# Left and right principal ideals.

Give an example of a ring $A$ and an element $x\in A$ such that $Ax \subsetneq xA$.

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Let $$A=\{\left(\begin{array}{cc} a & b \\ 0 & c \end{array}\right)\mid a,b,c\in\mathbb Z\}$$ and $$x = \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right).$$ Then $$Ax=\{\left(\begin{array}{cc} a & 0 \\ 0 & 0 \end{array}\right)\mid a\in\mathbb Z\}$$ and $$xA=\{\left(\begin{array}{cc} a & b \\ 0 & 0 \end{array}\right)\mid a,b\in\mathbb Z\}.$$ Obviously $Ax\subsetneq xA$.

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Take an $x$ that has a right inverse but no left inverse (such as the shift operator on infinite sequences) in a ring where such elements exist. Then $xA = A$ but $Ax$ does not contain $1$.

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Edit: Thanks to YACP, I was answering the question $x A \not \subseteq A x$. I'll leave this here in case you are interested.

Consider the matrix ring $A = M_{2 \times 2}(\mathbb{R})$. Let $$x = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)$$

Then $$\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right) \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)$$

But $\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)$ is not $a x$ for any $x \in A$.

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Probably. Can you please explain why? I might have shown the other way, that $x A \not\subseteq Ax$. I figure its ultimately the same thing. – muzzlator Mar 19 '13 at 17:58
Oh. I misread the question. I thought the $\subsetneq$ was $\not\subseteq$ I'll edit it saying so – muzzlator Mar 19 '13 at 18:01