Is it true that the only proper left ideals of $M_2(K)$, the ring of the matrices whose coefficients are in a field $K$, are $$ \left\{\begin{pmatrix}ah & ak \\ bh & bk\end{pmatrix}: a,b \in K\right\} $$ for all $h,k \in K$ not both $0$?

This is my attempt.
Proper ideals don't contain invertible elements, so the determinants of the elements in the ideals are all zeros. $2 \times 2$ matrices with null determinant are those with one of the two columns multiple of the other and one of the two rows multiple of the other, that is, the matrices of the form $$\begin{pmatrix}ah & ak \\ bh & bk\end{pmatrix}, \quad a,b,h,k \in K$$ Let $I$ be a proper left ideal. It contains a matrix $\bigr(\begin{smallmatrix}ah & ak \\ bh & bk\end{smallmatrix}\bigl)$ for some $a\neq 0,b\neq 0,h,k \in K$ and $h,k$ not both $0$ and thus, for all $x,y \in K$, $$\begin{pmatrix}x & 0 \\ 0 & y \end{pmatrix} \begin{pmatrix}ah & ak \\ bh & bk\end{pmatrix} = \begin{pmatrix} ahx & akx \\ bhy & bky\end{pmatrix}$$ which implies that $$\left\{\begin{pmatrix}ih & ik \\ jh & jk\end{pmatrix} : i,j \in K\right\} \subseteq I$$ The left hand side is itself a proper left ideal. Wlog $h\neq 0$. If $I$ contained $\bigr(\begin{smallmatrix}ah & al \\ bh & bl\end{smallmatrix}\bigl)$, with $l\neq k$, then it would also contain $$\begin{pmatrix}1& 0 \\ 0 & 0\end{pmatrix} \begin{pmatrix}ah & ak \\ bh & bk\end{pmatrix} +\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}ah & al \\ bh & bl\end{pmatrix} = \begin{pmatrix}ah & ak \\ bh & bl\end{pmatrix}$$ which is invertible, then $I=M_2(K)$, contradiction.

Is this correct?

  • $\begingroup$ Yes, any set of all matrices with a fixed proportionality between the two columns is a left ideal. $\endgroup$ – Arthur Jun 4 '16 at 12:57
  • $\begingroup$ I've added a little bit of work. Is it correct? $\endgroup$ – Human Jun 4 '16 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.