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I've been studying the decompositions of a finitely generated module over a PID (the structure theorem), and appyling them successfully to obtain decompositions of finite abelian groups (finding all groups of order $n$ up to isomorphism, for example).

However, I found this exercise which I haven't been able to solve:

Find the primary decomposition of the $\mathbb{Z}$-module $M$, and describe its elementary divisors, where $M=\mathbb{Z^2}/K$, and $K$ is the ideal generated by $(6,6)$, $(3,6)$, $(3,12)$.

By the theorem, I know $M \simeq \mathbb{Z}^r$ $ \oplus$ $(\bigoplus_i \mathbb{Z}/(q_i))$, where the $q_i$ are prime powers. Also, $M = T(M)$, so I know $r=0$. Any hints or help finding the $q_i$ would be appreciated, I don't know where I could start.

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  • $\begingroup$ Do you know how to use the Smith normal form of a matrix for finding the invariant factors of a f.g. module over a PID? $\endgroup$ – user26857 Dec 11 '15 at 17:50
  • $\begingroup$ Not really, in fact this is the first time I've heard of it. I searched around a bit, doesn't seem too complicated. For which matrix should I find the Smith normal form? I do know how to find the elementary divisors, after finding the primary factors. $\endgroup$ – Mauro Dec 11 '15 at 18:17
  • $\begingroup$ For the matrix having the rows (or columns) the three generators. $\endgroup$ – user26857 Dec 11 '15 at 18:26
  • $\begingroup$ Thanks! I'll look into this more, seems like it would be especially useful when dealing with a bigger module. $\endgroup$ – Mauro Dec 11 '15 at 18:52
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In this case inspection yields that

$-(6,6)+3(3,6)=(3,12)$, so the last generator is superfluous.

Then $(6,6)-(3,6)=(3,0)$, and then $(3,6)-(3,0)=(0,6)$.

So, the same subgroup is generated by $(3,0)$ and $(0,6)$, and $K=3\Bbb Z\times 6\Bbb Z$

Then the quotient $\frac{\Bbb Z\times \Bbb Z}{3\Bbb Z\times 6\Bbb Z}\cong\frac{\Bbb Z}{3\Bbb Z}\times \frac{\Bbb Z}{6\Bbb Z}$ via a likely map. You said you know how to take it from here, right?

Of course, behind the secenes this is just an ad-hoc Smith Normal form.

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  • $\begingroup$ Thanks, I think this is the way we were meant to do it, since we didn't cover the Smith normal form in class. Now that we have that, the elementary divisors must be 3, 3, and 2, right? $\endgroup$ – Mauro Dec 11 '15 at 18:56
  • $\begingroup$ Hmm, isn't $\Bbb Z_3\times \Bbb Z_6\cong \Bbb Z_3\times \Bbb Z_3 \times \Bbb Z_2 $? $\endgroup$ – Mauro Dec 11 '15 at 19:03
  • $\begingroup$ @Mauro Of course you're right. For some reason I thought I had to collapse them into two and get distinct prime powers. That isomorphism is correct. $\endgroup$ – rschwieb Dec 11 '15 at 19:04
  • $\begingroup$ Great then, thanks again! $\endgroup$ – Mauro Dec 11 '15 at 19:05

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