This is a question related to this:
"...if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is never free. Thus the only commutative rings $R$ for which every finitely generated module is free are those with no nontrivial ideals, and this is one of several equivalent ways to define a field."
Question 1: How do I show that if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is not free? Is this really true? I think the following is a counterexample: $R = \mathbb Z \times \mathbb Z , I = \langle (1, 0) \rangle $. Define $(a,b) + (c,d) = (a+c, b+d)$ and $(a,b) \cdot (c,d) = (ac, bd)$. Then $R/I = \langle (0,1 ) \rangle$ and the basis of $R/I$ is $\{(0,1)\}$.
Question 2: Can you give me an example of a finitely generated module that doesn't have a basis?
Question 3: What's the relationship between $R$ modules and ideals $I$ of $R$? Are all finitely generated $R$-modules of the form $R/I$? Are there no others?