About the $1-ab$ invertible implies $1-ba$ invertible problem I was trying to understand the method given here: 
https://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem?newreg=dfe597b83f2c47b0b5f581238ba5ec64
What is the statement of Krob's theorem? 
 A: The Theorem in question is Théorème VI.3 of https://doi.org/10.1007/BFb0083505.
Here $K$ is a ring, $A$ is an alphabet. We have the ring $K\langle A \rangle$ of non-commutative polynomials in the alphabet $A$ contained inside the ring $K\langle\!\langle A^* \rangle\!\rangle$ of non-commutative formal power series in the alphabet $A$.
For a power series $a$ with constant coefficient $0$, Krob defines
$$a^* = 1 + a + a^2 + a^3 + \cdots = (1 - a)^{-1}.$$
Krob defines the ring $\mathcal{PE}_K\mathcal{R}\mathrm{at}\langle A\rangle$ to be the ring generated by $K\langle A \rangle \subseteq K\langle\!\langle A^* \rangle\!\rangle$ under $^*$ (again: only power series with constant coefficient $0$ can be acted on by $^*$). So this includes things such as
$$ (a^* + b)^* = (1 - (1 - a)^{-1} - b)^{-1}.$$
Krob's Theorem says that every identity of non-commutative rational functions (meaning an identity in $\mathcal{PE}_K\mathcal{R}\mathrm{at}\langle A\rangle$) can be obtained from the two fundamental identities:
\begin{align*}
\mathrm{A}_g &: a^* = 1 + aa^* &\iff&& (1-a)^{-1} = 1 + a(1 - a)^{-1}, \\
\mathrm{A}_d &: a^* = 1 + a^*a &\iff&& (1-a)^{-1} = 1 + (1 - a)^{-1}a.
\end{align*}
($g$ and $d$ stand for gauche (left) and droit (right) respectively.)
The point is that if you have a rational identity that holds in every ring, it must come from an identity in $\mathcal{PE}_K\mathcal{R}\mathrm{at}\langle A\rangle$ (Proposition II.2) and therefore it must come from $\mathrm{A}_g$ and $\mathrm{A}_d$.
