I was just reading a bit about modules on wikipedia, which, as I understood it, are generalizations of vector spaces.

I read there exists some modules that do not have a basis, and I couldn't think of an example or why this happens (vs vector spaces: they all have some basis).

Could someone explain this?

  • 2
    $\begingroup$ Any module that has a basis is called a free module, and these are the modules that are closest to vector spaces. To find a module that isn't free, think of modules where every element can be zeroed by an element in the ring. Then, no non-empty set can be a basis since we can have a nontrivial linear combination where we annihilate one of the elements. $\endgroup$
    – Marcus M
    Sep 9, 2015 at 1:16
  • $\begingroup$ By "every element can be zeroed by an element in the ring" you mean a ring where every element is a zero divisor? $\endgroup$ Sep 9, 2015 at 1:17
  • $\begingroup$ possible duplicate math.stackexchange.com/questions/137442/… $\endgroup$
    – Nikos M.
    Sep 9, 2015 at 1:18
  • $\begingroup$ Not exactly. Every module, $M$, is associated with a ring $R$ where $R$ acts on $M$. Think of a module, $M$, and a ring $R$ so that for every $m \in M$, $\exists~r \in R$ with $r \neq 0$ and $rm = 0$. $\endgroup$
    – Marcus M
    Sep 9, 2015 at 1:18
  • 1
    $\begingroup$ Note: not every vector space has a basis, unless you assume the axiom of choice. $\endgroup$
    – vadim123
    Sep 9, 2015 at 1:19

3 Answers 3


Consider $\mathbb{Z}/2\mathbb{Z}$ over $\mathbb{Z}$. Why does it not have a basis?

  • $\begingroup$ Hmm wouldn't $\{1\}$ be a basis? $\endgroup$ Sep 9, 2015 at 1:18
  • 3
    $\begingroup$ Because a nontrivial linear combination of $\{1\}$ yields $0$. $\endgroup$
    – vadim123
    Sep 9, 2015 at 1:20
  • 4
    $\begingroup$ $\{1\}$ is a generating set. However, it fails the linearly independent condition. $\endgroup$
    – LASV
    Sep 9, 2015 at 1:23

Another example that you already knew: $\Bbb Q$ as a $\Bbb Z$-module.

  • $\begingroup$ I don't see why this doesn't have a basis, could you elaborate? $\endgroup$ Dec 17, 2015 at 16:00
  • 1
    $\begingroup$ Gladly. Any two elements are $\Bbb Z$-linearly dependent; therefore a basis, if any there were, would have at most one element. But a single element does not generate $\Bbb Q$ as a $\Bbb Z$-module. $\endgroup$
    – Lubin
    Dec 17, 2015 at 20:17
  • 1
    $\begingroup$ I see! Given $\frac a b, \frac c d \in \Bbb Q$ then $(bc)\frac a b+ (-ad)\frac c d=0;\, bc, -ad\in \Bbb Z$, for $bc,-ad$ not necesarilly 0, right? How do you show that a single element can't generate $\Bbb Q$? $\endgroup$ Dec 17, 2015 at 20:22
  • $\begingroup$ For that, I’d ask you to look at the picture. But if $\{a/b\}$ is your basis, then $a/2b$ is not in the $\Bbb Z$-span. $\endgroup$
    – Lubin
    Dec 17, 2015 at 20:25

I stumbled over the example regarding the pair $(M,\varphi)$ with $M$ being a $K$-Vectorspace and $\varphi\in\text{End}_K(M)$ and the corresponding $K[X]-$Module $M$.

If we choose a family $(x_n)$, who might be a linear independend family:

$$f(X)_0*x_0+f_1(X)*x_1+...+f_n(X)*x_n=0$$ then by choosing the local minimal polynomial of $x_i$ as $f_i(X)$ regarding $\varphi$, we get:

$$f_i(\varphi)(x_i)=0,\forall i\in\lbrace0,...,n\rbrace$$

concluding the family is linear dependend and especially no basis.

Please correct me if I am wrong.


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