2
$\begingroup$

Suppose that $R$ is an integral domain and $M$ is a finitely generated module. Does $M$ necessarily have finite rank?

DEFINITIONS: $M$ is finitely generated if there exists $x_1, \dots, x_n$ such that each element of $M$ can be written as a linear combination of $x_1, \dots, x_n$. We define the rank of $M$ to be $$ \sup_I |I| $$ where the supremum is taken over all $I\subseteq M$ where $I$ is a linearly independent set.

The result is true if we assume that $M$ is a free module. I am wondering if it also holds in this more general setting.

$\endgroup$
1
$\begingroup$

Yes.

Since $M$ is finitely generated, there exists a surjection $$ \pi: A^m \to M $$ for some integer $m$. Indeed, if $x_1, \ldots, x_m$ are generators of $m$, then one can take $\pi$ to send the $i$th elementary vector $e_i$ to $x_i$.

I claim that the rank of $M$ is at most $m$. Indeed, let $I$ be a linearly independent set in $M$ with $n$ elements. Let $M_I$ be the submodule generated by $I$; since $I$ is linearly independent, $M_I$ is a free module of rank $n$.

Now, consider the preimage $\pi^{-1}(M_I)$. It is a submodule of $A^m$, and $\pi$ induces a surjection $$ \pi: \pi^{-1}(M_I) \to M_I. $$ Since $M_I$ is free, this surjection admits a section $\sigma:M_I \to \pi^{-1}(M_I)$. In particular, $\sigma(M_I)$ is a free submodule of rank $n$ of $A^m$. Since $A$ is commutative, this implies that $n\leq m$.

$\endgroup$
  • $\begingroup$ Could you elaborate on how you obtain the section $\sigma$? $\endgroup$ – Quoka Apr 30 '18 at 17:18
  • $\begingroup$ @MathUser_NotPrime Any free module is projective, and any surjection onto a projective module admits a section, see for example this Wikipedia page. $\endgroup$ – Pierre-Guy Plamondon Apr 30 '18 at 17:28
  • $\begingroup$ Appreciate it. Thank you :-) $\endgroup$ – Quoka Apr 30 '18 at 17:37

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.