Suppose $G, G'$ are Grobner bases for the ideals $I, I' \subseteq F[x_1, \dots, x_n]$ respectively. Then is it the case that $GG'$ is a Grobner bases for the ideal $II'$?
My intuition on this is no, but I am unable to come up with a concrete counterexample.
My own attempt:I have done a bunch of scratchwork, a mess of polynomial multiplications and coefficients, but I don't think I have much to concretely show unfortunately.
Let $G = \{ x,y \}$ and $G' = \{ x,y+1 \}$ in $\mathbb{Q}[x,y]$ with the lexicographic monomial ordering.
The sets of generators $G$ and $G'$ are Groebner bases of $I=\langle G \rangle$ and $I' = \langle G' \rangle$ respectively.
The product set $G \cdot G' = \{ x^2, xy+x, xy, y^2+y \}$ is not a Groebner basis of the product ideal $I \cdot I'$.
Indeed, $x = (xy+x)-xy \in I \cdot I'$, but there exists no $g \in G \cdot G'$ with $LT(g) \mid LT(x) = x$.