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We know that for general modules over a commutative ring with $1$, you can't always extract a basis from a generating set.

This makes me think that maybe there should be free modules of infinite rank which could be finitely generated. Do such things exist?

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    This is one of the most posted question. Please use search and read [this](http://www.jstor.org/discover/10.2307/2316897?uid=3738920&uid=2129&uid=2&uid=70&uid=4&sid=21101516982237). You will find that the answer to your question is no.2012-12-04

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Suppose we had such a horrible thing, a surjection: $R^{\oplus n} \twoheadrightarrow R^{\oplus I}$where $I$ is something infinite (or just finite and $> n$). Pick any maximal ideal of $R$ and tensor up with $R/\mathfrak{m}$, it's right exact so we still have: $R^{\oplus n} \otimes R/\mathfrak{m} \twoheadrightarrow R^{\oplus I} \otimes R/\mathfrak{m}$But these are isomorphic to: $R/\mathfrak{m}^{\oplus n} \twoheadrightarrow R/\mathfrak{m}^{\oplus I}$a surjection of vector spaces.

So, can't happen!

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    Yeah another way to see it would be to write each generator in a finite generating set as a linear combination of basis vectors (finitely many in each case). We then deduce that every element in $M$ is a finite $R$-linear sum over this finite set of independent vectors; thus we've shown our basis was finite to begin with.2012-12-06