1
$\begingroup$

Based on the comments of rschwieb's answer in this question asked recently: Can we contruct a basis in a finitely generated module.

If $M=\langle e_1,\ldots,e_n\rangle$ is a finitely generated torsion free $R$-module. I'm trying to construct a free submodule $F$, i.e, isomorphic to $R^s$ for some $s$, finding a subset $S=\{e_1,\ldots,e_s\}$ such that $S$ is a maximal independent subset of $M$, then $S$ generates this free submodule $F$ of $M$ with basis $S$.

I'm asking that because I didn't understand why Peter Clark in his commutative algebra pdf wrote this:

Thanks in advance

$\endgroup$
1
$\begingroup$

Let's call $S$ the L.I. subsets of $\{x_1,\ldots,x_n\}$. Since $M$ is a torsion free, for $x_1\in M,r\in R$, we have $rx_1=0$ iff $r=0$. Then $S$ is non-empty.

Since $\{x_1,\ldots,x_n\}$ is finite, then it has a maximal element $\{x_1,\ldots, x_s\}$, $1\le s\le n$

$\endgroup$
  • $\begingroup$ I think this isn't quite enough for Prof. Clark's claim that the given $s$ is unique, but this can be obtained by tensoring with $R$'s fraction field and invoking the well-definedness of dimension for a vector space over a field. On the other hand, I don't see that Prof. Clark actually needs $s$'s uniqueness in the present proof. $\endgroup$ – Ben Blum-Smith Dec 17 '15 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.