The Stacks project

Lemma 15.66.20. Let $R' \to R$ be a surjective ring map whose kernel is a nilpotent ideal. Let $K' \in D(R')$ and set $K = K' \otimes _{R'}^\mathbf {L} R$. Let $a, b \in \mathbf{Z}$. Then $K$ has tor amplitude in $[a, b]$ if and only if $K'$ does.

Proof. One direction follows from Lemma 15.66.13. For the other, assume $K$ has tor amplitude in $[a, b]$ and let $M'$ be an $R'$-module. We have to show that $K' \otimes _{R'}^\mathbf {L} M'$ has nonzero cohomology only for degrees contained in the interval $[a, b]$.

Let $I = \mathop{\mathrm{Ker}}(R' \to R)$. Then $I^ n = 0$ for some $n$. If $IM' = 0$, then we can view $M'$ as an $R$-module and argue as follows

\[ K' \otimes _{R'}^\mathbf {L} M' = K' \otimes _{R'}^\mathbf {L} (R \otimes _ R^\mathbf {L} M') = (K' \otimes _{R'}^\mathbf {L} R) \otimes _ R^\mathbf {L} M' = K \otimes _ R^\mathbf {L} M' \]

which has nonvanishing cohomology only in the interval $[a, b]$ by assumption on $K$. If $I^{t + 1}M' = 0$, then we consider the short exact sequence

\[ 0 \to IM' \to M' \to M'/IM' \to 0 \]

By induction on $t$ we have that both $K' \otimes _{R'}^\mathbf {L} IM'$ and $K' \otimes _{R'}^\mathbf {L} M'/IM'$ have nonzero cohomology only for degrees in the interval $[a, b]$. Then the distinguished triangle

\[ K' \otimes _{R'}^\mathbf {L} IM' \to K' \otimes _{R'}^\mathbf {L} M' K' \otimes _{R'}^\mathbf {L} M'/IM' (K' \otimes _{R'}^\mathbf {L} IM')[1] \]

proves the same is true for $K' \otimes _{R'}^\mathbf {L} M'$ as desired. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.66: Tor dimension

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Your email address will not be published.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 0H75. Beware of the difference between the letter 'O' and the digit '0'.