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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?

YoTengoUnLCD
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    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. – Marcus M Sep 09 '15 at 01:16
  • By "every element can be zeroed by an element in the ring" you mean a ring where every element is a zero divisor? – YoTengoUnLCD Sep 09 '15 at 01:17
  • possible duplicate http://math.stackexchange.com/questions/137442/a-module-without-a-basis – Nikos M. Sep 09 '15 at 01:18
  • 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 $rm = 0$. – Marcus M Sep 09 '15 at 01:18
  • @NikosM. Hmm, I'm also asking why does this happen, not just for an example. – YoTengoUnLCD Sep 09 '15 at 01:19
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    Note: not every vector space has a basis, unless you assume the axiom of choice. – vadim123 Sep 09 '15 at 01:19
  • A basis $B$ is a subset of a module $M$ which (1) spans $M$ (every module trivially has a spanning set) (2) is linearly independent (this is where modules fail to have a basis). – Rachmaninoff Sep 09 '15 at 01:19

3 Answers3

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

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

Lubin
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  • I don't see why this doesn't have a basis, could you elaborate? – YoTengoUnLCD Dec 17 '15 at 16:00
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    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. – Lubin Dec 17 '15 at 20:17
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    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 $\Bbb Q$? – YoTengoUnLCD Dec 17 '15 at 20:22
  • 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. – Lubin Dec 17 '15 at 20:25
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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.

CoffeeArabica
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