Minimum number of generators of the product of ideals

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.

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 $$\mu(I\otimes J) = \mu(I)\mu(J)$$.