The Stacks project

Lemma 91.12.3. Let $A$ be a ring. Let $P = A[E]$ be a polynomial ring. Set $I = (e; e \in E) \subset P$. The maps $\text{Tor}_ i^ P(A, I^{n + 1}) \to \text{Tor}_ i^ P(A, I^ n)$ are zero for all $i$ and $n$.

Proof. Denote $x_ e \in P$ the variable corresponding to $e \in E$. A free resolution of $A$ over $P$ is given by the Koszul complex $K_\bullet $ on the $x_ e$. Here $K_ i$ has basis given by wedges $e_1 \wedge \ldots \wedge e_ i$, $e_1, \ldots , e_ i \in E$ and $d(e) = x_ e$. Thus $K_\bullet \otimes _ P I^ n = I^ nK_\bullet $ computes $\text{Tor}_ i^ P(A, I^ n)$. Observe that everything is graded with $\deg (x_ e) = 1$, $\deg (e) = 1$, and $\deg (a) = 0$ for $a \in A$. Suppose $\xi \in I^{n + 1}K_ i$ is a cocycle homogeneous of degree $m$. Note that $m \geq i + 1 + n$. Then $\xi = \text{d}\eta $ for some $\eta \in K_{i + 1}$ as $K_\bullet $ is exact in degrees $ > 0$. (The case $i = 0$ is left to the reader.) Now $\deg (\eta ) = m \geq i + 1 + n$. Hence writing $\eta $ in terms of the basis we see the coordinates are in $I^ n$. Thus $\xi $ maps to zero in the homology of $I^ nK_\bullet $ as desired. $\square$


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Your email address will not be published.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 08RE. Beware of the difference between the letter 'O' and the digit '0'.