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April 4th, 2010, 10:25 PM   #1
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matrix ring

Consider the matrix ring M_2(Z_2)
a.find the order of the ring, that is the number of elements in it.
so Z_2= {0,1}
this is a 2x2 matrix from M_2. So the orders of elements should be 4 which only has the elements 0 and 1.

b.List all units in the ring,
Since Z_2 has only 0,1 and 0 is not a unit because it is a 0 divisor of 2, therefore it leave 1 to be the unit.

is this correct?
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April 5th, 2010, 09:42 AM   #2
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Re: matrix ring

Quote:
Originally Posted by tinynerdi
Consider the matrix ring M_2(Z_2)
a.find the order of the ring, that is the number of elements in it.
so Z_2= {0,1}
this is a 2x2 matrix from M_2. So the orders of elements should be 4 which only has the elements 0 and 1.
The order of the ring is the number of elements. There are 4 entries in each matrix, and 2 choices for each entry...

Quote:
b.List all units in the ring,
Since Z_2 has only 0,1 and 0 is not a unit because it is a 0 divisor of 2, therefore it leave 1 to be the unit.
The ring of interest in this case is M_2(Z_2), not Z_2. We want to know which elements are units... i.e., which have an inverse. So, we want invertible 2x2 matrices. What does an invertible matrix look like?
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April 5th, 2010, 11:45 AM   #3
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Re: matrix ring

a. it has the order of 8.

b.invertible of 2x2 matrix is
(a b) = A^-1
(c d)
(d -b) x 1/(ad-bc)
-c a)
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April 5th, 2010, 01:53 PM   #4
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Re: matrix ring

Quote:
Originally Posted by tinynerdi
b.invertible of 2x2 matrix is
(a b) = A^-1
(c d)
(d -b) x 1/(ad-bc)
-c a)
Right, but we want something a little easier to work with... Just thinking back to anything you know about linear algebra, how can you determine if a matrix is invertible?
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April 5th, 2010, 02:53 PM   #5
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Re: matrix ring

well to invertible then the det(M_2(Z_2) does not = 0. is that what you mean?
Are we saying that the elements in the 2x2 matrix has any entries in the R? What does the Z_2 mean in this case if the entries in the M_2(Z_2) is not 0,1.
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April 5th, 2010, 04:15 PM   #6
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Re: matrix ring

Right; a matrix is invertible iff it has non-zero determinant. You can prove that this holds for the nxn matrices over any field*. Z_2 is a field, and so the condition holds. There's no reason to worry about elements of R.
So... there are two elements of Z_2. i.e. we want det(M)=1 for invertibility.


*The theorem actually says that a matrix over a (commutative?) ring is invertible iff its determinant is invertible in the ring. Every non-zero element in a field...

Also, I missed this the first time:
Quote:
it has the order of 8.
You have 2 choices for each. You have this 4 times. It should be 2*2*2*2 = 16
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April 6th, 2010, 11:49 AM   #7
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Re: matrix ring

Therefore the unit is 1 because it has an inverse. I know this is not part of the question.

But what if we work in Z_3 and Z_3 has {0,1,2}. So the det(M) can equal to 1 or 2, so the unit is 1 and 2.

And if we work in Z_4, {0,1,2,3} then the unit is 1 and 3. 2 is not a unit because it is
2 = 0 mod 4. It is therefore 0 divisor.
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April 6th, 2010, 04:10 PM   #8
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Re: matrix ring

I'm not completely following what you're trying to say with the last post. Everything is correct, but I don't see the relevance.

The only invertible element in Z_2 is 1, so a matrix A in M_2(Z_2) is invertible iff det(A)=1. Which elements of M_2(Z_2) have determinant 1?
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April 6th, 2010, 05:52 PM   #9
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Re: matrix ring

thanks.
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