# Finitely-Generated modules over a commutative ring with $1$.

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 $$1$$.

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.

• That is a (very useful) comment, but not an answer. May 26 at 2:33
• @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.
– KCd
May 26 at 2:34
• What means "OP"? May 26 at 3:33
• @Student2271 that is you: OP = original poster or original post.
– KCd
May 26 at 3:38