[Math] Left and Right Ideal Generated by Two Matrices.

abstract-algebraidealsnoncommutative-algebraring-theory

Let $R= {\rm Mat}_2(\Bbb R)$ be the ring (with $1$) of $2\times2$-matrices with entries in $\Bbb R$. Let $$M = \left\{\begin{pmatrix}1&0 \\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\},$$ i.e. a set of two matrices. What are the left and right ideal generated by $M$ in $R$?


I tried to write my answer by it won't accept it. I don't know how to write Matrices (and their product) in the right format.

Is this correct:
For left ideal: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\c&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&a\\0&c\end{pmatrix}$$

hence $L = \begin{pmatrix}a&a\\c&c\end{pmatrix}$.

For right ideal:

$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\0&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$

hence $R = \begin{pmatrix}a+c&b+d\\0&0\end{pmatrix}$.

Best Answer

Sorry, there are some issues here caused by your choice of notation. By multiplying the same $a,b,c,d$ matrix with both generators, you've overlooked that there's nothing wrong with multiplying the generators with two different matrices and adding, which will produce many more possiblities.

In the left ideal generated by those two things, a general element will look like this:

$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}+\begin{bmatrix}e&f\\g&h\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&e\\c&g\end{bmatrix}$.

Notice how we didn't recycle the $a,b,c,d$ matrix for both generators. Since you can pick $a,e,c,g$ to be whatever you want, you can see that the left ideal generated by these two things is the entire matrix ring.

Try to apply similar reasoning to the right ideal generated by these two things. You will get a different answer than the left ideal, and the answer you gave is not incorrect. It's just that you can express what the right ideal looks like more simply.