The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07M6. Beware of the difference between the letter 'O' and the digit '0'.