# Isomorphism of Tensor Product over a Group Ring.

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 $\mathbb{Q}$.

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