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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.2012-10-16
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    If $A = 0$ with $n \ge m$ (which you don't exclude), then $f_i = 1$ gives a non zero solution.2012-10-16
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    Simul-posted to (but quickly closed on) MO, http://mathoverflow.net/questions/109829/i-have-a-question-in-abstract-algebra-i-think-it-is-right-but-i-have-no-idea-how2012-10-17
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    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.2012-10-17
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    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 $\neq 0$?2012-12-03
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    @YACP, yes, posted 16 Oct, closed 17 Oct, deleted 19 Oct. On MO, people are expected to clean up their own questions, which OP made no effort to do. The customs at m.se are different.2013-02-12

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