Is there a ring with an element $x$ such that $2x=1$ which is non-central (there is some $y$ such that $xy \neq yx$)?
Suppose there is, then by taking the subring generated by $x$ and $y$, there had to be a ring with two generators that works, so in particular there is a quotient of the free ring on two generators that works. By the universal property, this means my question is equivalent to $X$ in $\mathbb{Z}\langle X,Y \rangle / (2X-1)$ having this property. This in turn is equivalent to $XY-YX$ not being in $(2X-1)$ in $\mathbb{Z}\langle X,Y \rangle$.
This seems obvious, but I don't really know how to prove it. Questions like this in general seem really obvious, but really difficult, which makes me wonder if there's some trick I'm missing.
I've been able to prove it for some similar problems, like proving $\sqrt{2}$ can be non-central. I can prove $XY-YX$ is not in $(X^2-2)$ by looking at the $XY$ term in any element of $(X^2-2)$. $aX^2b$ has no $XY$ term for any $a, b$ in $\mathbb{Z}\langle X,Y \rangle$, so the $XY$ term only depends on the '$-2$' part. But that would mean it would have to be a multiple of $2$, and the $XY$ term in $XY-YX$ isn't a multiple of $2$, so it can't be in the ideal. But I can't find a version of this proof that works for $(2X-1)$.
I would be happy with any proof, but I would also particularly like to know if my method can be generalised to $(2X-1)$.
Let $x$ satisfy $2x=1$. Then, for $y \in R$, we have $y = 1y = 2xy$ and $y = y1 = y2x$. Since $2$ commutes with everything, we have $2xy = 2yx$. so $xy-yx$ is $2$-torsion.
Now let $a \in R$ with $2a = a2 = 0$. Then, $a = a1 = a2x = 0$, so $a = 0$. Therefore, $xy = yx$.