Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to show that if $M$ is an $R$-module where $R$ is a field then: $M$ is artinian if and only if $M$ is finite dimensional.

$\Rightarrow:$ Assume for sake of contradiction that $M$ is not finite dimensional. Then let $B=\{x_i\}_I$ be a basis for $M$ where $I$ must be an infinite index set. Let $\{y_n:n\in \mathbb{N}\}$ be a subset of the basis.

Then I consider $M_0=M$ and $M_1=\langle B\backslash y_1\rangle$ and in general $M_n=\langle B\backslash\{y_1,\ldots,y_n\}\rangle$. I just furnished an infinite descending chaing of submodules contradiction the assumption that $M$ was artinian.

$\Leftarrow:$ It is clear since if $M_1$ properly contains $M_2$ then the dimension of $M_1$ must be greater than that of $M_2$, and the result follows since we cannot have an infinite descending chain of submodules since the dimension we started with was finite.

My question goes: I am assuming that $M$ has a basis (i.e., $M$ is free), is it true in general that if we are over a field our modules must be free? or does the way the question was phrased implies that we are assuming that we have a basis (when the say finite dimensional)?

Thanks.

share|improve this question
    
"Let $\{y_n\mid n\in\mathbb{N}\text{ and }y_n\in B\} be" what ? –  Arturo Magidin Mar 1 '12 at 20:54
1  
Your $\Rightarrow$ is not a proof by contradiction, it's a proof by contrapositive; write it that way! –  Arturo Magidin Mar 1 '12 at 20:55
3  
"Module over a field" = "vector space". The assertion that every vector space is a free module (that every vector space has a basis) is equivalent to the Axiom of Choice. (The assertion that every finitely generated vector space has a basis does not depend on the Axiom of Choice; it's true even in the absence of AC). –  Arturo Magidin Mar 1 '12 at 20:59
    
@ArturoMagidin: Oh ok, so using choice I get that $M$ is free, and is then my argument correct? –  Daniel Montealegre Mar 1 '12 at 21:08
    
Yes; I would rewrite the first part as contrapositive instead of contradiction. That every infinite set contains a denumerable subset also requires a fragment of AC, but since you are already assuming the whole thing, I wouldn't worry about it. –  Arturo Magidin Mar 1 '12 at 21:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.