# What is the difference between a module of finite rank and finitely generated module.

R is an integral domain and every module we talk about is an R-module. If a module is finitely generated then obviously every element of the module can be written as finite R-linear combination of the set of generators. I observe that this expression might not be unique due to which the module could fail to be free. What has this got to do with the rank? More specifically, how exactly is rank of a module defined.

• Do you know there is a definition, other than the smallest cardinality of a set of generators? Googling only turns up the definition of rank of a free module. – Thomas Andrews Apr 9 '14 at 4:57
• @ThomasAndrews $\dim_{{\rm Frac}(R)}({\rm Frac}(R)\otimes_RM)$ would make sense. – anon Apr 9 '14 at 4:59

• Let $\mathbf{R}$ be an integral domain. $\mathbf{I}$ be a non-principal ideal of $\mathbf{R}$. Then $\mathbf{I}$ is not free $\mathbf{R}$-module. But $\mathbf{I}$ has rank 1 since it is a non-trivial submodule of a rank 1 module(ie $\mathbf{R}$). – user2902293 Apr 14 '14 at 17:11