Is a left invertible element of a ring necessarily right invertible? 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.

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).
