Ideal in matrix ring

**hedi** · October 24th 2012, 08:40 AM

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

**johnsomeone** · October 24th 2012, 08:17 PM

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:

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

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

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

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

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

What of the other elementary matricies?

**hedi** · October 25th 2012, 05:32 AM

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.

**hedi** · October 25th 2012, 08:00 AM

In second thought there is no meaning for rank for matrices over rings.

**Deveno** · October 25th 2012, 09:05 AM

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 (2E_i2+E_i3)(diag(0,1,-3))(2E_2j+E_3j) =

(2E_i2-3E_i3)(2E_2j+E_3j) =

(4E_i2E_2j) + (2E_i2E_3j) + (2E_i32E_2j) - (3E_i3E_3j)

= 4E_ij + 0 + 0 - 3E_ij = E_ij.

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

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

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

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

this gives 10,077,696 elements of I (i think).

**hedi** · October 25th 2012, 10:11 AM

Thank you very much for your help,It was very helpfull.

**johnsomeone** · October 25th 2012, 12:56 PM

There's an observation that Deveno and I relied on that you should understand make an argument for.

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

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

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

$\text{Perhaps }I = 2R\supsetneq 6R? \text{ How do you know that can't happen?}$

**hedi** · October 25th 2012, 02:45 PM

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.

**johnsomeone** · October 25th 2012, 03:37 PM

Right. Since $X_0 = \text{ diag} \{0, -6, 18 \} = 6 \text{ diag} \{0, -1, 3 \}\subset 6R$ , have that $I = <X_0> \ \subset \ <6R> \ = \ 6R.$

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

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