Are submodules of free modules free?

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 нетривиальному $A$-линейному соотношению $c_1a + c_2b = 0$, 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.

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 non-zero 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 non-Noetherian. Maybe non-Noetherian 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?