3
$\begingroup$

Let $A$ be a abelian group generated by elements $\langle a_1,a_2,a_3\rangle$ and $B$ be a subgroup generated by $\langle b_1,b_2,b_3\rangle$ where $\begin{pmatrix} b_1\\ b_2\\b_3 \end{pmatrix}= \begin{pmatrix} a_1&a_2&0\\ a_1&0&a_3\\ 0&a_2&a_3 \end{pmatrix}\begin{pmatrix}\alpha\\ \beta\\ \gamma\end{pmatrix}$ and $\alpha,\beta, \gamma \in \mathbb Z$

How then might we express $A/B$ as $\bigoplus_i \mathbb Z_{m_i}$?

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    http://sierra.nmsu.edu/morandi/notes/SmithNormalForm.pdf2012-03-17
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    @DylanMoreland: Is the smith form always achievable? I am not so sure...2012-03-17
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    @DylanMoreland: Say $\alpha=5,\beta=3,\gamma=10$?2012-03-17
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    @alexb: Yes, the Smith Normal Form is always achievable.2012-03-17
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    @ArturoMagidin: Ah I see, yes, I see that the example I gave actually has an SNF... Is there a condition on the "domain" of the entries that makes it always possible? or is it really possible for all domains?2012-03-17
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    @alexb: Smith normal form exists whenever the entries are in Principal Ideal Domain. See the first sentence of the [Wikipedia entry](http://en.wikipedia.org/wiki/Smith_normal_form)2012-03-17
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    @ArturoMagidin: Thank you. What is the SNF of a 3-by-3 diagonal matrix with entries 3, 20, 30?2012-03-17
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    @Alexb: The Smith Normal form a diagonal matrix is the matrix itself.2012-03-17
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    @ArturoMagidin: but I thought they need to be some sort of a divides b divides c relation between them?2012-03-17
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    @alexb: Sorry; it's a diagonal matrix with $1$, $6$, and $60$. Why not follow the algorithm described in Wikipedia and get some practice?2012-03-17
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    @ArturoMagidin: Thanks. I have tried comprehending the algorithm in Wikipedia, but I found it quite confusing, I think it is because of notation... :(2012-03-17
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    @ArturoMagidin: Is it at all possible to show me how you got 0,6,60? I can't for the life of me get it to that form...the best I got was 1,30,602012-03-17
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    @Alexb: I'm doing it wrong in my head; it *should* be $1$, $30$, $60$. I lost a $5$ in my head while doing it, which is how I got $1$, $6$, $60$.2012-03-17
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    @Arturo, I hope you find that 5 that you lost in your head - you might need it some day.2012-03-19

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