Proving that an element in an algebra presentation is nonzero

Question

Let $F$ be a field, $F\langle x,y\rangle$ the free $F$-algebra on two generators (polynomials in two noncommuting variables), $A= F\langle x,y\,|\, xy\!=\!1\rangle= F\langle x,y\rangle/\langle\langle xy\!-\!1\rangle\rangle$ an algebra presentation, where $\langle\langle xy\!-\!1\rangle\rangle = \{\sum_i f_i(xy\!-\!1)g_i; f_ig_i\!\in\!F\langle x,y\rangle\}$ is the ideal of $F\langle x,y\rangle$, generated by $xy\!-\!1$.

In $A$: $yx$ is an idempotent, $x\!-\!yx^2$ is a nilpotent, $x$ is a left zero-divisor, $y$ is a right zero-divisor.

Questions: How can I prove the following claims?

1. $yx\neq0$ in $A$, or equivalently, $yx=\sum_i f_i(xy\!-\!1)g_i$ in $F\langle x,y\rangle$ cannot happen.
2. $yx\neq1$ in $A$, or equivalently, $yx\!-\!1=\sum_i f_i(xy\!-\!1)g_i$ in $F\langle x,y\rangle$ cannot happen.
3. $x\!-\!yx^2\neq0$ in $A$, or equivalently, $x\!-\!yx^2=\sum_i f_i(xy\!-\!1)g_i$ in $F\langle x,y\rangle$ cannot happen.
4. $x$ is not a right zero-divisor in $A$, or equivalently, $fx=\sum_i f_i(xy\!-\!1)g_i$ in $F\langle x,y\rangle$ implies $f=\sum_i \tilde{f_i}(xy\!-\!1)\tilde{g_i}$ for some $\tilde{f_i},\tilde{g_i}\!\in\!F\langle x,y\rangle$.
5. $y$ is not a left zero-divisor in $A$, or equivalently, $yf=\sum_i f_i(xy\!-\!1)g_i$ in $F\langle x,y\rangle$ implies $f=\sum_i \tilde{f_i}(xy\!-\!1)\tilde{g_i}$ for some $\tilde{f_i},\tilde{g_i}\!\in\!F\langle x,y\rangle$.

Answer 1

See "A ring element with a left inverse but no right inverse?" and also "Right inverse but no left inverse." That last question was one of mine.

Note (4) and (5) are immediate since $xy=1$ in $A$ (i.e., $x$ is a left-inverse of $y$ and $y$ is a right-inverse of $x$).

For the others, the trick is to map your algebra to a concrete algebra where computations are easier. The canonical example seems to be the endomorphism algebra of a countable dimension $F$-vector space (i.e., $V=\bigoplus_{i=1}^\infty Fe_i$). Map $x$ onto the "shift left" operator and $y$ onto the "shift right" operator. Then statement (3) takes the following form: "shifting left twice and then shifting right is not the same as shifting left once."