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. $\square$

There are also:

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

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).