence $0 \to K \to A^{\oplus n} \to M \to 0$ is split as $M$ is projective. In particular we see that $K$ is a finite $A$-module and hence $C$ is of finite presentation over $A$ as $C$ is a quotient of $A[x_1, \ldots , x_ n]$ by the ideal generated by $K \subset \bigoplus Ax_ i$. The computation of Lemma 15.9.12 shows that $\mathop{N\! L}\nolimits _{C/A}$ is homotopy equivalent to $(K \to M) \otimes _ A C$. Hence $\mathop{N\! L}\nolimits _{C/A}$ is quasi-isomorphic to $C \otimes _ A M$ placed in degree $0$ which means that $C$ is smooth over $A$ by Algebra, Definition 10.137.1. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).