# Smith Normal Form and quotient $\mathbb{Z}^{3}/M \mathbb{Z}^{3}$

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

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

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.