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I am learning modules and the Smith Normal Form, but I got stuck in the following: I found the Smith Normal Form of

$$M = \begin{pmatrix} 21 & 0 & 1 \\ 8& 4 & 1\\ 3& 8 & 1 \end{pmatrix}$$ to be $$SNF = \begin{pmatrix} 1 & 0 & 0 \\ 0& 1 & 0\\ 0& 0 & 32 \end{pmatrix}.$$

So what would be the quotient $\mathbb{Z}^{3}/M \mathbb{Z}^{3}$ isomorphic to as a $\mathbb{Z}$-module (i.e., an abelian group)? Should it be $\mathbb{Z}/32\mathbb{Z}$?

Thanks

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  • $\begingroup$ I found the SNF, but I do not understand what is the quotient isomorphic to? $\endgroup$ – Leonhard Leibniz Apr 21 '15 at 13:18
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    $\begingroup$ $\mathbb{Z}^{3}/M \mathbb{Z}^{3}\simeq \mathbb{Z}^{3}/(\mathbb Z\oplus\mathbb Z\oplus 32\mathbb Z)\simeq\mathbb Z/32\mathbb Z$. $\endgroup$ – user26857 Apr 21 '15 at 13:26
  • $\begingroup$ So it is right whatever I did? $\endgroup$ – Leonhard Leibniz Apr 21 '15 at 13:40
  • $\begingroup$ Yes, it is! ${}$ $\endgroup$ – user26857 Apr 21 '15 at 13:41
  • $\begingroup$ Nice, Thanks, just needed that line you added to completely understand $\endgroup$ – Leonhard Leibniz Apr 21 '15 at 13:41
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We have $$\mathbb{Z}^{3}/M \mathbb{Z}^{3}\simeq(\mathbb{Z}\oplus\mathbb Z\oplus\mathbb Z)/(\mathbb Z\oplus\mathbb Z\oplus 32\mathbb Z)\simeq\mathbb Z/32\mathbb Z,$$ so your guess is correct.

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