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Given an $m\times n$ (with $n>m)$ matrix $M$ over a polynomial ring $R=k[x_1,...,x_n]$, suppose that every column of $M$ is an $R$-linear combination of $m$ specified columns. I would like to explicitly find these linear combinations. Are there any programs that would allow me to do this?

Of course, I can do this manually (and I have so far), but I am dealing with matrices with over a 100 polynomials and it's taking up a lot of time.

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    What do you mean by "rank" here? Do you mean the rank over the fraction field of $R$?2011-04-24
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    The column space of the matrix is a submodule of $R^m$. Submodules of free modules are free. So the rank of this submodule is well defined, unless I am missing something.2011-04-24
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    @Milo: your second statement is false. See http://mathoverflow.net/questions/16953/are-submodules-of-free-modules-free . So what condition do you actually want? That the column space is all of $R^m$?2011-04-24
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    @Qiaochu: It states in Hungerford's book that submodules of free modules are free. I don't have it handy, but I will check what assumptions on the ring he makes. http://books.google.com/books?id=t6N_tOQhafoC&pg=PA339&dq=rank+of+a+matrix+over+a+ring&hl=en&ei=cma0TZCYBIugtwedtMXpDg&sa=X&oi=book_result&ct=result&resnum=4&ved=0CD8Q6AEwAw#v=onepage&q=rank%20of%20a%20matrix%20over%20a%20ring&f=false2011-04-24
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    In any case, without worrying about the definition of the rank, I know every column is an $R$-linear combination of $m$ specified columns. I would like to find the coefficients that govern these linear combinations hopefully using some software.2011-04-24
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    @Milo: the theorem Hungerford proves is restricted to the case that $R$ is a principal ideal domain.2011-04-24
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    @Qiaochu: OK. I have modified the question.2011-04-24
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    Free resolutions would be really boring to study if submodules of free modules were necessarily free.2011-04-24

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