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Thread: Tensor Product of Modules

  1. #1
    Senior Member slevvio's Avatar
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    Tensor Product of Modules

    Hello all, I was wondering if I could get some help proving the following statement:

    $\displaystyle \mathbb{Z} / 8 \otimes_{\mathbb{Z}} \mathbb{Z}_{\langle 2 \rangle} = \mathbb{Z} / 8$

    where $\displaystyle \mathbb{Z}_{\langle 2 \rangle}$ is the integers localised at the prime ideal <2>.

    Can anyone give me some hints as to how to calculate this?
    Any help would be appreciated.
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  2. #2
    MHF Contributor

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    Re: Tensor Product of Modules

    Quote Originally Posted by slevvio View Post
    Hello all, I was wondering if I could get some help proving the following statement:

    $\displaystyle \mathbb{Z} / 8 \otimes_{\mathbb{Z}} \mathbb{Z}_{\langle 2 \rangle} = \mathbb{Z} / 8$

    where $\displaystyle \mathbb{Z}_{\langle 2 \rangle}$ is the integers localised at the prime ideal <2>.

    Can anyone give me some hints as to how to calculate this?
    Any help would be appreciated.
    let's consider a more general case: let $\displaystyle R$ be a commutative ring with 1 and let $\displaystyle I$ be an ideal of $\displaystyle R.$ let $\displaystyle S$ be a multiplicatively closed subset of $\displaystyle R$ with $\displaystyle 1 \in S$ and $\displaystyle 0 \notin S.$ suppose that every element of $\displaystyle S$ is invertible modulo $\displaystyle I,$ i.e. $\displaystyle s + I$ is an invertible element of $\displaystyle R/I$ for all $\displaystyle s \in S.$ then

    $\displaystyle S^{-1}(R/I) \cong R/I,$

    because the natural $\displaystyle R$-module homomorphism $\displaystyle f : R/I \longrightarrow S^{-1}(R/I)$ defined by $\displaystyle f(x)=x/1$ is an isomorphism (why?). thus

    $\displaystyle (R/I) \otimes_R S^{-1}R \cong S^{-1}(R/I) \cong R/I.$
    Last edited by NonCommAlg; Dec 9th 2011 at 11:44 AM.
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