April 4th, 2010, 10:25 PM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  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? 
April 5th, 2010, 09:42 AM  #2  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: matrix ring Quote:
Quote:
 
April 5th, 2010, 11:45 AM  #3 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  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/(adbc) c a) 
April 5th, 2010, 01:53 PM  #4  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: matrix ring Quote:
 
April 5th, 2010, 02:53 PM  #5 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  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. 
April 5th, 2010, 04:15 PM  #6  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: matrix ring
Right; a matrix is invertible iff it has nonzero 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 nonzero element in a field... Also, I missed this the first time: Quote:
 
April 6th, 2010, 11:49 AM  #7 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  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. 
April 6th, 2010, 04:10 PM  #8 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  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? 
April 6th, 2010, 05:52 PM  #9 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: matrix ring
thanks. 

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