Number of generators of $I\otimes_R I$ Let $R$ be a polynomial ring $K[x_,y,z]$, where $K$ is a field. Let $I$ denote the ideal generated by $x,y,z$. Prove $R$-module $I\otimes_RI$ can be generated by $9$ elements and no fewer than $9$.
I am not sure how to prove the no fewer than $9$ part. I think it might be helpful for one considering the exact sequence
$0\longrightarrow I^2\longrightarrow I\longrightarrow I/I^2\longrightarrow0$
and maybe then take  $(\_ \otimes I)$ to get another right exact sequence. However, I am not sure how to continue.
 A: 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 $I\otimes I$.
