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Let $M$ be a $\mathbb{Z}$-module, and suppose we form a $3$ by $2$ presentation matrix $$A$$ for $\mathbb{Z}$, from a generating set $S$ of three elements, and a generating set for the relations of the elements of $S$ given by $T$. Each column of the matrix represents a generating relation in $T$. Then we know that applying elementary column/row operations to $A$ corresponds to changing the generating set or the set of relations to an equivalent generating set/ set of relations for the same module $M$. Suppose we use a sequence of such row/column operations to put $A$ into the following smith normal form:

$$\begin{matrix} 2 & 0 \\ 0 & 0 \\ 0 & 0  \end{matrix}$$

Then this shows $M \equiv \mathbb{Z} \oplus \mathbb{Z} / 2\mathbb{Z}$, but I am confused since this presentation matrix says that the relations on our three (new) generators is:

$2g_1 + 0g_2 + 0g_3 = 0$ and $0g_1 + 0g_2 + 0g_3 = 0$, so how exactly are we going from here to the isomorphism with that direct sum of modules?

Do we do so as follows?

$r_1g_1 + r_2g_2 + r_3g_3 = r_1'g_1 + r_2'g_2 + r_3'g_3$ if and only if $(r_1-r_1)'g_1 + ... + (r_3-r_3')g_3 = 0$, and so this occurs iff $r_1 - r_1 \in (2)$ and $r_2 = r_2'$, $r_3 = r_3'$, but then why isn't

$M = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} / 2\mathbb{Z}$?

Fri, 27 Nov 2020 20:04 GMT