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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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Alphonse
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  • As what structure do you consider $\mathbb{Z}^n$? An abelian group (a module over the integers) or as a ring? On the one hand, the former is natural given tags and what you link to, on the other hand the notion ideal makes sense only in the latter case. – quid Sep 01 '16 at 12:50
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    @WSL: $I= \left<(3,0), (0,2)\right>$ is generated by $(3,2)$. – martini Sep 01 '16 at 12:52
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    you're right. I was thinking about it as an abelian group. – Spencer Leslie Sep 01 '16 at 12:55
  • @martini Based on the linked questions, the OP is interested in abelian groups, not rings (why the word "ideal" was used is not clear though). – Najib Idrissi Sep 01 '16 at 12:55
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    @all : sorry, I was not clear. I was thinking to abelian groups, so that subgroups are not necessarily generated by only one element. – Alphonse Sep 01 '16 at 13:01
  • So for instance $\mathbb Z^n/\langle (a,...,a) \rangle$ and $\mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ are not isomorphic as rings (just to see the difference between abelian groups/rings)? – Alphonse Sep 01 '16 at 13:02
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    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. – quid Sep 01 '16 at 13:04
  • @quid: yes, thank you! We have $\Bbb Z^n/\langle (a,...,a)\rangle \cong (Z/aZ)^n$ as rings if $\langle . \rangle$ means "ideal generated by". But if it means "subgroup generated by", then $\Bbb Z^n/\langle (a,...,a)\rangle \cong (Z/aZ) \times Z^{n-1}$ as rings is right? – Alphonse Sep 01 '16 at 13:22
  • @Alphonse, as abelian groups, not as rings. – Spencer Leslie Sep 01 '16 at 13:26
  • @WSL: Thank you, ah yes because the first one has characteristic $a$, and the second one characteristic $0$ [do you agree?].$\tag*{}$ But if $\langle . \rangle$ means again "subgroup generated by", then is there an "easier" ring which $Z^n/\langle (a,...,a)\rangle$ is isomorphic to? – Alphonse Sep 01 '16 at 13:30
  • This is the 'wrong' question to start with. The one to start with is: Is the quotient of a ring with respect to an additive subgroup that is not an ideal even a ring (in a natural way)? The answer is typically "no" as the multiplication is not well defined, and I do not think this is an exception (but I did not check the details). – quid Sep 01 '16 at 13:56
  • @quid: ah yes, thank very much for pointing this out. – Alphonse Sep 01 '16 at 14:09
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    Here are some relevant posts: 1, 2, 3 – Viktor Vaughn Sep 01 '16 at 23:54
  • @Alphonse: related: http://math.stackexchange.com/questions/1864265 – Watson Sep 07 '16 at 12:31

2 Answers2

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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).

Spencer Leslie
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  • Thank you! So you are working with additive groups or with rings, just to be sure? – Alphonse Sep 01 '16 at 13:08
  • 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). – quid Sep 01 '16 at 13:08
  • @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…) – Alphonse Sep 01 '16 at 13:09
  • Related: http://snark.math.rpi.edu/Teaching/Smith_Normal_Form.html – Alphonse Sep 01 '16 at 13:24
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    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. – Spencer Leslie Sep 01 '16 at 13:24
  • Ok, apparently the Smith normal form of $A$ has two $2$ on the diagonal and one $0$, so $\Bbb Z^3/I \cong (\Bbb Z/2\Bbb Z)^2$ – Alphonse Sep 01 '16 at 13:33
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    Not quite. The $0$ indicates that the third coordinate is not affected in the quotient, so you get $\mathbb{Z}^3/I \cong \mathbb{Z}\oplus (\mathbb{Z}/2\mathbb{Z})^2$. – Spencer Leslie Sep 01 '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.

Pedro
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