# When nilpotency of an ideal implies vanishing of the tensor product?

I have some questions relating the nilpotency of an ideal $$I$$ (i.e. $$\exists m\in\mathbb{N}$$ s.t. $$I^m=0$$ ) over a (non-necessarily commutative) ring $$R$$ and the vanishing, for some $$n\in\mathbb{N}$$, of the tensor product $$I\otimes_R I\otimes_R...\otimes_R I=0$$ ($$n$$ times):

1. With the notations above, is it true in general that if there exists some $$m$$ s.t. $$I^m=0$$, then there exists some $$n$$ s.t. ($$n$$ times) $$I\otimes_R I\otimes_R...\otimes_R I=0$$ ?

I think I have managed to prove it if we assume furthermore that $$I$$ is flat.

Here it is a sketch of the proof:

Let $$m$$ be the smallest natural number s.t. $$I^m=0$$. Consider now, for any $$1\leq r\leq m-1$$ the short exact sequence $$\begin{equation} 0\rightarrow I^r\rightarrow R\rightarrow R/I^r\rightarrow 0 \end{equation}$$ given by inclusion and projection. Then tensoring with $$I$$ we get (after identifying $$I\otimes_R R$$ with $$I$$) the exact sequence $$\begin{equation} 0\rightarrow I^r\otimes_{R} I \rightarrow I\rightarrow R/I^r \otimes_{R} I\rightarrow 0 \end{equation}$$ because $$I$$ is flat. The map $$I^r\otimes_{R} I \rightarrow I$$, call it f, is given in this case by $$f(x\otimes a)=xa$$. So Imf$$=I^{r+1}$$ and since $$f$$ is a monomorphism we get $$I^r\otimes_{R} I\cong I^{r+1}$$. Since this happens for every $$1\leq r\leq m-1$$, it follows inductively that ($$m$$ times) $$I\otimes_{R}...\otimes_{R} I=I^m=0$$.

The fact that $$I$$ is flat has been used only to get, for every $$1\leq r\leq m-1$$, from $$(1)$$ the short exact sequence $$(2)$$; so I guess the same holds whenever tensoring with $$I$$ "preserves" those short exact sequences.

What about the general case, is it true?

1. In case it is not true, can we characterize the rings which satisfy that for every nilpotent ideal $$I$$ there exists some $$n$$ s.t. ($$n$$ times) $$I\otimes_R I\otimes_R...\otimes_R I=0$$, or at least give a sufficient condition for that to happen?

Thank you.

This doesn't hold for the ideal $$I=(x)$$ in $$k[x]/(x^2)$$ over some field $$k$$. All $$m$$-fold tensor powers of $$I$$ are 1-dimensional over $$k$$, even though the ideal is nilpotent.
You need something the vanishing of $$Tor_1^R(I/I^r,I)$$ for all $$r$$ for your argument to work, so the flatness of either of the argumebts is sufficient. It should also be close to being a necessary condition, but maybe you can play around and see for which $$r$$ this must jold.