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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.

Quoka
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1 Answers1

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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 $n\leq m$.

Pierre-Guy Plamondon
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