# Ideal in matrix ring

• Oct 24th 2012, 08:40 AM
hedi
Ideal in matrix ring
Hi,
I need help with the following problem:
consider the ring of 3x3 matrices with entries in the ring Z36.How many matrices are there in the two sided ideal generated by the matrix diag(o,-6,18)?
Attempt at a solution:
I computed the general matrix in the two sided ideal,but counting was complicated since different products of the parameters may be equal in Z36.
I will be gratefull for any help.
Hedi
• Oct 24th 2012, 08:17 PM
johnsomeone
Re: Ideal in matrix ring
Perhaps there's a more clever way, but the method you proposed should work. Remember that you have 18 free parameters.
To simplify things, I'd factor out a -6: A diag{0, -6, 18 }B = (-6) A diag {0, 1, -3} B. To get you started:

$\displaystyle \left( \begin{matrix} a_{1, 1} & a_{1, 2} & a_{1, 3} \\ a_{2, 1} & a_{2, 2} & a_{2, 3} \\ a_{3, 1} & a_{3, 2} & a_{3, 3} \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -3 \end{matrix} \right) \left( \begin{matrix} b_{1, 1} & b_{1, 2} & b_{1, 3} \\ b_{2, 1} & b_{2, 2} & b_{2, 3} \\ b_{3, 1} & b_{3, 2} & b_{3, 3} \end{matrix} \right) = \left (\begin{matrix} a_{1, 1} & a_{1, 2} & a_{1, 3} \\ a_{2, 1} & a_{2, 2} & a_{2, 3} \\ a_{3, 1} & a_{3, 2} & a_{3, 3} \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ b_{2, 1} & b_{2, 2} & b_ {2, 3} \\ -3b_{3, 1} & -3b_{3, 2} & -3b_{3, 3} \end{matrix} \right)$

$\displaystyle = \left(\begin{matrix} (a_{1, 2}b_{2,1} -3a_{1,3}b_{3, 1}) & (a_{1, 2}b_{2,2} -3a_{1,3}b_{3, 2}) & (a_{1, 2}b_{2,3} -3a_{1,3}b_{3, 3}) \\ (a_{2, 2}b_{2,1} -3a_{2,3}b_{3, 1}) & (a_{2, 2}b_{2,2} -3a_{2,3}b_{3, 2}) & (a_{2, 2}b_{2,3} -3a_{2,3}b_{3, 3}) \\ (a_{3,2}b_{2,1} -3a_{3,3}b_{3, 1}) & (a_{3, 2}b_{2,2} -3a_{3,3}b_{3, 2}) & (a_{3, 2}b_{2,3} -3a_{3,3}b_{3, 3}) \end{matrix} \right)$

If you can choose (I don't know that you can, but it seems possible at first glance) combinations of a's and b's so that the resulting matrix is anything

in $\displaystyle R = M_{3, 3}\left(\mathbb{Z}_{36}\right)$, then your ideal I would equal (-6)R, whose number you can count.

$\displaystyle \text{Suppose } A = \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right), B = \left( \begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \text{ so } a_{1,2} = b_{2,1} = 1, \text{ and all entries 0 otherwise.}$

$\displaystyle \text{Then } A \text{ diag}\{0, 1, -3 \} \ B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) = E_{1, 1}, \text{ so } -6E_{1, 1} \in I.$

What of the other elementary matricies?
• Oct 25th 2012, 05:32 AM
hedi
Re: Ideal in matrix ring
This ideal cannot be the whole ring because it's matrices are not invertible.From this reason it cannot contain all possible
matrices with zero entries except one 1.
• Oct 25th 2012, 08:00 AM
hedi
Re: Ideal in matrix ring
In second thought there is no meaning for rank for matrices over rings.
• Oct 25th 2012, 09:05 AM
Deveno
Re: Ideal in matrix ring
it appears that I = (-6)R = 6R. in Z36, <6> = <-6> (since 6 and -6 are additive inverses, and both generate the same additive subgroup of (Z36,+)).

to prove this, it suffices to show that we can always pick 4 elements a,b,c,d in Z36 with:

ab - 3cd = 1. a = b = 2, and c = d = 1 will serve.

thus (2Ei2+Ei3)(diag(0,1,-3))(2E2j+E3j) =

(2Ei2-3Ei3)(2E2j+E3j) =

(4Ei2E2j) + (2Ei2E3j) + (2Ei32E2j) - (3Ei3E3j)

= 4Eij + 0 + 0 - 3Eij = Eij.

for example, with i = 3, j = 1:

$\displaystyle \begin{bmatrix}0&0&0\\0&0&0\\0&2&1 \end{bmatrix} \begin{bmatrix}0&0&0\\0&1&0\\0&0&-3 \end{bmatrix} \begin{bmatrix}0&0&0\\2&0&0\\1&0&0 \end{bmatrix} =$

$\displaystyle \begin{bmatrix}0&0&0\\0&0&0\\0&2&-3 \end{bmatrix}\begin{bmatrix}0&0&0\\2&0&0\\1&0&0 \end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\1&0&0 \end{bmatrix}$.

yes, we can have "duplicates", because in Z36, ab = ac doesn't imply b = c. but the relevant fact here is that the map:

k --> 6k is a surjective homomorphism from Z36 to <6>.

since 6Eij is in our ideal for any choice of i and j, then any Z36-linear combination of the 6Eij are also in the ideal (by closure), and so I = 6R.

this gives 10,077,696 elements of I (i think).
• Oct 25th 2012, 10:11 AM
hedi
Re: Ideal in matrix ring
• Oct 25th 2012, 12:56 PM
johnsomeone
Re: Ideal in matrix ring
There's an observation that Deveno and I relied on that you should understand make an argument for.

$\displaystyle \text{Since you can show that }6E_{i, j} \in I \ \forall \ i, j \in \{1, 2, 3 \},$

$\displaystyle \text{ it follows that } 6R \subset I.$

$\displaystyle \text{But how do you then justify that }I = 6R \ ?$

$\displaystyle \text{Perhaps }I = 2R\supsetneq 6R? \text{ How do you know that can't happen?}$
• Oct 25th 2012, 02:45 PM
hedi
Re: Ideal in matrix ring
I is the two sided ideal in the ring of matrices over Z36 generated by the matrix 6diag(0,1,-3),so the entries of all matrices in I are elements of Z36 that are multiples of 6.
• Oct 25th 2012, 03:37 PM
johnsomeone
Re: Ideal in matrix ring
Right. Since $\displaystyle X_0 = \text{ diag} \{0, -6, 18 \} = 6 \text{ diag} \{0, -1, 3 \}\subset 6R$, have that $\displaystyle I = <X_0> \ \subset \ <6R> \ = \ 6R.$

So once you've also shown that $\displaystyle 6R \subset I$, you'll have that $\displaystyle I = 6R.$