Lemma 92.12.3. 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
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)