Let $M$ be a finitely generated module over a commutative ring $A$.

If there exists a natural number $n$ such that for any prime ideal $P$ of $A$, $M_P$ is a free $A_P$-module of rank $n$, then $M$ is said to be of constant finite rank.

I have known every finitely generated projective module of constant finite rank over a semilocal ring is free.

I want to know whether there exists a finitely generated module of constant finite rank which is not free or not.

If $M$ can be generated by $n$ elements $x_1,\cdots,x_n\in M,$ and $\dim_{A_P}(M_P)=n$ for any prime ideal $P$, we cannot find such examples. One way goes as follows: Consider the following map $$\phi\colon A^n\rightarrow M,\ e_i\mapsto x_i,$$ where $e_i$ is the standard basis of $A^n.$ By localization, $\phi$ is an isomorphism. When $A$ is a seimilocal ring, we can prove $M$ can be generated by $n$ elements.

  • 1
    $\begingroup$ Why not take a projective module which is not free? $\endgroup$
    – Mohan
    Commented Jul 20, 2022 at 18:54
  • $\begingroup$ @Mohan cannot gurantee a projective module has constant finite rank $\endgroup$
    – qinxs
    Commented Jul 21, 2022 at 2:00

1 Answer 1


If $A$ is an integral domain , any finitely generated projective $A$-module $M$ has constant rank, equal to $\dim_{Frac(A)}(M\otimes_A Frac(A)),$ since for any prime ideal $P$, we have a canonical injection $A_P\to Frac(A)$.

Hence any example of a non free finitely generated projective module over an integral domain will do.

For $A$, you can take for instance a non principal Dedekind domain and for $P$ a non principal ideal of $A$.

  • $\begingroup$ This makes perfect sense, thanks. $\endgroup$
    – qinxs
    Commented Jul 21, 2022 at 15:48

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