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?

For future readers: a ring in which $xy=1$ implies $yx=1$ is called Dedekind-finite. As Henning Makholm's answer shows, not all rings are Dedekind-finite, but there are several kinds of rings which are always Dedekind-finite, e.g. commutative rings, finite rings, domains, etc. — Joe, Apr 11 '23 at 20:42

score 9 · Accepted Answer · answered Mar 17 '16 at 20:56

9

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 '16 at 20:56

hmakholm left over Monica

286,031

score 4 · Answer 2 · answered Mar 17 '16 at 20:55

4

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 $yx \ne 1$.

answered Mar 17 '16 at 20:55

martini

84,101

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

2 Answers2

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