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Let $A=\mathbb{Z}[\mu_6]$ where $\mu_6$ is the 6th unit root of 1. Let $J$ be an ideal of $A$ with $A/J=\mathbb{Z}/N\mathbb{Z}$.

It is said that by the structure theorem for modules over a PID, there exists a basis $(u,v)$ of $A$ over $\mathbb{Z}$ such that $(u,Nv)$ is a basis of $J$ of $\mathbb{Z}$.

My question is how the structure theorem is applied here? I cannot see the logic here.

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As an $\mathbb Z$-module $A$ is free of rank $2$, and $J$ is, in particular, a submodule of $A$. Then $J$ is also free as $\mathbb Z$-module and by the structure theorem we get that there exists a basis $u,v$ of $A$ and $d_1,d_2$ positive integers with $d_1\mid d_2$ such that $d_1u,d_2v$ is a basis for $J$. This shows that $A/J\simeq \mathbb Z/d_1\mathbb Z\times\mathbb Z/d_2\mathbb Z$. Furthermore, the structure theorem says that this decomposition is unique. Then we get $d_1=1$ and $d_2=N$.

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Can you give me a reference for the structure theorem you state? I looked up wiki and the statement there seems not the same with the statement given by you (involving basis of $J$)...I want to see a precise description of the theorem you give. – hxhxhx88 Jan 12 '13 at 14:45
How can you see there exists a basis $u,v$ of $A$ and $d_1,d_2$ positive integers with $d_1∣d_2$ such that $d_1u,d_2v$ is a basis for $J$ at first? – hxhxhx88 Jan 12 '13 at 16:23
Look at Theorem M.4.12; it is proved all that you want to know about the structure theorem. – user26857 Jan 13 '13 at 0:21
good, thank you! – hxhxhx88 Jan 13 '13 at 1:44

I don't think you need the structure theorem for modules over a PID here. Saying that $A/J=\Bbb Z/N\Bbb Z$ amounts to the fact that the composite morphism $\Bbb Z\to A\to A/J$ is surjective and has kernel $NZ$. This means that $N$ is the least positive integer in $J$, and that every element of $A$ can be written as the sum of an element of $J$ and an integer in $\{0,\ldots,N-1\}$. In particular this holds for $\mu_6$, which together with $1$ forms a $\Bbb Z$-basis of $A$, so let $u\in A$ be such that $\mu_6=u+i$ with $i\in\{0,\ldots,N-1\}$, and $v=1$. Clearly $(u,v)=(u,1)$ is a $\Bbb Z$-basis of $A$. Also $(u,Nv)=(u,N)$ is a $\Bbb Z$-basis of $J$, since for $k,l\in\Bbb Z$ the general element $ku+lv$ of $A$ lies in $J$ if and only if $lv=l$ lies in $J$, which happens if and only if $l$ is a multiple of $N$.

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sorry I don't understand your last step, why clearly $(u,Nv)$ is a basis of $J$? – hxhxhx88 Jan 12 '13 at 15:03
@hxhxhx88: See my modified last sentence. – Marc van Leeuwen Jan 12 '13 at 19:49
OK, I understand, thank you! – hxhxhx88 Jan 13 '13 at 1:24

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