What is $\Bbb Z^n/(a_1, \dots, a_n)$ or $\Bbb Z^n / I$ isomorphic to?
I would like to know:
- What is $\Bbb Z^n/\langle(a_1, \dots, a_n)\rangle$ isomorphic to, as abelian group?
- 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!
Solution 1:
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).
Solution 2:
The algorithm and theorem you want to look at is the Smith normal form, which works in general for principal ideal domains.