I have the following situation, that is much alike the Chinese Remainder Theorem. Let $\phi_d(\alpha)$ be the $d^{th}$ cyclotomic polynomial in the variable $\alpha$ (I'm not specifying the coefficient ring on purpose here). Now for the different powers of the same prime $p$ the cyclotomic polynomials are coprime when the coefficient ring is $\mathbb{Q}$ but they are only relatively prime when the coefficient ring is $\mathbb{Z}$. Now we have the obvious map:

\begin{align*} \mathbb{Z}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)&\rightarrow\prod^k_{i=1}\mathbb{Z}[\alpha]/\phi_{p^i}(\alpha)\\ \alpha&\mapsto(\alpha,\ldots,\alpha) \end{align*}

As @RobertBruner kindly pointed out here this map is 1-1 but not onto. But of course if we tensor with $\mathbb{Q}$ this map is an isomorphism and if we set $P=\prod^k_{i=1}\phi_{p^i}(\alpha)$ the inverse is given by

\begin{align*} \prod^k_{i=1}\mathbb{Q}[\alpha]/\phi_{p^i}(\alpha)&\rightarrow\mathbb{Q}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)\\ (x_i)&\mapsto\sum^k_{i=1}x_i\cdot\frac{P}{\phi_{p^i}(\alpha)}\cdot p_i \end{align*}

where $p_i$ is some polynomial to be determined. Now I would think that, for the general case, it would be pretty tedious to actually get these polynomials. And in fact I am only interested the denominators of the coefficients, i.e. I would like to know the smallest subring of $\mathbb{Q}$ that suffices as coefficient-ring in order that the above map be an isomorphism.

I have done a few calculations and all of them suggest that all occurring denominators are powers of $p$. In fact the highest denominator occurring always was $p^{k-1}$. Though this is all speculating...

Does anyone have an idea how compute these denominators or at least to show that they are $p$-powers? I have thought about using the matrix of the map, considered as a map of free $\mathbb{Z}$-modules, however, its determinant increases rapidly and with no visible pattern when $k$ increases.

Finally there is reason why the denominators should be as proposed as this is part of the calculation of $ku_*(B\mathbb{Z}/p^k)$ as pointed out in the thread mentioned above. This in turn is a torsion module and its torsion is related to $p$.

PD: This is surely a follow-up of this thread and I apologize if this is regarded as wrong cross-posting. But I think this addresses a different question and different kinds of mathematics, so a new thread seemed appropriate.

I have the following situation, that is much alike the Chinese Remainder Theorem. Let $\phi_d(\alpha)$ be the $d^{th}$ cyclotomic polynomial in the variable $\alpha$ (I'm not specifying the coefficient ring on purpose here). Now for the different powers of the same prime $p$ the cyclotomic polynomials are coprime when the coefficient ring is $\mathbb{Q}$ but they are only relatively prime when the coefficient ring is $\mathbb{Z}$. Now we have the obvious map:

\begin{align*} \mathbb{Z}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)&\rightarrow\prod^k_{i=1}\mathbb{Z}[\alpha]/\phi_{p^i}(\alpha)\\ \alpha&\mapsto(\alpha,\ldots,\alpha) \end{align*}

As @RobertBruner kindly pointed out here this map is 1-1 but not onto. But of course if we tensor with $\mathbb{Q}$ this map is an isomorphism and if we set $P=\prod^k_{i=1}\phi_{p^i}(\alpha)$ the inverse is given by

\begin{align*} \prod^k_{i=1}\mathbb{Q}[\alpha]/\phi_{p^i}(\alpha)&\rightarrow\mathbb{Q}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)\\ (x_i)&\mapsto\sum^k_{i=1}x_i\cdot\frac{P}{\phi_{p^i}(\alpha)}\cdot p_i \end{align*}

where $p_i$ is some polynomial to be determined. Now I would think that, for the general case, it would be pretty tedious to actually get these polynomials. And in fact I am only interested the denominators of the coefficients, i.e. I would like to know the smallest subring of $\mathbb{Q}$ that suffices as coefficient-ring in order that the above map be an isomorphism.

I have done a few calculations and all of them suggest that all occurring denominators are powers of $p$. In fact the highest denominator occurring always was $p^{k-1}$. Though this is all speculating...

Does anyone have an idea how compute these denominators or at least to show that they are $p$-powers? I have thought about using the matrix of the map, considered as a map of free $\mathbb{Z}$-modules, however, its determinant increases rapidly and with no visible pattern when $k$ increases.

Finally there is reason why the denominators should be as proposed as this is part of the calculation of $ku_*(B\mathbb{Z}/p^k)$ as pointed out in the thread mentioned above. This in turn is a torsion module and its torsion is related to $p$.

PD: This is surely a follow-up of this thread and I apologize if this is regarded as wrong cross-posting. But I think this addresses a different question and different kinds of mathematics, so a new thread seemed appropriate.

I have the following situation, that is much alike the Chinese Reminder Theorem. Let $\phi_d(\alpha)$ be the $d^{th}$ cyclotomic polynomial in the variable $\alpha$ (I'm not specifying the coefficient ring on purpose here). Now for the different powers of the same prime $p$ the cyclotomic polynomials are coprime when the coefficient ring is $\mathbb{Q}$ but they are only relatively prime when the coefficient ring is $\mathbb{Z}$. Now we have the obvious map:

\begin{align*} \mathbb{Z}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)&\rightarrow\prod^k_{i=1}\mathbb{Z}[\alpha]/\phi_{p^i}(\alpha)\\ \alpha&\mapsto(\alpha,\ldots,\alpha) \end{align*}

As @RobertBruner kindly pointed out here this map is 1-1 but not onto. But of course if we tensor with $\mathbb{Q}$ this map is an isomorphism and if we set $P=\prod^k_{i=1}\phi_{p^i}(\alpha)$ the inverse is given by

\begin{align*} \prod^k_{i=1}\mathbb{Q}[\alpha]/\phi_{p^i}(\alpha)&\rightarrow\mathbb{Q}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)\\ (x_i)&\mapsto\sum^k_{i=1}x_i\cdot\frac{P}{\phi_{p^i}(\alpha)}\cdot p_i \end{align*}

where $p_i$ is some polynomial to be determined. Now I would think that, for the general case, it would be pretty tedious to actually get these polynomials. And in fact I am only interested the denominators of the coefficients, i.e. I would like to know the smallest subring of $\mathbb{Q}$ that suffices as coefficient-ring in order that the above map be an isomorphism.

I have done a few calculations and all of them suggest that all occurring denominators are powers of $p$. In fact the highest denominator occurring always was $p^{k-1}$. Though this is all speculating...

Does anyone have an idea how compute these denominators or at least to show that they are $p$-powers? I have thought about using the matrix of the map, considered as a map of free $\mathbb{Z}$-modules, however, its determinant increases rapidly and with no visible pattern when $k$ increases.

Finally there is reason why the denominators should be as proposed as this is part of the calculation of $ku_*(B\mathbb{Z}/p^k)$ as pointed out in the thread mentioned above. This in turn is a torsion module and its torsion is related to $p$.

PD: This is surely a follow-up of this thread and I apologize if this is regarded as wrong cross-posting. But I think this addresses a different question and different kinds of mathematics, so a new thread seemed appropriate.

I have the following situation, that is much alike the Chinese Remainder Theorem. Let $\phi_d(\alpha)$ be the $d^{th}$ cyclotomic polynomial in the variable $\alpha$ (I'm not specifying the coefficient ring on purpose here). Now for the different powers of the same prime $p$ the cyclotomic polynomials are coprime when the coefficient ring is $\mathbb{Q}$ but they are only relatively prime when the coefficient ring is $\mathbb{Z}$. Now we have the obvious map:

\begin{align*} \mathbb{Z}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)&\rightarrow\prod^k_{i=1}\mathbb{Z}[\alpha]/\phi_{p^i}(\alpha)\\ \alpha&\mapsto(\alpha,\ldots,\alpha) \end{align*}

As @RobertBruner kindly pointed out here this map is 1-1 but not onto. But of course if we tensor with $\mathbb{Q}$ this map is an isomorphism and if we set $P=\prod^k_{i=1}\phi_{p^i}(\alpha)$ the inverse is given by

\begin{align*} \prod^k_{i=1}\mathbb{Q}[\alpha]/\phi_{p^i}(\alpha)&\rightarrow\mathbb{Q}[\alpha]/\left(\prod^k_{i=1}\phi_{p^i}(\alpha)\right)\\ (x_i)&\mapsto\sum^k_{i=1}x_i\cdot\frac{P}{\phi_{p^i}(\alpha)}\cdot p_i \end{align*}

where $p_i$ is some polynomial to be determined. Now I would think that, for the general case, it would be pretty tedious to actually get these polynomials. And in fact I am only interested the denominators of the coefficients, i.e. I would like to know the smallest subring of $\mathbb{Q}$ that suffices as coefficient-ring in order that the above map be an isomorphism.

I have done a few calculations and all of them suggest that all occurring denominators are powers of $p$. In fact the highest denominator occurring always was $p^{k-1}$. Though this is all speculating...

Does anyone have an idea how compute these denominators or at least to show that they are $p$-powers? I have thought about using the matrix of the map, considered as a map of free $\mathbb{Z}$-modules, however, its determinant increases rapidly and with no visible pattern when $k$ increases.

Finally there is reason why the denominators should be as proposed as this is part of the calculation of $ku_*(B\mathbb{Z}/p^k)$ as pointed out in the thread mentioned above. This in turn is a torsion module and its torsion is related to $p$.

PD: This is surely a follow-up of this thread and I apologize if this is regarded as wrong cross-posting. But I think this addresses a different question and different kinds of mathematics, so a new thread seemed appropriate.