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Let $R$ be a unitary ring and let $a,b \in R$ such that $ab=1$. Does it imply that $a$ is invertible?

The definition of invertible element requires that $ab = ba = 1$, so I guess it doesn't imply that a is invertible, but I didn't manage to find a unitary ring in which this happens.

rschwieb
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user91015
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1 Answers1

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Right, it doesn't imply invertibility. Consider the Hilbert space $\ell^2(\mathbb{N},\mathbb{C})$ of all square summable complex sequences, and let $R = \mathcal{B}(\ell^2)$ the ring of (continuous) endomorphisms of $\ell^2$. Let $a = D \colon (x_0,x_1,\dotsc) \mapsto (x_1,x_2,\dotsc)$ the "drop" operator, and $b = S \colon (x_0,x_1,\dotsc) \mapsto (0,x_0,x_1)$ the shift operator. We have $ab = D\circ S = \operatorname{id}_{\ell^2}$, but $a$ is not invertible (it's not injective).

Daniel Fischer
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    You bet me to the finish line... Basically just the fact that left-inverse and right-inverse are not the same thing. Also all this square summable stuff is a red herring. You can take $\mathbb{Z}^\mathbb{N}$. – Thomas Oct 31 '13 at 18:26
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    I was preparing the same thing with the endomorphism ring of a free $\mathbb Z$-module with a countable basis, but you beat me to it. – Lubin Oct 31 '13 at 18:26
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    @Thomas Sure, works in any sequence space. I just felt like using a Hilbert space today ;) – Daniel Fischer Oct 31 '13 at 18:35