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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.

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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.

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  • $\begingroup$ Thank you for the extensive answer. I had already known the first issue and this makes perfectly sense. I could also figure out the second issue and it seems likely to me that the analogous statements will hold for different primes. I will still have to figure out what the index means to the third issue precisely. This should also point into the same direction as your other answer. The sanity check works and it should work for the same number-theoretic reason as third. The fifth issue doesn't tell me anything at the moment, so I will have to do some investigation on that. $\endgroup$ – Felix Springer May 22 '13 at 13:06
  • $\begingroup$ I guess, there is another typo or mistake in the formula for $A_n$. When $j=0$ the second part of $A_n=(Z/p^{j+2})^s\oplus(Z/p^{j+1})^{p-1-s}$ should not be there. I've almost figured it out for $p=3$ (still need a general argument) and it seems that the pattern should be like this for higher $p$. $\endgroup$ – Felix Springer Jun 3 '13 at 9:06
  • $\begingroup$ Formulating the question as $ku_{2i-1} = A^i$ for appropriate A, the key step is determining the periodicity, i.e., an integer $d$ and diagonal matrix $D$ such that SNF(A^{i+d}) = D*SNF(A^i). This reduces the work to a (small) finite number of cases. $\endgroup$ – Robert Bruner Jun 3 '13 at 14:43
  • $\begingroup$ I meant $ku_{2i-1} = Coker(A^i)$, of course. And $SNF(A^{i+d})) = D * SNF(A^i)$ should have been in math mode. $\endgroup$ – Robert Bruner Jun 4 '13 at 15:20
  • $\begingroup$ do you have reference for the stuff on Horn's inequalities? Everything I can find is on the comparison of the eigenvalues of matrices A,B and A+B. To me this seems to be quite different. $\endgroup$ – Felix Springer Jun 5 '13 at 11:57
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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.

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  • $\begingroup$ Oooh, nice! This makes my answer moot, of course :) $\endgroup$ – Dylan Wilson May 21 '13 at 18:59
  • $\begingroup$ @Dylan: Thanks for your comment and the answer that has been deleted meanwhile too. The failure of the 1-1 correspondence above made me wonder and fail ;-) $\endgroup$ – Felix Springer May 22 '13 at 13:08

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