I am attempting to solve part (a) of Ch 8 Problem 2.4 from Artin's algebra textbook. I have already solved part (b), showing that $R/I$ is a free R-module iff $I=\{0\}$ or $I=R$.
Let $I$ be an ideal of a ring $R$.
(a) Under what circumstances is $I$ a free R-module?
(b) Under what circumstances is the quotient ring $R/I$ a free R-module?
Now, it seems to me that if $I$ is finitely generated, that is if $I=(i_1,i_2,...,i_n)$, then $I$ will have some subset of $\{i_1,...i_n\}$ as a basis, hence $I$ will be a free R-module. Correct me if I am wrong. I haven't proven this yet, but this is my intuition.
But not all ideals are finitely generated. Can I still find (or at least prove that there exists) some basis for $I$ even in the case that it is not finitely generated? If so, how do I do this?
$I$ is a free $R$-module if and only if there is some $a\in R$ such that $I=(a)$ and either $a=0$ or $a$ is not a divisor of $0$.
In fact, no two distinct elements of $R$ can be linearly independent over $R$, since $a_2\cdot\underline{a_1}+(-a_1)\cdot\underline{a_2}=0$. So, a basis for $I$ must either be empty, or a singleton. Moreover, the definition of linear independence applied to a singleton $\{a\}$ becomes: "there can't be $d\in R\setminus\{0\}$ such that $da=0$".