# Chinese Remainder Theorem backwards

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.

## 2 Answers

Since $h_k := \prod_{i=1}^k \phi_{p^i} = (X^{p^k}-1)/(X-1)$, you are trying to compute the exponent of the cokernel of the inclusion $$\mathbf{Z}[X]/(h_k) \rightarrow \prod_{i=1}^k \mathbf{Z}[\zeta_{p^i}]$$ that is of finite index (since it becomes an isomorphism upon tensoring with $\mathbf{Q}$). We will prove that the exponent of the cokernel is always exactly $p^{k-1}$, as you guessed from examples. First we show it divides $p^{k-1}$, and then we revisit the argument to get the exact equality. The key input to the proof is an understanding of the basic ramification theory of cyclotomic integer rings (as in any basic book on algebraic number theory), together with a little bit of visualization coming from algebraic geometry inside the affine line over $\mathbf{Z}$.

To show that the exponent of the cokernel divides $p^{k-1}$ we proceed by induction on $k$, the case $k=1$ being clear. For $k \ge 2$, consider the natural map $$\mathbf{Z}[X]/(h_k) \rightarrow (\mathbf{Z}[X]/(h_{k-1})) \times_{\mathbf{F}_p[\varepsilon]/(\varepsilon^{p^{k-1}}-1)} \mathbf{Z}[\zeta_{p^k}]$$ where the fiber product ring has first projection $X \mapsto 1 + \varepsilon$ and second projection $\zeta_{p^k} \mapsto 1+\varepsilon$. A separate induction on $k$ using ramification theory of $p$-power cyclotomic fields shows that this map of rings is an isomorphism; see Lemmas 2.2 and 2.3 in the paper "Finite-order automorphisms of a certain torus" in Michigan Math Journal. (In algebro-geometric terms, this inductively describes how ${\rm{Spec}}(\mathbf{Z}[X]/(h_k))$ is built via "gluing along closed subschemes" inside the affine line over $\mathbf{Z}$ via the various closed subschemes given by ${\rm{Spec}}(\mathbf{Z}[\zeta_{p^i}])$ for $i \le k$.)

This ring isomorphism onto the fiber product is compatible with the initial inclusion we are trying to study. So taking $k \ge 2$, by induction if we multiply an element $$(z_1,\dots, z_{k-1}) \in \prod_{i=1}^{k-1} \mathbf{Z}[\zeta_{p^i}]$$ by $p^{k-2}$ then we get something inside $\mathbf{Z}[X]/(h_{k-1})$. Choosing an additional element $z_k \in \mathbf{Z}[\zeta_{p^k}]$, the obstruction to having $$(p^{k-2}z_1,\dots,p^{k-2}z_{k-1},p^{k-2}z_k) \in \mathbf{Z}[X]/(h_k)$$ is whether or not the images of $(p^{k-2}z_1,\dots,p^{k-2}z_{k-1})$ and $p^{k-2}z_k$ in $\mathbf{F}_p[\varepsilon]/(\varepsilon^{p^{k-1}}-1)$ coincide. Maybe they do, or maybe they don't, but certainly if we multiply both sides by $p$ then their images in $\mathbf{F}_p[\varepsilon]/(\varepsilon^{p^{k-1}}-1)$ coincide (in fact, are each 0). Hence, we conclude that multiplying throughout by $p^{k-1}$ puts us inside the fiber product ring which coincides naturally with $\mathbf{Z}[X]/(h_k)$. This completes the induction on $k$, so indeed the exponent always divides $p^{k-1}$.

Now we revisit the induction to show that the exponent is exactly $p^{k-1}$. For $k=1$ this is obvious, so assume $k > 1$ and by induction we can find $(z_1,\dots,z_{k-1})$ such that no $p^i$ with $0 \le i < k-2$ multiplies $(z_1,\dots,z_{k-1})$ into $\mathbf{Z}[X]/(h_{k-1})$. Hence, for any $z_k \in \mathbf{Z}[\zeta_{p^k}]$ it follows that the least $i \ge 0$ such that $p^i(z_1,\dots,z_k) \in \mathbf{Z}[X]/(h_k)$ is either $i=k-2$ or $i=k-1$. We therefore just need to find $z_k$ so that the elements $(p^{k-2}z_1,\dots,p^{k-2}z_{k-1}) \in \mathbf{Z}[X]/(h_{k-1})$ and $p^{k-2}z_k \in \mathbf{Z}[\zeta_{p^k}]$ have distinct images in $\mathbf{F}_p[\varepsilon]/(\varepsilon^{p^{k-1}}-1)$. By design our element in $\mathbf{Z}[X]/(h_{k-1})$ is not divisible by $p$ in here, so its image under the first projection of the fiber product is nonzero (as that projection is readily seen to be exactly "reduction modulo $p$"). Thus, if $k > 2$ then we win for any $z_k$ (as $p^{k-2}z_k$ has vanishing image under the second projection in the fiber product). Finally, to backstep and deal with $k=2$, in that case we just have to pick $z_2 \in \mathbf{Z}[\zeta_{p^2}]$ so that its image in $\mathbf{F}_p[\varepsilon]/(\varepsilon^p-1)$ avoids a specified nonzero element. But that second projection in the fiber product is visibly surjective and there is more than one nonzero element in $\mathbf{F}_p[\varepsilon]/(\varepsilon^p-1)$ (almost a close call when $p=2$), so a suitable $z_k$ can always be found.

QED

• thanks for your answer too. I have worked through everything apart from the last paragraph so far and this part already gives what I had been after so far and I get the upper bound that Emil's proof couldn't supply. So I guess this already answers my question though strictly speaking I had asked for a strict bound. I will read through rest now and see if and how this can be of even greater help. [Edit:] alright, this last part was really straight forward. I will figure out, if I can actually use this extra bit of precision. – Felix Springer Jul 3 '14 at 14:06

If $q\ne p$ is a prime, the cyclotomic polynomials $\phi_{p^i}$ are coprime over the finite field $\mathbb F_q$. An identity $$\tag{*}1=\sum_{i\le k}f_i(x)\prod_{j\ne i}\phi_{p^j}(x)$$ with $f_i\in\mathbb F_q[x]$ can be lifted to one with $f_i(x)\in\mathbb Z_q[x]$ using Hensel’s lemma. If we further demand $\deg(f_i)<\deg(\phi_{p^i})$, these polynomials are unique, hence the unique such solution of $(*)$ in $\mathbb Q[x]$ must have coefficients in $$\mathbb Q\cap\bigcap_{q\ne p}\mathbb Z_q=\mathbb Z[p^{-1}].$$

• thanks Emil. This is very helpful. Indeed I will need a little more to go on. However this gives me the existence of an upper bound already. – Felix Springer Jul 2 '14 at 14:16
• Alternatively, to show that the inclusion $\mathbf{Z}[X]/(\prod_{i=1}^k \phi_{p^i}) \rightarrow \prod_{i=1}^k \mathbf{Z}[X]/(\phi_{p^i})$ has no prime $\ell \ne p$ dividing its index it suffices to check equality modulo $\ell$. That in turn is just the Chinese Remainder Theorem for the separable polynomial $(X^{p^k}-1)/(X-1) = \prod_{i=1}^k \phi_{p^i}$ over $\mathbf{F}_{\ell}$! – user27920 Jul 2 '14 at 14:18