# Quotient rings and free modules

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?

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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.
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.)
Thanks, and I got myself Atiyah-MacDonald. So "R is a field" iff "R has non non-trivial ideals" iff "every finitely generated $R$-module is free"? –  Rudy the Reindeer May 13 '12 at 13:22
@Clark: for $R$ commutative, yes. –  Qiaochu Yuan May 13 '12 at 15:58