If an ideal is free then it has rank 1.

Question

Suppose we consider an ideal $I \subset R$ as an $R-$submodule. If $I \neq 0$ and it's free, then it's free of rank 1.

I was wondering whether it's enough to say that $I$ is isomorphic to $R^n$, but it's a submodule of $R$ and $R$ is free of rank 1, so $I$ is free of rank 1 is as well. Or am I overlooking some important details?

Answer 1

Note: the proposition is only true when $R$ is commutative. There exist rings whose right ideals are free but not rank $1$.

Hint: Suppose $a,b$ are two basis elements. Then what would $(a)b+(-b)a=$?

I was wondering whether it's enough to say that $I$ is isomorphic to $R^n$, but it's a submodule of $R$ and $R$ is free of rank 1, so $I$ is free of rank 1 is as well.

It turns out that the statement that "free submodules of free modules of finite rank have lesser or equal rank" is true for commutative rings, however it is notorious for being a little tricky.

It is not true for some rings which aren't commutative, though.