Minimum number of generators of the product of ideals

Question

Let $k$ be an algebraically closed field, let $I, J \subseteq k[x_1,\dots, x_n, y_1,\dots, y_m]$ be ideals, and let $i,j$ be the minimum number of generators of $I$ and $J$, respectively. I have two questions:

1. Is the minimum number of generators of the product ideal $IJ$ at least $\max\{i,j\}$?

2. If $I$ is generated by polynomials that depend only on $x_1,\dots, x_n$ and $J$ is generated by polynomials that depend only on $y_1,\dots, y_m$, then is the minimum number of generators of the product ideal $IJ \subseteq k[x_1,\dots, x_n, y_1,\dots,y_m]$ equal to $i j$?

Question two is a repost from MSE.

Answer 1

1. No. $k[x,y]$ we can find an ideal $I$ with $m$ generators for any $m\geq 5$ but $I^2$ has $9$ generators. See this paper.

2. Yes, at least in the graded case. There is probably a better way to show this, but from the top of my head here is one approach: show that $R/I, R/J$ are Tor-independent. It follows that the natural map from $I\otimes_R J\to IJ$ is an isomorphism. Finally, it is easy to see that $\mu(I\otimes J) = \mu(I)\mu(J)$.