Considering the product ideal $IJ = \{ \sum_{i=1}^n a_ib_i | a_i \in I, b_i \in J \forall i\}$, I've always seen it written that the more naive notion $IJ = \{ ij | i \in I, j \in J\}$ is not an ideal because in general it is not additively closed and is thus not a subring. I am unable to come up with a proof or any examples.
Any ideas?
I think I figured out an example thanks to Mike's comment and another post found here: Defining the Product of Ideals
Let $\mathbb{R}[x,y]$ be a two-variable polynomial ring over $\mathbb{R}$, and consider the ideal $(x,y)$. Notably $x^2$ is in $(x,y).(x,y) = \{ p(x,y)q(x,y) | p(x,y) \in (x,y) \text{ and } q(x,y) \in (x,y)\}$ as is $y^2$, but $x^2+y^2$ is not in $(x,y).(x,y)$ (as it cannot be written as the product of two polynomials, it is irreducible). So then, in general, $I.J = \{ ij | i \in I, j \in J\}$ is not an ideal.