Hi

consider the ring of 3x3 matrices with entries in Z36.

How many different matrices are there in the two sided ideal generated by the matrix diag(0,30,6)?

Thanks for any help,

Hedi

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- Oct 21st 2012, 08:27 AMhediideal in a matrix ring
Hi

consider the ring of 3x3 matrices with entries in Z36.

How many different matrices are there in the two sided ideal generated by the matrix diag(0,30,6)?

Thanks for any help,

Hedi - Oct 22nd 2012, 04:57 AMBingkRe: ideal in a matrix ring
You'll have to evaluate the two-sided ideal, meaning left (or right) multiply $\displaystyle \begin{pmatrix} 0 & 0 & 0 \\ 0 & 30 & 0 \\ 0 & 0 & 6 \end{pmatrix}$ by $\displaystyle \begin{pmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix}$ with $\displaystyle a_i \in \bbmath{Z}_{36}$.

You'll probably get something like $\displaystyle \begin{pmatrix} 0 & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & b \end{pmatrix}$ where $\displaystyle a,b \subseteq \bbmath{Z}_{36}$. You'll have to figure out what exactly $\displaystyle a$ and $\displaystyle b$ can be

So, if you get something like $\displaystyle a= 0, 10, 20, \text{or } 30$ and $\displaystyle b = 5, \text{or } 25$, then the ideal will have $\displaystyle 4 * 2 = 8$ elements - Oct 22nd 2012, 05:07 AMhediRe: ideal in a matrix ring
Thanks.

- Oct 22nd 2012, 09:14 AMDevenoRe: ideal in a matrix ring
i think you are vastly underrating the complexity involved here. the matrix ring is not commutative, and has 1,296 elements (its not "small"). furthermore neither AM nor MA where A is an arbitrary matrix in our ring and M is the diagonal matrix diag(0,30,6), will be diagonal.

what we need are all matrices of the form AM + MB + CMD, where A,B,C,D are arbitrary matrices in our matrix ring over Z36.

some things that may help:

30 = -6 (mod 36), so diag(0,20,6) = (6I)*(diag(0,-1,1)). note 6I commutes with the entire ring.

in Z36, <30> = <6> = {0,6,12,18,24,30}.

see if you can prove that any entry but the top-left can be any multiple of 6 (in Z36), and that the top-left entry is always 0. this gives us 6*35 = 210 possible members of the ideal. - Oct 22nd 2012, 04:23 PMhediRe: ideal in a matrix ring
The general form of a matrix in the two sided ideal involved products of parameters,so the counting is not so simple.In Z36 2x6=8x6

for example,so different products may be equal in Z36. - Oct 25th 2012, 09:08 AMDevenoRe: ideal in a matrix ring
it appears i made a mistake, because (as was shown in a later thread) E

_{11}can be written as AMB.

this shows that our ideal is 6R.

i also badly calculated the size of the ring, and the ideal:

R has 36^{9}, elements, therefore the ideal has 6^{9}elements. - Oct 25th 2012, 09:30 AMhediRe: ideal in a matrix ring
Yes,this is the answer i ultimately got.

Thank's a lot.