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. 



Quotients in Sums of Rings
There are several issues to address here, but let me first point out that the formula for $B\_n$ at the top of p. 71 of our Memoir has a typo: it should say
$B\_n = (Z/p^{j+1})^s \oplus (Z/p^j)^{p^2-p-s} $ where $0 < s \leq p^2-p$ and 
$n = 2j(p^2-p)+2s+2p-3$. The point is that there should be $(p^2-1) - (p-1) = p^2-p$ summands: there are $p^2-1$ in all, and $A\_n$ has already accounted for $p-1$ of them.


Second, the image of $Z[x]/(1+x+x^2+x^3)$ in $Z[x]/(1+x) \times Z[x]/(1+x^2)$ has index 2
under the map you describe: $a+bx+cx^2$ maps to $(a-b+c, a-c+bx)$, and $a-b+c \equiv a-c+b$ mod 2. The Chinese Remainder theorem is a theorem about PIDs and $Z[x]$ is not a PID, so a bit more care is needed.


Third, the groups $ku\_{2i-1}BC\_4$ are (writing a+b for $Z/a \oplus Z/b$, etc.):
4, 8+2, 16+2+2, 32+4+2, 64+4+4, etc. For $BC\_9$ you'd get 9, 9+9, 27+9+3, 27+27+3+3, 81+27+3+3+3, 81+81+3+3+3+3, 243+81+3+3+3+3+3, 243+243+3+3+3+3+3+3, 729+243+9+3+...+3, etc.
The formulas for A\_n and B\_n on p.71 of the Memoir are just saying this in general.


Fourth, as a sanity check, $ku\_1BC\_n = Z[x]/(1+x+\cdots+x^{n-1},1-x)$ does give $ku\_1BC\_n = Z/n$.


Fifth, I don't recall how I worked this out (it has been 10 years!) but it is an instance of an interesting general question about Smith Normal Form: what invariants of an integer matrix are needed to predict the Smith Normal Form of powers of that matrix? (Any PID will do here, not just the integers.) The work on Horn's inequalities tells us the possible values of SNF(AB) in terms of SNF(A) and SNF(B). The closer det(A) and det(B) are to being relatively prime, the narrower the range of possibilities, until SNF(AB) = SNF(A)SNF(B) in the case where they are relatively prime. (This is a triviality: an extension whose subgroup and quotient group have relatively prime orders must split.) With SNF(A^i), we are at the opposite extreme. Nonetheless, someone who understands the work on Horn's inequalities might be able to give a precise answer. It'd be interesting to see.


