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Math Help - module homomorphism, ring, subset

  1. #1
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    module homomorphism, ring, subset

    Background: Let R be a ring, S \subseteq R a multiplicatively closed subset, and let M, N be R-modules. Let \phi : M \rightarrow N be an R-module homomorphism, and let \phi_s : S^{-1}M \rightarrow  S^{-1}N denote the S^{-1}R-module homomorphism, and let \phi_s(\frac{m}{s})=\frac{\phi(m)}{s}. You may assume without proof that \phi_s is a well-defined S^{-1}R-module homomorphism.

    Prove:
    (a) Prove that if \phi is injective, then \phi_s is injective.
    (b) Prove that if \phi is surjective, then  \phi_s is surjective.
    (c) Prove that if M is a finitely generated free R-module with basis \{ b_1,\ldots, b_m  \} then S^{-1}M is a finitely generated free S^{-1}R-module with basis \{  \frac{b_1}{1},\ldots, \frac{b_m}{1}   \}.
    (d) Now suppose R is an integral domain and that m, n are positive integers. Prove that if R^m \cong R^n, then m=n. (Hint: consider S=R-\{0\}.)
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  2. #2
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    Quote Originally Posted by Erdos32212 View Post
    Background: Let R be a ring, S \subseteq R a multiplicatively closed subset, and let M, N be R-modules. Let \phi : M \rightarrow N be an R-module homomorphism, and let \phi_s : S^{-1}M \rightarrow S^{-1}N denote the S^{-1}R-module homomorphism, and let \phi_s(\frac{m}{s})=\frac{\phi(m)}{s}. You may assume without proof that \phi_s is a well-defined S^{-1}R-module homomorphism.

    Prove:
    (a) Prove that if \phi is injective, then \phi_s is injective.
    (b) Prove that if \phi is surjective, then  \phi_s is surjective.
    (c) Prove that if M is a finitely generated free R-module with basis \{ b_1,\ldots, b_m \} then S^{-1}M is a finitely generated free S^{-1}R-module with basis \{ \frac{b_1}{1},\ldots, \frac{b_m}{1} \}.
    (d) Now suppose R is an integral domain and that m, n are positive integers. Prove that if R^m \cong R^n, then m=n. (Hint: consider S=R-\{0\}.)
    (a): if \phi_S(m/s)=0, then \frac{\phi(m)}{s}=0. thus there exists t \in S such that t \phi(m)=0. hence: \phi(tm)=t\phi(m)=0. but \phi is injective. thus tm=0, i.e. \frac{m}{s}=0.

    (b) if x=\frac{n}{s} \in S^{-1}N, then since \phi is surjective, there exists m \in M such that \phi(m)=n. thus: x=\frac{\phi(m)}{s}=\phi_S(m/s). hence \phi_S is surjective.

    (c) let x=\frac{v}{s} \in S^{-1}M. then v=\sum_{i=1}^m r_ib_i, \ r_i \in R. thus: x=\sum_{i=1}^m \frac{r_i}{s}\frac{b_i}{1}, which proves that I=\{\frac{b_i}{1} : i=1,2, \cdots m \} generates S^{-1}M as S^{-1}R module, because \frac{r_i}{s} \in S^{-1}R.

    to prove that the set I is linearly independent, we assume that \sum_{i=1}^m \frac{r_i}{s_i}\frac{b_i}{1}=0, which after clearing denominators gives us: \sum_{i=1}^n t_ir_ib_i=0, where t_i \in S. since \{b_i: \ i=1,2, \cdots, m\}

    is assumed to be a basis for M, it's a set of linearly independent elements of R. thus t_i r_i = 0, \ i = 1, 2, \cdots , m. hence \frac{r_i}{s_i}=0, \ i=1,2, \cdots , m.

    (d) let S=R- \{0 \}, which is obviously multiplicatively closed because R is a domain. now Q(R)=S^{-1}R is the field of fractions of R. on the other hand we have:

    \bigoplus_{i=1}^m Q(R)=\bigoplus_{i=1}^m S^{-1}R \simeq S^{-1}R^m \simeq S^{-1}R^n \simeq \bigoplus_{i=1}^n S^{-1}R=\bigoplus_{i=1}^n Q(R). \ \ \ \ (1)

    now since Q(R) is a field, the LHS and RHS of (1) are two Q(R)-vector spaces. since they are isomorphic, their dimensions must be equal. (recall from linear algebra!) thus m=n.
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