I define $\mathcal{I}^k(U):=\mathcal{I}(U)^k$ with the obvious restriction maps, for any natural number $k>1$. Now, I suspect that $\mathcal{I}^k$ is not a sheaf in general, since any obvious proof does not seem to go through. Here is my first attempt:
Let $$s_i=\sum_j\prod_{h=1}^k\alpha_{ijh}\in\mathcal{I}^k(U_i),\ \ \alpha_{ijh}\in\mathcal{I}\ \mathrm{for\ all}\ i,j,h$$ and let $$s_i\vert_{U_i\cap U_j}=s_j\vert_{U_i\cap U_j}$$ for all $i,j$, so if $U=\bigcup_i U_i$, then we may glue the $s_i$ to get a unique $s\in\mathcal{I}(U)$. However, I suspect $s$ need not be in $\mathcal{I}^k(U)$. Are there examples of $\mathcal{I}$ such that this fails?
I'm asking this because I'm trying to understand the definition of the sheaf of $\mathcal{O}$-modules, $\mathcal{I}/\mathcal{I}^2$.
