Is there a right semihereditary domain which isn't right Ore? I do not have a lot of examples in my head for semihereditary domains at all, and I haven't been able to see how to resolve this question: > Is there a right semihereditary domain which isn't right Ore? Reminders: * A **right semihereditary ring** is one in which all finitely generated right ideals are projective as modules. * A **right Ore domain** is a domain where the nonzero elements satisfy the Ore conditions. However, if I remember correctly these conditions are equivalent to the domain having finite right uniform dimension. My "obvious" examples of semihereditary domains are Bezout domains, but those are all Ore!

Free algebras provide an example:

Let $k$ be a field, and let $R=k\langle x,y\rangle$ be the free associative, unital $k$-algebra in two (non-commuting) indeterminates. Then every left and right ideal of $R$ is free, and so $R$ is necessarily hereditary on both sides. However, the right ideal $X$ generated by $x$ and the right ideal $Y$ generated by $y$ have zero intersection, since the elements of $X$ are precisely the $k$-linear combinations of monomials beginning with $x$ and the elements of $Y$ are precisely the $k$-linear combinations of monomials beginning with $y$; $R$ is a $k$-direct sum $k \oplus X \oplus Y$. If $S=R\setminus\\{0\\}$, then $xS \cap yR = (X \setminus \\{0\\}) \cap Y = (X \cap Y) \setminus \\{0\\} = \\{0\\} \setminus \\{0\\} = \varnothing$ so the set of non-zero-divisors of $R$ is not a right permutable set, and $R$ is not a right Ore domain (or a left Ore domain).