# Chinese remainder theorem and isomorphism

Suppose that there are different primes $p_1$ and $p_2$ and the group $C_{p_1}$. I would like to decompose the following tensorproduct and thought of using the chinese remainder theorem to get something like: $$\mathbb{Z}_{p_2} \otimes _\mathbb{Z} \mathbb{Z}G \cong \mathbb{Z}_{p_2} \bigoplus\limits_{?} \mathbb{Z}[\theta_{p_1}^?]$$ where $\theta_{p_1}$ is a primitive root of unity. But I'm not sure over what one should sum. Thank you very much for each hint.

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Write $p=p_1$ and $\ell=p_2$. Note that $\mathbb ZG=\mathbb Z[X]/(X^p-1)$. So $$\mathbb Z_\ell\otimes \mathbb ZG= \mathbb Z_\ell[X]/(X^p-1).$$ The polynomial $X^p-1$ is separable in $\mathbb F_\ell[X]$ and decomposes into a product of pairwise coprime irreducible factors $h_1(X)\dots h_n(X)$. As $\mathbb Z_\ell$ is complete, Hensel lemma says that $$X^p-1=H_1(X)\dots H_n(X)$$ with $H_i(X)\in\mathbb Z_\ell[X]$ reducing to $h_i(X)$ mod $\ell$. Hence $$\mathbb Z_\ell\otimes \mathbb ZG=\oplus_i \mathbb Z_\ell[X]/(H_i(X))=\oplus_i \mathbb Z_\ell[\theta_i]$$ where $\theta_i$ is a root of $H_i(X)$. We can take $h_1(X)=X-1$, hence $H_1(X)=X-1$ and the first fact in the above decomposition is just $\mathbb Z_\ell$. The other $\theta_i$'s are primitive $p$-th roots of unit because they are roots of $X^p-1$ and are different from $1$.

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