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I would like to know:

  1. What is $\Bbb Z^n/\langle(a_1, \dots, a_n)\rangle$ isomorphic to, as abelian group?
  2. More generally, if $I$ is a subgroup of $\Bbb Z^n$, then would you proceed to find $\Bbb Z^n/I$? Is there any algorithm? For instance for $I=\langle(4,0,2),(2,-2,0)\rangle$ or $J=\langle(-2,4,0,2),(2,-2,0,1)\rangle$?

My aim is to know how to compute a quotient of $\Bbb Z^n$, which has the form $$\Bbb Z^m \oplus \bigoplus_{i=1}^s \Bbb Z/p_i^{r_i} \Bbb Z$$ because it is finitely generated.

I am aware of this particular case, and of this one, and also maybe this one.

Thank you for your help!

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    $\begingroup$ @WSL: $I= \left<(3,0), (0,2)\right>$ is generated by $(3,2)$. $\endgroup$ – martini Sep 1 '16 at 12:52
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    $\begingroup$ you're right. I was thinking about it as an abelian group. $\endgroup$ – WSL Sep 1 '16 at 12:55
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    $\begingroup$ @all : sorry, I was not clear. I was thinking to abelian groups, so that subgroups are not necessarily generated by only one element. $\endgroup$ – Alphonse Sep 1 '16 at 13:01
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    $\begingroup$ No they are not. As a ring (assumin coordinatwise mult of course) the former would be $(Z/aZ)^n$. As a ring you can treat the coordinates completely separately. $\endgroup$ – quid Sep 1 '16 at 13:04
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    $\begingroup$ Here are some relevant posts: 1, 2, 3 $\endgroup$ – André 3000 Sep 1 '16 at 23:54
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The algorithm you want is called Smith normal form. This allows to to compute the quotients as follows:

Take for example your subgroup $I=\langle (4,0,2),(2,−2,0)\rangle$. Then, we can view this as $I = A\mathbb{Z}^3$, where $$A = \left(\begin{array}{ccc} 4&0&2\\ 2&-2&0\\0&0&0\end{array}\right).$$

Apply the algorithm to put $A$ in Smith normal form and you can easily read off the quotient. This also applies to 1).

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  • $\begingroup$ Thank you! So you are working with additive groups or with rings, just to be sure? $\endgroup$ – Alphonse Sep 1 '16 at 13:08
  • $\begingroup$ It may make sense to add that one does not get it precisely in the form in OP though. There is an extra step. SNF will give the finite part with ascending chain of divisors (not prime powers). $\endgroup$ – quid Sep 1 '16 at 13:08
  • $\begingroup$ @quid: You're right, but I don't care about this actually ;-) I don't need prime powers. So that's fine for me (sorry again for being a bit unprecise…) $\endgroup$ – Alphonse Sep 1 '16 at 13:09
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    $\begingroup$ I am viewing $\mathbb{Z}^n/J$ as a module over $\mathbb{Z}$; that is, as an abelian group. The reason I used that language is that this result (the form of the quotient you wrote as well as the solution via Smith normal form) holds in the more general context of modules over a PID, of which $\mathbb{Z}$ is an example. $\endgroup$ – WSL Sep 1 '16 at 13:24
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    $\begingroup$ Not quite. The $0$ indicates that the third coordinate is not affected in the quotient, so you get $\mathbb{Z}^3/I \cong \mathbb{Z}\oplus (\mathbb{Z}/2\mathbb{Z})^2$. $\endgroup$ – WSL Sep 1 '16 at 13:54
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The algorithm and theorem you want to look at is the Smith normal form, which works in general for principal ideal domains.

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