What is $\mathbb{Z}^2/\mathbb{Z}v_1+\mathbb{Z}v_2$?

Question

Let $v_1 = (2,3)^T, v_2 = (4,5)^T \in \mathbb{Z}^2$. Let $H =\mathbb{Z}v_1+\mathbb{Z}v_2$. What is the quotient group $\mathbb{Z}^2/H$? Describe it.

To start with, I consider an easier version: $\mathbb{Z}^2 / \mathbb{Z} v$ where $v = (3,5)^T$. I see that we can define $h: \mathbb{Z}^2 \to \mathbb{Z}_3 \oplus\mathbb{Z}_5$ by $h(x,y) = (x \mod 3, y \mod 5)$. But I am having trouble with this question.

I naively considered the following. For any vector $(a,0)^T$, it is either in $(0,0)^T + H$ or in $(1,0)^T + H$. For any vector $(0,b)^T$, it is either in $(0,0)^T + H$ of $(0,1)^T + H$ since $\gcd(3,5) = 1$. But what about any general $(a,b)^T$?

See if you find this helpful. Also Smith normal form. – Viktor Vaughn Nov 08 '16 at 01:24 — Viktor Vaughn, Nov 08 '16 at 01:24
@SpamIAm That should work. Thanks! – 3x89g2 Nov 08 '16 at 02:51 — 3x89g2, Nov 08 '16 at 02:51
Feel free to post your solution once you figure it out! – Viktor Vaughn Nov 08 '16 at 04:17 — Viktor Vaughn, Nov 08 '16 at 04:17

lhf · Accepted Answer · 2016-11-08T09:59:10.193

1

Note that $2v_1-v_2=e_2$ and so $H=\mathbb Z 2e_1 + \mathbb Z e_2$. This makes it easy to compute $\mathbb Z^2 / H$.

This corresponds to the following reduction to Smith normal form: $$ \pmatrix{ 2 & 3 \\ 4 & 5} \to \pmatrix{ 2 & \hphantom{-} 3 \\ 0 & -1} \to \pmatrix{ 2 & 0 \\ 0 & 1} $$

edited Nov 08 '16 at 09:59

answered Nov 08 '16 at 01:30

lhf

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What is $\mathbb{Z}^2/\mathbb{Z}v_1+\mathbb{Z}v_2$?

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