TITLE: Quotients in Sums of Rings
QUESTION [8 upvotes]: Suppose we are given a commutative ring $R$ with a unit. Suppose that $R$ is the direct product of two rings $R\cong R_1\times R_2$. It's straightforward to show that any ideal $I\subset R$ maps to an ideal $I_1\times I_2\subset R_1\times R_2$ by the above isomorphism. It is, however, not straightforward at all to give a proper description of $I_1$ and $I_2$. 
To be precise, my problem is the following, arising from the book "The Connective K-Theory of Finite Groups" by Bruner and Greenlees.
Let $C_n$ be the cyclic group with $n$ elements; then in odd degrees $2i-1$ we have 
$ku_{2i-1}(BC_n)\cong \left(\mathbb{Z}[\alpha]/(1+\alpha+\ldots+\alpha^{n-1})\right)/(1-\alpha)^i$
I am especially interested in the case where $n=p^k$ for $p$ an odd prime and $k\geq2$. By the Chinese remainder theorem we have an isomorphism 
$$\mathbb{Z}[\alpha]/(1+\alpha+\ldots+\alpha^{p^k-1})\cong\prod\limits_{\begin{array}{c}d\mid p^k\\ d>1\end{array}}\mathbb{Z}[\alpha]/\Phi_d(\alpha)$$
where $\Phi_d(\alpha)$ is the $d^{th}$ cyclotomic polynomial.
If I don't make any silly mistakes the isomorphism should be given by mapping the residue class of $\alpha$ to the tuple which has the residue class of $\alpha$ at every single entry. However, if I take $p=2$ and $k=2$ the first component of $ku_{4j+2s+1}(BC_4)$ would result to be $\mathbb{Z}/p^j$ instead of $\mathbb{Z}/p^{j+1}$.
It would be great if anyone could give me hint or a reference how to solve this issue.

REPLY [6 votes]: Pardon me for not editing my previous answer, but this is really a different answer, not an improvement on my previous answer, which addresses different aspects of the (neighborhood of) the question.
The fact that $Z[x]/(fg) \longrightarrow Z[x]/(f) \times Z[x]/(g)$ is not iso here makes the attempt to replace $Z[x]/(1+x+\cdots+x^{n-1})$ by $\prod_d Z[x]/(\Phi_d(x))$ fail!
Take the simplest case, n=4. Let $R = Z[x]/(1+x+x^2+x^3)$, $R_1=Z[x]/(1+x)$ and $R_2 = Z[x]/(1+x^2)$.  The map $1-x : R \longrightarrow R$ and the map  $1-x : R_1 \times R_2 \longrightarrow R_1 \times R_2$ do not have the same cokernel!  The cokernel of the former is Z/4, but the coker of the latter is $Z/2 \times Z/2$ !  The cokernel of $R \longrightarrow R_1 \times R_2$ is $Z/2$, but the map induced on this coker by 1-x is trivial.  The snake lemma's ker-coker sequence is amusing here, and if I knew how to make mathJax display commutative diagrams, I'd have used fewer words to say all this.
This shows that Smith Normal Form is not invariant under monomorphisms!  Lovely example.