Left ideals in simple rings

[gerben]

I read the following on the wikipedia page about simple rings (http://en.wikipedia.org/wiki/Simple_ring):

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:

$$\begin{pmatrix} 0 &1 &1\\ 0 &0 &0\\ 0 &0 &0 \end{pmatrix}$$

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:

$$\begin{pmatrix} 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0 \end{pmatrix}$$

Can anybody help me understand what the wikipedia page is trying to say, or where I am seeing things wrong?

 

[gerben]

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:

$$\begin{pmatrix} 1 &1 \\ 0 &0 \\ \end{pmatrix}$$

 

[arkajad]

I think you can show that for any left ideal I there is a unique vector subspace L of $$\mathbf{R}^n$$ such that $$I=\{A\in I:\,AL=0\}$$. Then, for any k-dimensional subspace of $$\mathbf{R}^n$$ 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

$$A=\begin{pmatrix} 1 &1 \\ 0 &0 \\ \end{pmatrix}$$

$$SAS^{-1}=\begin{pmatrix} 1 &0 \\ 1 &0 \\ \end{pmatrix}$$
where
$$S=\begin{pmatrix} 1 &1 \\ 1 &-1 \\ \end{pmatrix}$$

But I am not an expert.

 

[gerben]

I think you can show that for any left ideal I there is a unique vector subspace L of $$\mathbf{R}^n$$ such that $$I=\{A\in I:\,AL=0\}$$.

Yes, I see that there is a subspace of $$\mathbf{R}^n$$ 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 $$\mathbf{R}^n$$ 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.

 

[blue_raver22]

I would like to clarify that the statement on the wikipedia page is correct, but it requires some additional context and understanding of the concepts involved. Let me break it down for you:

Firstly, a simple ring is a ring that has no non-trivial two-sided ideals. This means that the only ideals of a simple ring are the zero ideal and the entire ring itself. In other words, there are no proper subrings that are also ideals.

Next, the statement is specifically referring to left ideals in the ring of matrices with entries in a division ring. A left ideal is a subset of the ring that is closed under left multiplication by any element of the ring. In other words, if you take any element of the left ideal and multiply it on the left by any element of the ring, the result will still be in the left ideal.

Now, let's look at the specific example you mentioned. The ring of 3 by 3 matrices over the real numbers is not a simple ring, as it has non-trivial two-sided ideals. However, the statement on the wikipedia page is talking about the ring of matrices with entries in a division ring, which is a more specific case.

In this case, the statement is saying that every left ideal in this ring can be described as a set of matrices where certain columns have zero entries. This is because in a division ring, every non-zero element has an inverse, and this property allows us to manipulate the matrices in a way that the left ideal will always have this form.

So, while your example is correct for the general case of a ring of matrices over the real numbers, it does not hold for the more specific case of a division ring.

I hope this clarifies the statement for you and helps you understand the concept better.