The product-beating conjecture for non-degenerate genus-one systems

Let $L=(L_1,\ldots,L_t)$ be a non-degenerate genus-one translation-invariant system of equations over $\mathbb{F}_q$, where each $L_j$ is a translation-invariant equation in $s$ variables. A set $A\subseteq\mathbb{F}_q^n$ is $L$-free if the only simultaneous solutions of $L_j(x_1,\ldots,x_s)=0$ for $1\leq j\leq t$ in $A$ are diagonal solutions with $x_1=\cdots=x_s$. The system has genus one when there is no nonempty proper subset $I\subsetneq [s]$ such that $\sum_{i\in I}c_{i,j}=0$ for every $1\leq j\leq t$, where $L_j(x_1,\ldots,x_s)=\sum_{i=1}^s c_{i,j}x_i$. The product-beating conjecture. If $A\subseteq\mathbb{F}_q^n$ is $L$-free and has size $c^n$, then there exists an $L$-free set $B\subseteq\mathbb{F}_q^m$ for some $m>n$ such that

$$|B|>c^m.$$

This would extend the corresponding product-beating result from a single translation-invariant equation to systems of equations; the non-degeneracy assumption is necessary because a degenerate system makes the whole space $L$-free. The claim is presented as an expectation for systems and no resolution is supplied here.