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Math Help - Proof: Z/nZ (x) Q = 0

  1. #1
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    Proof: Z/nZ (x) Q = 0

    Hello,
    I have to proof that:  \mathbb Z/n\mathbb Z \otimes  \mathbb Q = 0
    I proofed it as follows but I don't know if it is correct:

    {\text{Let }}a \in \mathbb{Z},{\text{ }}b \in \mathbb{Q}.

      (a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\
      = (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}}) <br />
= abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\
       = (0 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \cdot (1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \hfill \\

    Please correct me, if there are any mistakes.
    Thank you!

    -Lisa-
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  2. #2
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    Its correct (if it is \otimes_{\mathbb{Q}}).
    If it is \otimes_{\mathbb{Z}}, then you should rewrite b as a fraction of integers, since you factor the fraction b out of the tensor product. But that is an easy modification.
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  3. #3
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    Quote Originally Posted by lisa View Post
    Hello,
    I have to proof that:  \mathbb Z/n\mathbb Z \otimes \mathbb Q = 0
    I proofed it as follows but I don't know if it is correct:

    {\text{Let }}a \in \mathbb{Z},{\text{ }}b \in \mathbb{Q}.

     (a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\
     = (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}}) <br />
= abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\
     = (0 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \cdot (1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \hfill \\

    Please correct me, if there are any mistakes.
    Thank you!

    -Lisa-
    it's correct. you can also write: (a + n \mathbb{Z}) \otimes r = (a + n\mathbb{Z}) \otimes (n \cdot n^{-1}r)=(na + n\mathbb{Z}) \otimes n^{-1}r = 0 \otimes n^{-1}r = 0.
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  4. #4
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    Smile

    Hi,

    thank you for your answer.
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  5. #5
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    Quote Originally Posted by vemrygh View Post

    Its correct (if it is \otimes_{\mathbb{Q}}).
    it has to be \otimes_{\mathbb{Z}} because \mathbb{Z}/n\mathbb{Z} is not a \mathbb{Q} module.
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