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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.

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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 $ac=0$.

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

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