Let $L_1, \ldots, L_m \in \mathbb{Z}[x_1, \ldots, x_n]$ be polynomials of the form $L_i = l_{i1} \cdot l_{i2} \ldots \cdot l_{ik}$, where every $l_{ij}$ is an integer linear form.

Assume that magnitudes of all coefficients of all $l_{ij}$ are bounded by some integer $H$. (So, every $l_{ij}$ has form $A_1 \cdot x_1 +\ldots + A_n \cdot x_n$, where every $A_i$ is an integer such that $|A_i| \le H$.)

Suppose that polynomials $L_1, \ldots, L_m$ are linearly dependent over $\mathbb{Z}$, i.e. there exists $B_1, \ldots, B_m \in \mathbb{Z}$ (not all equal to zero) such that $B_1\cdot L_1 + \ldots + B_m\cdot L_m \equiv 0$.

Is it true that these coefficients $B_1, \ldots, B_m$ can be chosen in such a way so that every $|B_i| \le H^{\text{poly}(n,m,k)}$ for some polynomial $\text{poly}(n,m,k)$?

This question is motivated by studing depth-$3$ arithmetic circuits.


Yes. The coefficients of the $L_j$ are $O(k! H^k)$ (the $k!$ is a bound for how many different products can contribute to the same term, a multinomial coefficient with $k$ on top). The $B_i$ are given as the kernel of a matrix whose entries are the coefficients of $L_j$. That kernel can be computed by Cramer's rule, giving $m \times m$ determinants, so the $B$'s are $O((m!) (k!)^m H^{mk}) = O(H^{m k} e^{m k \log k + m \log m})$.

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