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Given a matrix ring $K = \left\{ \begin{bmatrix}0 & 0\\ a & b\end{bmatrix} \,\Biggm|\, a, b \in \mathbb{R}\right\}$ and element $x = \begin{bmatrix}0 & 0\\ 1 & 1\end{bmatrix} $.

In the book it says that $x$ is the right divisor of zero but I think that this is a mistake. I need someone to confirm with better knowledge.

For example, right divisor is defined as when we take $A \in K$ and $A \neq 0$ then $A \times x = 0$ which makes sense but then it follows: $\begin{bmatrix}0 & 0\\ A_1 & A_2\end{bmatrix} \times \begin{bmatrix}0 & 0\\ 1 & 1\end{bmatrix} = \begin{bmatrix}0 & 0\\ A_2 & A_2\end{bmatrix} \neq 0$.

Therefore, $x$ is not the right divisor of the ring.

Can someone confirm this?

Thanks.

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    $\begingroup$ I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$? $\endgroup$ – Arturo Magidin Jan 14 at 23:23
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No: how do you know that $$\begin{bmatrix}0 & 0\\ A_2 & A_2\end{bmatrix} \neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1\neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $A\neq 0$ such that $A\cdot x=0$, so this is not required to hold for arbitrary $A$.

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  • $\begingroup$ Thanks for clarifying, now makes sense. $\endgroup$ – mcdonald reversed Jan 14 at 23:35

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