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Thread: Proof: Z/nZ (x) Q = 0

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

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

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

    $\displaystyle (a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\$
    $\displaystyle = (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}})
    = abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\ $
    $\displaystyle = (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 $\displaystyle \otimes_{\mathbb{Q}}$).
    If it is $\displaystyle \otimes_{\mathbb{Z}}$, then you should rewrite $\displaystyle b$ as a fraction of integers, since you factor the fraction $\displaystyle b$ 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: $\displaystyle \mathbb Z/n\mathbb Z \otimes \mathbb Q = 0$
    I proofed it as follows but I don't know if it is correct:

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

    $\displaystyle (a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\$
    $\displaystyle = (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}})
    = abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\ $
    $\displaystyle = (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: $\displaystyle (a + n \mathbb{Z}) \otimes r = (a + n\mathbb{Z}) \otimes (n \cdot n^{-1}r)=(na + n\mathbb{Z}) \otimes n^{-1}r = 0 \otimes n^{-1}r = 0.$
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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 $\displaystyle \otimes_{\mathbb{Q}}$).
    it has to be $\displaystyle \otimes_{\mathbb{Z}}$ because $\displaystyle \mathbb{Z}/n\mathbb{Z}$ is not a $\displaystyle \mathbb{Q}$ module.
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