1

I am trying to find all the left ideals of the subring R of $M_2(\mathbb{Q})$, giving a complete list without redundancies and then justify that all left ideals appear exactly once (?).

To do so I started by stating that $\mathbb{Q}$ is closed under multiplication because it is a field, and then trying to see whether multiplication of elements of R by elements of linear combinations of the basis fall into the subspace generated by the linear combination and therefore it would be an ideal. For the moment being, I found:

\begin{equation} \{ \langle(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix})\rangle, \langle(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix})\rangle, \langle(\begin{smallmatrix} 1 & 1 \\ 0 & 0 \end{smallmatrix})\rangle, \langle(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix})\rangle, \langle(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}), (\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix})\rangle, \langle(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}), (\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix})\rangle \} \end{equation} Also, trivially $\{0\}$ and $R$. I think there aren't any duplicates, but I'm not sure if I'm missing any or how to justify that all left ideals appear exactly once, apart from stating the obvious. Also, how would I describe all $\mathbb{Q}$-subalgebras? Thank you.

Alex A.G.
  • 143
  • Since instead of 1 it works for any number, I think the correct matrixes would be with any $\mathbb{Q}$. – Alex A.G. Jan 23 '24 at 12:57
  • The solutions at the links contain detailed descriptions of what the left and right ideals of a "triangular" matrix ring are, such as this one. – rschwieb Jan 23 '24 at 15:07

0 Answers0