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I'm trying to solve the following problem.

Let $m,n \in \mathbb{N}, m|n$. Prove that $\mathbb{Z[D_{n}]}/ \langle R^{m} - 1 \rangle \sim \mathbb{Z[D_{m}]}$.

I'm trying to prove it using a hands on approach exhibiting the isomorphism between both rings without much luck. I can't think of any way of doing it. The hint given by the notes is that I should find a morphism and it's inverse, so I'm led to believe it shouldn't be that difficult but it is for now.

Any hint so that I can solve it will be more than helpful. Thanks a lot!

hedphelym
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    Use $\langle x\rangle$ for $\langle x\rangle$. – Shaun Jul 09 '18 at 21:35
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    Hint: A map $$\mathbb{Z}[\mathbb{D}_n]/\langle R^m -1 \rangle \to \mathbb{Z}[\mathbb{D}_m]$$ is uniquely determined by the image of the generators of $\mathbb{D}_n.$ What is a reasonable choice for the image of these generators? – Aurel Jul 09 '18 at 22:13
  • @Aurel Thanks, I was suspecting that was the case as I knew the converse to be true. I will try to make a reasonable morphism. – hedphelym Jul 09 '18 at 22:28
  • @Aurel So I was thinking that I should send s to s and then as $\langle R \rangle$ is like a copy of $\mathbb{Z}_n$ then it suffices to show where I send R. – hedphelym Jul 09 '18 at 22:50
  • So that the morphism $ \phi $ that sends $R^i$ to $R^j$ where $i=j \mod m$. It can be clearly seen that $\langle R^m -1 \rangle \subset Ker( \phi ) $ – hedphelym Jul 09 '18 at 23:14
  • I'm having trouble proving the other inclusion. Could someone give me a hint or it isn't this the way to go. – hedphelym Jul 10 '18 at 01:39

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Let $m | n$ be two positive integers. Consider the morphism

$$ \newcommand{\zd}[1]{\mathbb{Z}[\mathbb{D}_{#1}]} \begin{align} g : \zd{n}& \to\zd{m} \\ & 1 \mapsto 1 \\ & r \mapsto \rho \\ & s \mapsto \sigma \end{align} $$

with $r,s$ and $\rho,\sigma$ the generators of the corresponding dihedral groups. This mapping is clearly surjective, and we also have $\langle r^m-1 \rangle \subset \ker g$, since

$$ g(a(r^m-1)) = g(a)(\rho^m-1) = g(a)(1-1) = 0. $$

Via the first isomorphism theorem it would suffice to see the other inclusion, so that $\ker g = \langle r^m -1 \rangle$ and therefore $\zd{n}/\langle r^m-1\rangle \simeq \zd{m}$.

Let $x = \sum_{s=1}^na_sr^s + bs\in \zd{n}$. Now, if

$$ 0 = g(x) = \sum_{s=1}^na_s\rho^s + b\sigma\in \zd{m}, $$

then $b = 0$ and $\sum_{s=1}^na_s\rho^s = 0$. Noting $n = mk$, we get

$$ 0 = \sum_{s=1}^na_s\rho^s = \sum_{i=1}^m\sum_{j=0}^{k-1}a_{mj+i}\rho^{mj+i} = \sum_{i=1}^m\left(\sum_{j=0}^{k-1}a_{mj+i}\right)\rho^{i} $$

and so

$$ \sum_{j=0}^{k-1}a_{mj+i} = 0 \quad (\forall i). \tag{$\star$} $$

Hence, we have that

$$ \begin{align} x &= \sum_{i=1}^m\sum_{j=0}^{k-1}a_{mj+i}r^{mj+i} = \sum_{i=1}^mr^i\sum_{j=0}^{k-1}a_{mj+i}r^{mj} = \sum_{i=1}^mr^i\left[\sum_{j=1}^{k-1}a_{mj+i}r^{mj} + a_i\right]\\ & \stackrel{(\star)}{=} \sum_{i=1}^mr^i\left[\sum_{j=1}^{k-1}a_{mj+i}r^{mj} -\sum_{j=1}^{k-1}a_{mj+i}\right] = \sum_{i=1}^mr^i\sum_{j=1}^{k-1}a_{mj+i}(r^{mj}-1) \\ & = \sum_{j=1}^{k-1}(r^{mj}-1)\sum_{i=1}^mr^ia_{mj+i}. \end{align} $$

To see that $x \in \langle r^m -1 \rangle$, it suffices to see that each summand is in this ideal, and moreover we can reduce this to proving $(r^{mj}-1) \in \langle r^m -1\rangle$ for each $j \in \{1,\dots,k-1\}$. In effect, by the 'difference of powers' equality,

$$ r^{mj}-1 = (r^m)^j - 1^j = (r^m-1)\sum_{l=0}^{j-1}(r^m)^l \in \langle r^m-1 \rangle. $$

qualcuno
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