Let $\mathbb{Q}$ be the rationals and $\mathbb{Z}$ integers. Let further $p$ be prime and $t\in \mathbb{Z}$ such that $p \mid t$. Then $\mathbb{Z}_{(p)}$ is the local ring. Let $G < H$ be groups, then $\mathbb{Z}_{(p)}[G] < \mathbb{Q}[H]$ are grouprings.

Now I am looking for an isomorphic $\mathbb{Z}_{(p)}[G]$-module to the following tensor product

\begin{align} \mathbb Q[H] \otimes_{\mathbb{Z}_{(p)}[G]} \mathbb{Z}/t\mathbb{Z} \end{align}

knowing that $t$ ist invertible in $\mathbb{Q}$.

  • $\begingroup$ You want an explicit description of this tensor product? $\endgroup$ – BIS HD Oct 24 '13 at 11:41
  • $\begingroup$ @BISHD an isomorphism: $Q[H] \otimes_{\mathbb{Z}_{(p)}[G]} \mathbb{Z}/t\mathbb{Z} \cong ?$ $\endgroup$ – R2D2 Oct 24 '13 at 11:45
  • $\begingroup$ Isn't it just $0$ ? $\endgroup$ – Cantlog Oct 24 '13 at 11:52
  • $\begingroup$ I hope so. But I don't how to show, since I am not well practiced with such isomorphisms. $\endgroup$ – R2D2 Oct 24 '13 at 11:55

Take $q \in {\mathbb Q}[H]$ and $a \in {\mathbb Z}_{(p)}$. Then $q \otimes a = (t \frac{1}{t} q) \otimes a = (\frac{1}{t} q) \otimes ta = \frac{1}{t} q \otimes 0 = 0.$ Therefore ${\mathbb Q}[H] \otimes_{{\mathbb Z}_{(p)}[G]} \mathbb Z/t\mathbb Z = 0$.

Note neither the groups $G$ and $H$, nor the localization at $(p)$ play a role in this argument.


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