The Stacks project

Lemma 23.7.3. Let $R' \to R$ be a surjection of Noetherian rings whose kernel has square zero and is generated by one element $f$. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Let $\sum r_ j f_ j = 0$ be a relation in $R[x_1, \ldots , x_ n]$. Assume that

  1. each $r_ j$ has coefficients in the annihilator $I$ of $f$ in $R$,

  2. for some lifts $r'_ j, f'_ j \in R'[x_1, \ldots , x_ n]$ we have $\sum r'_ j f'_ j = gf$ where $g$ is not nilpotent in $S/IS$.

Then $S$ does not have finite tor dimension over $R$ (i.e., $S$ is not a perfect $R$-algebra).

Proof. Choose a Tate resolution $R \to A \to S$ as in Lemma 23.6.9. Let $\xi \in H_2(A/IA)$ and $\theta : A/IA \to A/IA$ be the element and derivation found in Lemmas 23.7.1 and 23.7.2. Observe that

\[ \theta ^ n(\gamma _ n(\xi )) = g^ n \]

in $H_0(A/IA) = S/IS$. Hence if $g$ is not nilpotent in $S/IS$, then $\xi ^ n$ is nonzero in $H_{2n}(A/IA)$ for all $n > 0$. Since $H_{2n}(A/IA) = \text{Tor}^ R_{2n}(S, R/I)$ we conclude. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 23.7: Application to complete intersections

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Your email address will not be published.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 09PU. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 23.7.3 as pdf