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. $\square$

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