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Let $R:=\mathbb{Z}[X_1,X_2,\dots,X_{mn}]$. Suppose $A=(f_{ij})$ is a $m\times n$ matrix with entries in $R$ such that

(1)there is no zero column in $A$;

(2)for each $i,j$, either $f_{ij}=0$ or $f_{ij}=X_k$ for some $k\in \{1,2,…,mn\}$;

(3)if $f_{ij}\neq 0$ and $f_{i'j'} \neq 0$, then $f_{ij} \neq f_{i'j'}$.

Is it true that if there exists nonzero $f_1,f_2,\dots,f_m$(here $f_i\neq 0$ for every $i$) in $R$ such that $(f_1\ f_2\ \dots\ f_m)A=0$, then we must have $m>n$?

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I took a stab at texing it because it was nearly unreadable. @woa please do your best to help clarify the question, and check to make sure I've been faithful to what you intended. – rschwieb Oct 16 '12 at 16:32
If $A = 0$ with $n \ge m$ (which you don't exclude), then $f_i = 1$ gives a non zero solution. – Joel Cohen Oct 16 '12 at 18:30
Simul-posted to (but quickly closed on) MO,… – Gerry Myerson Oct 17 '12 at 6:40
I have edited the title to make it more informative. I hope I haven't missed the point of the question. If anyone has a better title, go for it. – Gerry Myerson Oct 17 '12 at 21:53
The problem reduces easily to square matrices, and what we have to prove is the following: for such square matrices can we have a $R$-linear combination of their rows with all coefficients $\neq 0$? – user26857 Dec 3 '12 at 22:34

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