4
$\begingroup$

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.

  • 0
    @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.2011-10-29
  • 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.2011-10-29
  • 0
    Dear @Jyrki, sorry for the elementary question, but why does the ideal $(a+bi)$ have index $a^2+b^2$ in $\mathbf{Z}[i]$?2011-10-29
  • 0
    @yunone I suggest that you pose this as a new question, so that we can give a proper answer.2011-10-29
  • 0
    @Phira, ok, I will do that. I wasn't sure if it was a one line explanation or something more.2011-10-29
  • 0
    @yunone I'm fairly sure that I've seen that question answered here. Need to search. The easiest argument is to draw a picture, and observe that the points of the ideal form a grid of squares if side length $\sqrt{a^2+b^2}$, but that may not be the most convincing way to prove it.2011-10-29
  • 0
    @Jyrki All right, I'll hold off from posting the question for now and look too. If you find it, please let me know. Otherwise I'll go ahead and post a question about it later.2011-10-29
  • 0
    @yunone: [A related question](http://math.stackexchange.com/q/23358/11619) - not quite what I promised, but there's relevant general discussion there also.2011-10-29
  • 0
    @Jyrki, thanks!2011-10-29
  • 0
    @Phira I've just posted a quick question about it.2011-10-29

1 Answers 1