Uniqueness of IdealsDate: 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! |
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