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?
Best Answer
If you want to learn more about modules (at least over commutative rings) you should pick up a book on commutative algebra, e.g. Atiyah-MacDonald. Anyway:
1: If $r \in R/I$ and $i \in I$ then by definition $ir = 0$, so $r$ cannot be part of a basis. Thus no subset of $R/I$ is linearly independent.
2: I just did!
3: Every ideal is a module; in fact, ideals are precisely the submodules of $R$ (as a module over itself). If $M$ and $N$ are two modules then $M \oplus N$ is also a module, so there are finitely generated modules of the form $R^n \oplus (R/I)^{m_I} \oplus (R/J)^{m_J} \oplus ... $. For PIDs these are all finitely generated modules by the structure theorem but in general modules can be complicated. (Every f.g. module is a quotient of $R^n$ but the structure of submodules of $R^n$, thus of quotients, can in general be complicated.)