I checked a lot of examples of non-commutative rings that came to my mind, but they weren't helpful. In particularly it's not the case for ring of matrices because of the multiplicity of the determinant. Any hints?
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2 Answers
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Take the ring of linear transformations on the space of infinite real sequences, and let $y$ be the shift-right operator, and $x$ be the shift-left operator.
answered Mar 17, 2016 at 20:56
hmakholm left over Monicahmakholm left over Monica
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Let $R = \mathrm{End}_{\mathbf Z}(\mathbf Z[X])$. Let $x,y \in R$ be given by $$ y(X^i) = X^{i+1}, \qquad x(X^{i+1}) = X^i, x(1) = 0 $$ Then $$ xy(X^i) = X^i ,\quad i \in \mathbf N $$ hence $xy = 1$, but $$ yx(1) = y(0) = 0 $$ hence $yx \ne 1$.
