# Thread: Question about Modules over rings

1. ## Question about Modules over rings

Hi! I would like some help with this question I need to prove
let R be a non zero commutative ring and we are required to prove that any linearly independent subset of Rsquared has at most 2 elements I wanted to use the theorem where every submodule M of Rtothe N has a basis with at most N elements BUT this is only true for PIDs

2. Originally Posted by gtkc Hi! I would like some help with this question I need to prove:

let R be a non zero commutative ring with identity and we are required to prove that any linearly independent subset of $\displaystyle \color{red}R^2=R \oplus R$ has at most 2 elements.

I wanted to use the theorem where every submodule M of $\displaystyle \color{red}R^N$ has a basis with at most N elements BUT this is only true for PIDs.
next time spend more time on writing your question or your question will probably be ignored. i'll prove a more general fact: suppose $\displaystyle M$ is a free $\displaystyle R$ module with a basis, say, $\displaystyle B=\{x_1, \cdots , x_n \}$.

let $\displaystyle A=\{y_1, \cdots , y_k \}$ be a linearly independent subset of $\displaystyle M.$ i'll prove that $\displaystyle k \leq n.$ choose any maximal ideal $\displaystyle m$ of $\displaystyle R.$ then $\displaystyle F=R/m$ is a field and thus $\displaystyle V=M/mM$ is a vector space over $\displaystyle F.$

the claim is that $\displaystyle \bar{B}=\{x_1+mM, \cdots , x_n + mM \}$ is a basis for $\displaystyle V.$ clearly $\displaystyle V=Span \bar{B}.$ to prove that the elements of $\displaystyle \bar{B}$ are linearly independent, suppose $\displaystyle \sum_{j=1}^n(r_j + m)(x_j + mM)=0,$ for some
$\displaystyle r_j \in R.$ then $\displaystyle \sum_{j=1}^n r_jx_j \in mM.$ thus $\displaystyle \sum_{j=1}^n r_jx_j=\sum_{j=1}^n m_jx_j,$ for some $\displaystyle m_j \in m.$ this gives us $\displaystyle \sum_{j=1}^n (r_j - m_j)x_j = 0$ and hence $\displaystyle r_j=m_j \in m.$ thus $\displaystyle r_j + m = 0,$ for all $\displaystyle 1 \leq j \leq n,$ which proves the claim.

so we've proved that $\displaystyle \dim_F V = n.$ similarly we have that the elements of $\displaystyle \bar{A}=\{y_1+mM, \cdots , y_k + mM \} \subset V$ are linearly independent over $\displaystyle F.$ but, from linear algebra, we know that the number

of elements of any linearly independent set in a vector space is at most equal to the number of elements of a basis of that vector space. therefore $\displaystyle k =|\bar{A}| \leq \dim_FV = n.$

3. Im sorry I'm not sure how the code here works.. my friend and i was discussing your solution and we werent sure exactly how y to yk was linearly independent in this line as y is only a linearly independent subset of M and perhaps there could be linear dependence relations with mM

4. Originally Posted by gtkc Im sorry I'm not sure how the code here works.. my friend and i was discussing your solution and we werent sure exactly how y to yk was linearly independent in this line as y is only a linearly independent subset of M and perhaps there could be linear dependence relations with mM
as i said it's exactly the same as the proof of linear independence of elements of $\displaystyle \bar{B}.$ read my proof again!

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