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Doing homology computations inevitably involves computing quotients of $\Bbb Z$-modules, and I am not familiar with any way of doing this.

Some examples I have been working with are:
$$\frac{\Bbb Z \oplus \Bbb Z \oplus \Bbb Z}{\langle (2,1,0)\rangle} \text{ and } \frac{\Bbb Z_2 \oplus \Bbb Z}{\langle (1,2) \rangle}.$$

I think in the first case the quotient is isomorphic to $\Bbb Z\oplus \Bbb Z$. In the second case I cannot tell. Would anyone know how to solve problems like this in general?

Open Season
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For the first case, you can notice that $\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ has a basis (linearly independent spanning set) of $\{(2,1,0), (1,1,0), (0,0,1)\}$. Then taking your quotient is just reducing the size of the basis by 1, so the result is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$ as you said.

For the second case, you can think of $\mathbb{Z}_2 \oplus \mathbb{Z}$ as $$\frac{\mathbb{Z} \oplus \mathbb{Z}}{\langle (2,0) \rangle}$$ so your given quotient is $$\frac{\mathbb{Z} \oplus \mathbb{Z}}{\langle (2,0), (1,2) \rangle}$$ Here, $\{(1,1), (1,2)\}$ is a basis for $\mathbb{Z} \oplus \mathbb{Z}$, so to calculate the quotient, first eliminate the second basis element. Now expressed in terms of the new basis, $(2,0) = 4(1,1) - 2(1,2)$ so the quotient is $$\frac{\langle (1,1) \rangle}{4\langle (1,1) \rangle} \cong \mathbb{Z}_4$$

Adam Venis
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  • I think its just a typo. Should be $(1,2)$. – Open Season Mar 19 '16 at 23:15
  • Thanks for the help Adam. I see how to apply this trick in general now. Play around with the bases until we get something explicit. – Open Season Mar 19 '16 at 23:17
  • One other thing to take away from this is that "playing around with the bases" can be done in a regular fashion, using row and/or column reduction of matrices. – Lee Mosher Mar 19 '16 at 23:39