I'm looking for a (necessarily noncommutative) ring with $1$ which contains elements $a$, $b$, and nonzero element $c$, such that $ab=1$ and $ac=0$.

The only noncommutative rings I know of are matrix rings and the quaternions, but I know that this property does not hold in either of them, since the latter is a skew field and since in the former left inverses are also right inverses.


Let $F$ be a field and let $V$ be the vector space $F^\Bbb{N}$, the space of infinite sequences of elements of $F$ indexed by natural numbers $\Bbb{N}$.

Let $R$ be the ring of endomorphisms of $V$. We consider $R$ as acting on $V$ as right operators (thus if $a,b \in R$, then $ab$ means do $a$, then $b$).

Let $a$ be the right shift operator $(x_1, x_2, \ldots) \mapsto (0, x_1, x_2, \ldots)$.

Let $b$ be the left shift operator $(x_1, x_2, \ldots) \mapsto (x_2, x_3, \ldots)$.

Then $ab = 1$.

Note that $a$ is not surjective. Now let $c$ be any non-zero endomorphism of $V$ sending the image of $a$ to 0 (in a vector space, we can always find a nonzero endomorphism sending a proper subspace to 0). Then $c \ne 0$ but $ac=0$.

  • $\begingroup$ Thanks! I should have considered infinite dimensional vector spaces... $\endgroup$ – Nishant Mar 26 '14 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.