Let $(R,\mathfrak m)$ be a Noetherian local ring and $I$, $J$ two ideals of $R$ such that $I$ is generated by an $R/J$-sequence (this means $I=(x_1,\dots,x_t)$ where $x_1,\dots,x_t$ is an $R/J$-sequence). Show that for all $n\geq 1$, $\operatorname{Tor}_1^R(R/I^n,R/J)=0$.
I tried by induction on $n$. If $n=1$ (see exercise 1.2.12(a), Cohen-Macaulay Rings, Bruns-Herzog). Let $n>1$… I can't continue by induction.
I guess to continue the proof the following sequence will be useful $$0\rightarrow I^n/I^{n-1}\rightarrow R/I^n\rightarrow R/I^{n-1}\rightarrow 0.$$
Best Answer
It is well known that $\operatorname{Tor}_1^R(R/I,R/J)\simeq I\cap J/IJ$ for any two ideals $I,J$ of $R$. In our case $\operatorname{Tor}_1^R(R/I^n,R/J)\simeq I^n\cap J/I^nJ$ and we want to show that $I^n\cap J=I^nJ$. Take $z\in I^n\cap J$. Since $I=(x_1,\dots,x_t)$ we can write $z=f(x_1,\dots,x_t)$ where $f\in R[X_1,\dots,X_t]$ is homogeneous of degree $n$. Then $\bar 0=\bar z=\bar f(\bar x_1,\dots,\bar x_t)$, where $\bar{a}$ denotes the residue classe of $a$ modulo $J$. Since $\bar f(\bar x_1,\dots,\bar x_t)=\bar 0\in I^{n+1}(R/J)$, by Bruns and Herzog, Theorem 1.1.7, we get that the coefficients of $\bar f$ are in $I(R/J)$. Therefore there exists a homogeneous polynomial $g$ of degree $n+1$ such that $\bar g(\bar x_1,\dots,\bar x_t)=\bar f(\bar x_1,\dots,\bar x_t)$. Continuing as before we find that the coefficients of $\bar f$ are in $I^m(R/J)$ for all $m\ge 1$. By Krull Intersection Theorem we have $\bigcap_{m\ge 1} I^m(R/J)=(\bar 0)$, so $\bar f=\bar 0$. Now we deduce easily that $z\in I^nJ$: since $\bar f=\bar 0$ the coefficients of $f$ belong to $J$, so each term of $f(x_1,\dots,x_t)$ belong to $I^nJ$, and therefore $z\in I^nJ$.
Remark. Since the case $n=1$ holds for any finite module $M$ instead of $R/J$, I've tried to prove the general case for this setting, but I've failed. However, if $I$ is generated by an $M$-regular sequence which is also an $R$-regular sequence, then $\operatorname{Tor}_i^R(R/I^n,M)=0$ for all $i\ge 1$, $n\ge 1$.