5
$\begingroup$

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?

$\endgroup$
7
  • 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

9
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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
8
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.