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Math Help - Two more questions involving direct products

  1. #1
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    Two more questions involving direct products

    1. Is Z x Z cyclic? [Here Z means (Z, +).]

    2. Prove the Chinese remainder theorem: If m_1,.....,m_n are positive integers such that  m_i and m_j are relatively prime for i not equal to j, and k_1,....,k_n are any integers, then there is an integer x such that x is congruent to k_i(mod m_i) for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for Zm_1 X......X Zm_n.)
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Janu42 View Post
    1. Is Z x Z cyclic? [Here Z means (Z, +).]
    no, however, it is generated by two elements, (1,0) and (0,1), i.e, <(1,0),(0,1)> = Z x Z
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  3. #3
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    Quote Originally Posted by Janu42 View Post
    2. Prove the Chinese remainder theorem: If m_1,.....,m_n are positive integers such that  m_i and m_j are relatively prime for i not equal to j, and k_1,....,k_n are any integers, then there is an integer x such that x is congruent to k_i(mod m_i) for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for Zm_1 X......X Zm_n.)
    If m_1,...,m_n are pairwise coprime then it means G=\mathbb{Z}_{m_1}\times ... \times \mathbb{Z}_{m_n} is a cyclic group, furthermore, ([1]_{m_1},...,[1]_{m_n}) is a generator for this group. Let k_1,...,k_n\in \mathbb{Z} and consider ([k_1]_{m_1}, ... , [k_n]_{m_n}) \in G. Therefore, there exists x\in \mathbb{Z} so that x([1]_{m_1},...,[1]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n}), but that means ([x]_{m_1}, ... , [x]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n}). Therefore, [x] = [k_j]_{m_j} for all 1\leq j\leq n. Thus, x\equiv k_j(\bmod m_j).
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