Is there a ring which satisfies $xy=1$ and $yx\neq 1$

Question

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?

Answer 1

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.

Answer 2

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 $yx \ne 1$.