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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 $\mathbb Z^3/K\simeq\mathbb Z$. –  user26857 Jul 1 '13 at 22:44

1 Answer 1

up vote 1 down vote accepted

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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Thanks Bill Cook. Do you have any suggestion of a good place to read about such module isomorphisms and how these structure questions can be solved? –  Buble Oct 28 '11 at 21:49
    
Back in the day, I learned this stuff from Artin's Algebra, but I can't say that's the best text to learn from. I rather like Dummit and Foote's Abstract Algebra text. –  Bill Cook Oct 28 '11 at 21:53
    
Oh wait. Maybe Dummit and Foote relegate this technique to a bunch of homework problems (see their chapter on classifying finitely generated modules over a PID). Artin does everything in the context of Euclidean domains -- so maybe Artin is a better choice. –  Bill Cook Oct 28 '11 at 21:54
    
en.wikipedia.org/wiki/… –  Bill Cook Oct 28 '11 at 21:56
    
Great, thanks for the comments! –  Buble Oct 28 '11 at 21:57

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