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I'm working on a problem asking me to determine the structure of $\mathbb{Z}^3/K$ where $K$ is generated by $(2,1,-3)$ and $(1,-1,2)$. I suspect as a $\mathbb{Z}$-module.

My first guess was that $\mathbb{Z}^3/K$ is isomorphic to some finitely generated $\mathbb{Z}$-module $M$ with generators $x_1,x_2,x_3$ maping to the standard basis elements $e_1,e_2,e_3$, but that was hard to work with.

I then set up a matrix $ \begin{pmatrix} 2 & 1 & -3 \\ 1 & -1 & 2\end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0\end{pmatrix}. $

Does this just mean $\mathbb{Z}^3/K\cong \mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$? Is this the correct thing to do, if so, why does it work?

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    Actually \begin{pmatrix} 2 & 1 & -3 \\ 1 & -1 & 2\end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix} and $\mathbb Z^3/K\simeq\mathbb Z$.2013-07-01

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The matrix operations are equivalent to quotient module isomorphisms. [Actually row operations just replace relations with equivalent relations. Column operations require isomorphisms.]

Thus you have $\mathbb{Z}^3/K \cong \mathbb{Z}^3/L$ where $L$ is generated by $(1,0,0)$ and $(0,3,0)$.

So your new relations say

$(1,0,0)=(0,0,0)$

and

$(0,3,0)=(0,0,0)$

(mod 1 in the first coord and mod 3 in the second).

Therefore, your module is isomorphic to $\mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}$.

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    This answer would be correct if the SNF of the given matrix would be correct, but it isn't.2013-07-02