How to solve $xa=yb$ for $x,y$ in a ring I'm probably missing something obvious, but does the equation $xa=yb$ necessarily have nontrivial solutions (where $x,y$ are not both zero) in a nonzero ring (i.e. $1\ne0$)? If one of them is zero, then $x=1,y=0$ or $x=0,y=1$ is a solution, so we can assume that both $a$ and $b$ are nonzero. If it is a commutative ring, then $x=b$, $y=a$ is a solution, and if it is a division ring, then $x=ba^{-1}$, $y=1$ is a solution. But for a plain non-commutative ring, is there always a solution to this equation?
 A: Take the free abelian group over the free monoid on generators $\{a,b\}$. The elements of this group look like $1+ab+baab+aa$, which is to say a commutative sum of finite strings of $a,b$ (with $1$ representing the empty string). It is easy to verify that this is a ring, where the multiplication operation distributes in the obvious way as $(aa+bba)ba=aaba+bbaba$ and similarly for left-distribution.
Then any term $xa$ is a sum of strings that end in $a$, while $yb$ is a sum of strings that end in $b$. Since no string can end in both $a$ and $b$, this implies that $x$ and $y$ are both empty sums, i.e. $x=y=0$. Thus there are no nontrivial solutions to this equation in this ring.
A: For another example, let $\mathcal H$ be an infinite-dimensional separable Hilbert space, and consider the ring $\mathcal B(\mathcal H)$ of bounded linear operators on $\mathcal H$.   Let $U$ be a closed linear subspace of $\mathcal H$ such that both $U$ and $U^\perp$ are infinite-dimensional.  Then $U$ and $U^\perp$ are both isomorphic to $\mathcal H$, so there are bounded linear operators $A$ and $B$ on $\mathcal H$ such that $A(U) = \{0\}$ and $A(U^\perp) = \mathcal H$, while
$B(U) = \mathcal H$ and $B(U^\perp) = \{0\}$.  If $X A = Y B$, then
$X(\mathcal H) = X A(U^\perp) = Y B(U^\perp) = \{0\}$ and similarly
$Y(\mathcal H) = \{0\}$, so $X = Y = 0$.
For a nice concrete example of this, take $\mathcal H = \ell^2$ (the square-summable sequences), with $U = \{x \in \ell^2: x_i = 0 \ \text{for even }i\}$,
$U^\perp = \{x \in \ell^2: x_i = 0 \ \text{for odd }i\}$,
$(A x)_i = x_{2i}$, $(B x)_i = x_{2i-1}$.
