A module without a basis

Many answers! ${\mathbb{Q}}$ is a module over ${\mathbb{Z}}$, and has no basis. Indeed, if a ring $R$ is not a field, then it has a module that has no basis: let $I$ be a proper ideal of $R$ which is different from $\{0\}$. Then $R/I$ is a module that has no basis.

Take $M = \mathbb Z / n \mathbb Z$ over $\mathbb Z$ for $n > 1$.

More generally, let $G$ be a finite abelian group, $|G|=n$. Let $g \in G$. Then $ng = 0$. By definition, a free module $M$ over a ring $R$ is a module with a basis. A basis of a module $M$ is a subset $B \subset M$ such that it is linearly independent and you can write every $m \in M$ as a finite linear combination of elements in $B$ (with coefficients in $R$).

If you choose $R = \mathbb Z$ and $G$ a finite abelian of order $n$, then for every finite linear combination of elements of $G$ you can get $0$ hence no subset of $G$ can be a basis hence $M = G$ cannot be free.

With modules, it is often useful to consider the slightly more general notion of "independent set":

Definition. Let $M$ be an $R$-module. We say that $m_1,\ldots,m_r\in M$ are "independent" if and only if for all $a_i\in R$, if $a_1m_1+\cdots+a_rm_r=0$, then $a_im_i=0$ for all $i$.

(This is weaker than "$R$-linearly independent": for example, $\{1+n\mathbb{Z}\}$ would be an independent set of $\mathbb{Z}/n\mathbb{Z}$ as a $\mathbb{Z}$-module, but not linearly independent).

Even with this weaker notion, it is not true that every module over a ring has independent spanning sets.

For example, $\mathbb{Q}$ does not have an independent spanning set over $\mathbb{Z}$, since any finitely generated subgroup is cyclic, hence not isomorphic to $\mathbb{Q}$, and any subset with at least two elements is not independent (given $\frac{a}{b}$ and $\frac{c}{d}$, take $bc(\frac{a}{b})-ad(\frac{c}{d})$, which equals zero though the separate summands do not).

Or consider any domain that is not a principal ideal ring, and let $I$ be an ideal that is not principal. Any two nonzero elements of $I$ are not independent (if $a,b\in I$, then $ba-ab=0$, but $ab\neq0$), so the only independent subsets of $I$ are the empty set and singletons, but $I$ is not generated by a single element because $I$ is not principal. For instance, take $(2,1+\sqrt{-5})$ in $\mathbb{Z}[1+\sqrt{-5}]$.