Let $A$ be a commutative ring with $1$. Suppose that every module $M$ over $A$ is finitely.generated with rank $n$ and every submodule $N \leq M$ has rank $r \leq n$. Show $A$ is a principal ideal domain. I do not how having a finite base is related to being a module over a commutative ring with $1$.


1 Answer 1


The premise "Suppose that every module $M$ over $A$ is finitely generated..." doesn't make sense since you can't demand all modules are finitely generated: consider $M = \bigoplus_{n \geq 1} A$, which is not finitely generated (except for the silly case where $A$ is the zero ring).

What you meant to say is "Suppose every module $M$ over $A$ that is finitely generated...". In other words, it is an assumption on all finitely generated $A$-modules, not on all $A$-modules. And when you say "... every submodule $N \subset M$ has rank $r \leq n$" did you mean "... every submodule $N \subset M$ is free of rank $r \leq n$"?

If you are assuming every finitely generated $A$-module is free and their submodules are also free of rank no greater than the rank of the starting module, then just look at the example of the $A$-module $A$ itself (free of rank $1$), whose submodules are its ideals to deduce that (i) all ideals are principal and (ii) $A$ is an integral domain.

  • $\begingroup$ That is a (very useful) comment, but not an answer. $\endgroup$
    – Rob Arthan
    May 26 at 2:33
  • $\begingroup$ @RobArthan I edited the end of my answer. While I am not giving a full answer to the question, I am directing the OP to the only cases worth thinking about among all $A$-modules. And I think that qualifies as a reasonable answer, as the OP can now think about the relevant situation. $\endgroup$
    – KCd
    May 26 at 2:34
  • $\begingroup$ What means "OP"? $\endgroup$ May 26 at 3:33
  • $\begingroup$ @Student2271 that is you: OP = original poster or original post. $\endgroup$
    – KCd
    May 26 at 3:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .