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I'm looking at the structure $D^3/N$ where $D=\mathbb{Z}[i]$ and $N$ is generated $(1,3,6)$, $(2+3i,-3i,12-18i)$, and $(2-3i,6+9i,-18i)$. Apparently $D^3/N$ is finite of order $352512$, but I don't see how.

I took the relation matrix for $N$, and reduced it as $$ \begin{bmatrix} 1 & 3 & 6 \\ 2+3i & -3i & 12-18i \\ 2-3i & 6+9i & -18i \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & -6+6i & -12-36i \\ 0 & 18i & -12 \end{bmatrix}. $$ Looking at the $2\times 2$ minor, $$ \begin{bmatrix} -6+6i & -12-36i \\ 18i & -12 \end{bmatrix} \longrightarrow \begin{bmatrix} -6+6i & 0 \\ 18i & -24+66i \end{bmatrix} $$ but subtracting $(-2+4i)$ times the first column from the second column. I don't see a way to reduce it further to a normal form.

I know that $\mathbb{Z}[i]$ is a principal ideal domain, so $D^3/N$ would be isomorphic to the direct sum of the quotients of the cyclic modules generated by the invariant factors. But if the invariant factors are elements of $\mathbb{Z}[i]$, I don't really know how many elements are in $\mathbb{Z}[i]/(a+bi)$. How can I get to the desired conclusion? Thank you.


Here's my computation. $$ \begin{bmatrix} 1 & 3 & 6 \\ 2+3i & -3i & 12-18i \\ 2-3i & 6+9i & -18i \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 3 & 6 \\ 4 & 6+6i & 12-36i \\ 2-3i & 6+9i & -18i \end{bmatrix}. $$ by adding the third row to the second. Then $$ \begin{bmatrix} 1 & 3 & 6 \\ 0 & -6+6i & -12-36i \\ 2-3i & 6+9i & -18i \end{bmatrix}. $$ by subtracting $4$ times the first row from the second. Then $$ \begin{bmatrix} 1 & 3 & 6 \\ 0 & -6+6i & -12-36i \\ 0 & 18i & -12 \end{bmatrix} $$ by subtracting $2-3i$ times the first row from the third. I then clear the first row by subtracting $3$ and $6$ times the first column from the others.

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@Phira I don't think I made any errors in the calculation. I've added it so maybe another pair of eyes can see if there's an error. –  Tipani Oct 29 '11 at 0:13
1  
The determinant of that 3x3 matrix is $-576+144i$, which is also the product of the invariant factors. The order of the quotient module is simply the squared norm of that, and is, indeed $$576^2+144^2=17\cdot144^2=352512.$$ This follows from the fact that the ideal of $\mathbf{Z}[i]$ generated by $a+bi$ is of index $a^2+b^2$. All the invariant factors give rise to three summands of that form (one of them is trivial), so the quotient module is the direct sum of two cyclic submodules - one for each non-trivial invariant factor. –  Jyrki Lahtonen Oct 29 '11 at 11:19
    
Dear @Jyrki, sorry for the elementary question, but why does the ideal $(a+bi)$ have index $a^2+b^2$ in $\mathbf{Z}[i]$? –  yunone Oct 29 '11 at 11:27
    
@yunone I suggest that you pose this as a new question, so that we can give a proper answer. –  Phira Oct 29 '11 at 11:29
    
@Phira, ok, I will do that. I wasn't sure if it was a one line explanation or something more. –  yunone Oct 29 '11 at 11:30

1 Answer 1

I will divide your matrix $\begin{pmatrix} -6+6i & -12-36i \\ 18i & -12 \end{pmatrix}$ by 6 for simpler calculation:

$\begin{pmatrix} -1+i & -2-6i \\ 3i & -2 \end{pmatrix}$

You have $(-1+i)(1+i) +3i (-i)=1$, so the invertible matrix you are looking for is $\begin{pmatrix} 1+i& -i \\ -3i & -1+i \end{pmatrix}$. (I am following here the description of the wikipedia article, obviously, you do not have to write this matrix if you are already familiar with the algorithm. We do not have to divide by the gcd, since the gcd is 1 here.)

Now you replace the first row by $(1+i)row1+(-i) row2$ and row 2 by $(-3i) row1 +(-1+i) row2$.

You obtain $\begin{pmatrix} 1 & * \\ 0 & -16+4i \end{pmatrix}$.

I don't calculate the star entry because it can clearly be eliminated anyway.

So the Smith normal form is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\cdot 4\cdot (-4+i) \end{pmatrix}$.

Note that you should use the norm of the determinant given in the comment to the OP by Jyrki Lahtonen if you just want to get the number of elements, but the calculation of the normal form gives you also the sizes of the two cyclic submodules.

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