2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Actually $ \begin{pmatrix} 2 & 1 & -3 \\ 1 & -1 & 2\end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}$ and $\mathbb Z^3/K\simeq\mathbb Z$. $\endgroup$ – user26857 Jul 1 '13 at 22:44
0
$\begingroup$

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 $\mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Buble Oct 28 '11 at 21:49
  • $\begingroup$ 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. $\endgroup$ – Bill Cook Oct 28 '11 at 21:53
  • $\begingroup$ 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. $\endgroup$ – Bill Cook Oct 28 '11 at 21:54
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Bill Cook Oct 28 '11 at 21:56
  • $\begingroup$ Great, thanks for the comments! $\endgroup$ – Buble Oct 28 '11 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.