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Math Help - Question about Modules over rings

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by gtkc View Post

    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 \color{red}R^2=R \oplus R has at most 2 elements.

    I wanted to use the theorem where every submodule M of \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 M is a free R module with a basis, say, B=\{x_1, \cdots , x_n \}.

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

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

    so we've proved that \dim_F V = n. similarly we have that the elements of \bar{A}=\{y_1+mM, \cdots , y_k + mM \} \subset V are linearly independent over 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 k =|\bar{A}| \leq \dim_FV = n.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by gtkc View Post
    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 \bar{B}. read my proof again!
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