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Uniqueness of Ideals

Date: 09/23/2003 at 15:20:29
From: Coralie
Subject: Modern algebra

How can I prove that in Mn(Q) (the ring of n*n matrices over the 
rational numbers Q), (0) and Mn(Q) are the only ideals?

I have trouble seing exactly what is an ideal. I know the 
definition, but it's not really concrete...

I think that I have to check that another ideal can't exist. This 
ideal can only have non-invertible matrices, because if it has an 
invertible matrix, it's a unit and then the ideal is Mn(Q) itself. 

But I can't manage to proove that such an ideal does not exist...


Date: 09/24/2003 at 03:46:39
From: Doctor Jacques
Subject: Re: Modern algebra

Hi Coralie,

Assume that J is a proper non-zero ideal of Mn(Q). This means that, 
if A and B are matrices of J, the following matrices also belong to J:

  M*A
  A*M
  A (+/-) B

where M is any matrix in the ring.

Consider the matrix E_mn, where all elements are 0 except e[m,n], 
which is equal to 1.

If you multiply a matrix A on the left by E_mn, the resulting matrix 
will have all its rows 0, except for row m, which will be equal to 
then nth row of A.

In the same way, if you multiply A on the right by E_mn, the 
resulting matrix will have all its column 0, except for column n, 
which will be the same as column m of A.

Pick any non-zero matrix A in J. A contains an element different from 
0, say a_ij. By multiplying A by a scalar matrix (1/a_ij), which 
leaves us in the ideal, we can assume that a_ij = 1. Consider now the 
matrices A_mn defined by:

  A_mn = E_mi * A * E_jn

The multiplication on the left ensures that only row m is non-zero, 
and the multiplication on the right ensures that only column n is non-
zero. In fact, the only element different from 0 is at position 
(m,n), and equal to a_ij = 1.

Now, all those matrices A_mn are in J, and so is any sum of those 
matrices.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 09/26/2003 at 22:36:35
From: Coralie
Subject: Thank you (Modern algebra)

Thank you very much!  I found your explanation quite helpful.
It's really a good job you're doing for students and others, helping
without giving the solution too easily.

Thank you again!
Associated Topics:
College Modern Algebra

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