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 ?


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.


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.


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


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 $\mathbb Z$.


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

  • $\begingroup$ Can you say why ? $\endgroup$ – Belgi Apr 26 '12 at 21:47
  • $\begingroup$ What don't you understand, or is it clear now? $\endgroup$ – Brett Frankel Apr 26 '12 at 21:48
  • $\begingroup$ Why such $x,y$ exist and how can we prove that $\langle p,q \rangle \neq \mathbb{Q}$ $\endgroup$ – Belgi Apr 26 '12 at 21:50
  • $\begingroup$ @Belgi: Because the subgroup generated by $p$ and $q$ cannot contain any rational whose denominator is greater than the product of the denominators of $p$ and $q$. $\endgroup$ – Arturo Magidin Apr 26 '12 at 21:51
  • $\begingroup$ $x$ and $y$ exist because we can multiply $p$ and $q$ each by some integer to clear denominators, and then multiply the resulting numbers by each other (so that they are now the same), and finally change the sign on one of the two numbers in question. $\endgroup$ – Brett Frankel Apr 26 '12 at 21:54

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