Number of generators of $I\otimes_R I$

Solution 1:

To provide some broader insight behind the solution discussed in the comments, very frequently the way to prove statements like this is to reduce to linear algebra. Understanding modules over a general ring is hard, but over a field, it is easy since we have the full theory of linear algebra available. An enormous number of complicated problems about modules over rings are solved by doing a lot of reductions that eventually bottom out with a statement about modules over fields which is easy.

In particular, in this case, you want to show a certain module cannot be generated by fewer than $9$ elements. If you were working over a field, this would be easy: you would just have to show its dimension is at least $9$. So, the idea is, you find a way to change the problem into one about a module over a field. In this case, you can do that by tensoring with the residue field $R/I\cong K$, so you can consider $R/I\otimes (I\otimes I)$ (or $I/I^2\otimes I$, to relate this to the exact sequence you were looking at). Since $R/I\otimes I\cong I^2/I$ is just a $3$-dimensional vector space over $R/I$, $R/I\otimes(I\otimes I)$ is a tensor product of two copies of this $3$-dimensional vector space, which is $9$-dimensional. So, $R/I\otimes (I\otimes I)$ cannot be generated (as an $R/I$-module, or equivalently as an $R$-module) by fewer than $9$ elements, and thus neither can $I\otimes I$.