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

Printable View

- October 21st 2012, 09:27 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 - October 22nd 2012, 05:57 AMBingkRe: ideal in a matrix ring
You'll have to evaluate the two-sided ideal, meaning left (or right) multiply by with .

You'll probably get something like where . You'll have to figure out what exactly and can be

So, if you get something like and , then the ideal will have elements - October 22nd 2012, 06:07 AMhediRe: ideal in a matrix ring
Thanks.

- October 22nd 2012, 10:14 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. - October 22nd 2012, 05:23 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. - October 25th 2012, 10:08 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. - October 25th 2012, 10:30 AMhediRe: ideal in a matrix ring
Yes,this is the answer i ultimately got.

Thank's a lot.