# Triviality of Ideals in M(n,R).

1. Oct 31, 2014

### WWGD

Hi All,
I am trying to show that the only ideals in M (2, $\mathbb R$) , the ring of 2x2 matrices with Real entries are the trivial ones.

I have a proof, but I am being kind of lazy rigorizing it. We know we cannot have any matrix in GL(n,$\mathbb R$), because we can then get the identity and we end up with the whole ring. Basically then, we take any
non-invertible matrix m and we show we can find matrices A,A' in M (2, $\mathbb R$) so that Am + Am' is invertible. Is there a way of tightening this?

I thought of using the result that maybe either GL(n, $\mathbb R$ ), or maybe
M (2,$\mathbb R$) acts transitively on the left on M (2, $\mathbb R$) by multiplication. Is this result true?

Thanks.

2. Oct 31, 2014

### jbunniii

Certainly $GL(2,\mathbb{R})$ does not act transitively on $M(2,\mathbb{R})$. For if $A$ is not invertible, and $M$ is invertible, then $X = MA$ is not invertible (if it were, then $A = M^{-1}X$ would also be invertible). Therefore, if $A$ is not invertible and $B$ is invertible, then $A$ and $B$ are not in the same orbit under this action.

$M(2,\mathbb{R})$ is not a group with respect to multiplication, so you have to clarify what you mean by "$M(2,\mathbb{R})$ acts on the left".

I don't know of an especially clean proof; the elementary one is straightforward: if $A \in M(2,\mathbb{R})$ has a nonzero entry in row $i$, column $j$, then we can premultiply and postmultiply by appropriate elements of $M(2,\mathbb{R})$ to obtain a matrix which has the same nonzero entry at $i,j$ and which has zeros everywhere else. If this nonzero entry is, say, $a$, then we can multiply (on the left or right) by $a^{-1}I$ to get a matrix with $1$ at $i,j$ and $0$ elsewhere. Then we can move this $1$ wherever we like by multiplying on the left by a row-swapping matrix or on the right by a column-swapping matrix. Therefore, $\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$ are in the ideal generated by $A$, and as a result, so is $I$. So the ideal contains a unit, and is therefore the whole ring.

Last edited: Oct 31, 2014
3. Oct 31, 2014

### WWGD

Thanks, sorry, I meant "act" in an informal sense, meaning given any A,B there is C with AC=B , and I meant on M(n,R)- GL(n, R).

4. Oct 31, 2014

### jbunniii

If you were to make an argument using group actions, I think you would have to consider actions by multiplication on the left and on the right, because $M(2,\mathbb{R})$ does have nontrivial one-sided ideals. For example, the set of all matrices in $M(2,\mathbb{R})$ with zeros in the second column is a left ideal.