Left ideals in simple rings

  • Thread starter gerben
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  • #1
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Main Question or Discussion Point

I read the following on the wikipedia page about simple rings (http://en.wikipedia.org/wiki/Simple_ring): [Broken]
Let D be a division ring and M(n,D) be the ring of matrices with entries in D. It is not hard to show that every left ideal in M(n,D) takes the following form:

{M ∈ M(n,D) | The n1...nk-th columns of M have zero entries},

for some fixed {n1,...,nk} ⊂ {1, ..., n}.
I do not see why this is the case. Take the ring of 3 by 3 matrices over the real numbers and the left ideal, J, generated by:

[tex]
\begin{pmatrix}
0 &1 &1\\
0 &0 &0\\
0 &0 &0
\end{pmatrix}
[/tex]

then J is not equal to S = {M ∈ M(3,ℝ) | The 1st column of M has zero entries},
since for example the following matrix is in S but not in J:

[tex]
\begin{pmatrix}
0 &1 &0\\
0 &0 &1\\
0 &0 &0
\end{pmatrix}
[/tex]

Can anybody help me understand what the wikipedia page is trying to say, or where I am seeing things wrong?
 
Last edited by a moderator:

Answers and Replies

  • #2
508
1
This is a bit earlier on the same wikipedia page:
the full matrix ring over a field does not have any nontrivial ideals (since any ideal of M(n,R) is of the form M(n,I) with I and ideal of R), but has nontrivial left ideals (namely, the sets of matrices which have some fixed zero columns).
Can somebody shed some light on what is so important about these zero columns? I would think that the number of independent columns is important. I just do not see why zero columns are necessary at all.

Isn't the ideal generated by the following matrix, without any zero columns, also a nontrivial left ideal:

[tex]
\begin{pmatrix}
1 &1 \\
0 &0 \\
\end{pmatrix}
[/tex]
 
  • #3
1,444
4
I think you can show that for any left ideal I there is a unique vector subspace L of [tex]\mathbf{R}^n[/tex] such that [tex]I=\{A\in I:\,AL=0\}[/tex]. Then, for any k-dimensional subspace of [tex]\mathbf{R}^n[/tex] you can choose a basis such that the first k vectors are in L. This will put your ideal into a canonical form that you are looking form .

For your matrix

[tex] A=\begin{pmatrix}
1 &1 \\
0 &0 \\
\end{pmatrix}[/tex]

[tex]SAS^{-1}=\begin{pmatrix}
1 &0 \\
1 &0 \\
\end{pmatrix}[/tex]
where
[tex]S=\begin{pmatrix}
1 &1 \\
1 &-1 \\
\end{pmatrix}[/tex]

But I am not an expert.
 
Last edited:
  • #4
508
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I think you can show that for any left ideal I there is a unique vector subspace L of [tex]\mathbf{R}^n[/tex] such that [tex]I=\{A\in I:\,AL=0\}[/tex].
Yes, I see that there is a subspace of [tex]\mathbf{R}^n[/tex] that is contained in the kernel of every M ∈ I. This subspace is the orthogonal complement of the row space of matrix A ∈ I that has maximum number of independent rows.

Then, for any k-dimensional subspace of [tex]\mathbf{R}^n[/tex] you can choose a basis such that the first k vectors are in L. This will put your ideal into a canonical form that you are looking form .
Ah yes I guess that was the idea: on an appropriate basis there will be zero columns.

Thank you very much.
 
Last edited:

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