Let $G$ be the free abelian group on $S = \{a,b,c,d,e,f\},$ let $V$ be the module over $\mathbb{Z}$ with basis $\{uv : u, v \in S\}$ (thus, $\dim V = 21$). A basic element of $V$ is one which is the product of $2$ elements of $G.$ For example, $be, ad, cf, (a+b)(d+e)=ad+ae+bd+be$ are basic elements.
How can we find basic elements $v_1, \dots, v_5 \in V$ such that $\{ad, ae+bd, af+be+cd, bf+ce, cf\} \subseteq \text{Span} \{v_1, \dots, v_5\}$?
The motivation is in the title of the question. I have found a basic $6$ element set: $\{(a+c)(d+f), (a+b)(d+e), (b+c)(e+f), be,ad,cf\},$ and it seems like some basic $5$ element set might work. Because $\mathbb{Z}$ is not a field, I can't think of a way to use Wolfram-Alpha or Mathematica to solve the problem via change of basis.