A module without a basis

It is well known that every vector space have a basis, after studying module theory I read in Wikipidia that not all modules have a basis.

Can someone please give an example of a module without a basis ?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $\mathbb{Z}[1+\sqrt{-5}]$.

Solution 4:

Another example, extracted from Musili's Introduction to Rings and Modules:

Any abelian group $M$ which has a nontrivial element of finite order cannot be free as a module over $\mathbb Z$.

In fact, say $B$ is a basis for $M$ over $\mathbb Z$. Let $a\in M\backslash\{0\}$ be such that $|a|=n$. Note that $n>1$. We have $$a=z_1b_1+z_2b_2+...+z_mb_m$$ for some $z_i\in \mathbb Z$ and $b_i\in B$, with $i=1,2,...,m$. Then, $$0=na=nz_1b_1+nz_2b_2+...+nz_mb_m.$$ By linearity of $B$, $$z_1=z_2=...=z_m=0 \Rightarrow a=0,$$ a contradiction. In particular,

Any finite abelian group is not free as a module over $\mathbb Z$.

Solution 5:

$\mathbb{Q}$ as a $\mathbb{Z}$-module (abelian group). If $p$ and $q$ are nonzero rational numbers, there exist integers $x$ and $y$ such that $px+qy=0$