# Tensor Product of Modules

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• Dec 9th 2011, 06:47 AM
slevvio
Tensor Product of Modules
Hello all, I was wondering if I could get some help proving the following statement:

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

where $\mathbb{Z}_{\langle 2 \rangle}$ 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.
• Dec 9th 2011, 08:49 AM
NonCommAlg
Re: Tensor Product of Modules
Quote:

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

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

where $\mathbb{Z}_{\langle 2 \rangle}$ 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 $R$ be a commutative ring with 1 and let $I$ be an ideal of $R.$ let $S$ be a multiplicatively closed subset of $R$ with $1 \in S$ and $0 \notin S.$ suppose that every element of $S$ is invertible modulo $I,$ i.e. $s + I$ is an invertible element of $R/I$ for all $s \in S.$ then

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

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

$(R/I) \otimes_R S^{-1}R \cong S^{-1}(R/I) \cong R/I.$