Are all submodules of free modules free? I would like a reference to a proof or counterexample please.
closed as offtopic by Ricardo Andrade, YCor, Karl Schwede, Chris Godsil, Andreas Blass Feb 22 at 14:03This question appears to be offtopic. The users who voted to close gave this specific reason:



Вот общий пример: неглавной идеал в кольце $A$. Кольцо $A$  свободный $A$модуль. Идеал в кольце  подмодуль, а он тоже свободный $A$модуль только в случае, что он главной идеал: ненулевые элементы $a$ и $b$ в кольце удовлетворяют нетривиальное $A$линейное соотношение $c_1a + c_2b = 0$, где $c_1 = b$ и $c_2 = a$, поэтому если существует базис, то мощность является одним. 


No, for a general ring, yes for PIDs. Take as a counter example the ring of integers mod 4 as a module over it self. 


To answer ding8191's question: let $A$ be any valuation ring of height $1$ which is not a discrete valuation ring. That is to say, let $F$ be a field and suppose that there is a function $v : F \to \mathbb{R} \cup \{ \infty \}$ such that $v(xy) = v(x) + v(y), v(1) = 0, v(0) = \infty$ and $v(x + y) \geq \min $ $\{ v(x), v(y) \}$ (thus $v$ is a valuation), such that $v(F)$ is not a discrete subgroup of $\mathbb{R}$, and let $A = \{ x \in F :v(x)\geq 0\}$ (the valuation ring of $F$). Then the maximal ideal $\mathfrak{m}$ in this ring is a direct limit of principal ideals and is therefore flat. If $\mathfrak{m}$ was projective, then it would have to be free by Kaplansky's Theorem (all projectives over any local ring are free), but since $\mathfrak{m}$ is contained in $F$ it has to have rank at most $1$, hence $\mathfrak{m}$ is principal. But this would force the valuation on $A$ to be discrete. You can find this material in Chapter VI of Bourbaki's "Commutative Algebra" (Hermann, 1972). In particular, Lemma $1$ of $\S 3.6$ says that if $A$ is a valuation ring (in a more general sense than I defined above), then every torsionfree $A$module is flat, and Proposition $9$ of the same subsection implies that $A$ is a discrete valuation ring whenever $A$ is a valuation ring of height $1$ with principal maximal ideal. To give a concrete example, consider the field $F$ of Puiseux series and let $A$ be its valuation ring. 


For any (unitary commutative) ring $A$, the following are equivalent: (i) submodules of free $A$modules are free, (ii) any ideal is free as $A$module, (iii) the ring $A$ is a principal ideal domain. The proof is already clear form the above discussion. Therefore, if counterexamples exist, conterexamples must exists as ideals of the ring. As for the counterexamples already given here, (1) the ideal $(X,Y)$ of the polynomial ring $k[X,Y]$ is not even flat, as noted by Robin Chapman, (2) the ideal $2\mathbb{Z}/4\mathbb{Z}$ in the ring $\mathbb{Z}/4\mathbb{Z}$ is not flat, since it is generated by a nonzero nilpotent element, (3) the ideal $\mathbb{Z}\times 0$ in the ring $\mathbb{Z}\times \mathbb{Z}$ is projective but not free, similar result holds for the ideal $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/3\mathbb{Z}$ in the ring $\mathbb{Z}/6\mathbb{Z}$. I am trying to find an ideal $I$ of a ring $A$ such that $I$ is flat but not projective as $A$module. Obviously, such a conterexample esists only if the ring $A$ is nonNoetherian. Maybe nonNoetherian prufer domains can work. Could someone give me such an example, i.e., to find an ideal $I$ of a ring $A$ such that $I$ is flat but not projective as $A$module? 


Another example: let $R=k[x,y]$ where $k$ is a field. Then $R$ is a free module over itself, and the ideal $I$ of $R$ generated by $x$ and $y$ is not only not free over $R$, it is not even flat over $R$. 


For a ring R, if e is an idempotent element diferent from zero and one, then Re is projective module which isn't free. 

