MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

4 added 40 characters in body

There are lots of examples to be found in the theory of tilted algebras.

A tilted algebra $B$ is an algebra of the form $\operatorname{End}_A(T)$ with $A$ an hereditary finite dimensional algebra (over an algebraically closed field, say) and $T$ a tilting $A$-module. A$-module. Then$B$and$A$have equivalent derived categories, but$B$is usually not hereditary (that is, its global dimension is usually bigger than$1$; one does have$\operatorname{gldim}B\leq2$, though, so it does not blow up too much). A concrete example: let$A$be the path algebra of the quiver $$\bullet \leftleftarrows \bullet \leftarrow \bullet$$ Number the vertices$1$,$2$and$3$from left to right, and let$T$be the direct sum of the simple$S(1)$and the indecomposable injective modules$I(1)$and$I(3)$, which is a tilting module. Then$B=\operatorname{End}_A(T)$is the algebra given by the same quiver, but bound by the relations given by the two paths of length two. In particular,$B$is not hereditary. You will find all this discussed at length in Assem, Simson and Slowroński's book Elements of the Representation Theory of Associative Algebras, Vol. 1. If one considers more generally tilting complexes as opposed to just modules, then the difference between the global dimensions of the algebras involved can be made as large as you want. 3 added 6 characters in body There are lots of examples to be found in the theory of tilted algebras. A tilted algebra$B$is an algebra of the form$\operatorname{End}_A(T)$with$A$an hereditary finite dimensional algebra (of over an algebraically closed field, say) and$T$a tilting module.$A$-module. Then$B$and$A$have equivalent derived categories, but$B$is usually not hereditary (that is, its global dimension is usually bigger than$1$; one does have$\operatorname{gldim}B\leq2$, though, so it does not blow up too much). A concrete example: let$A$be the path algebra of the quiver $$\bullet \leftleftarrows \bullet \leftarrow \bullet$$ Number the vertices$1$,$2$and$3$from left to right, and let$T$be the direct sum of the simple$S(1)$and the indecomposable injective modules$I(1)$and$I(3)$, which is a tilting module. Then$B=\operatorname{End}_A(T)$is the algebra given by the same quiver, but bound by the relations given by the two paths of length two. In particular,$B$is not hereditary. You will find all this discussed at length in Assem, Simson and Slowroński's book Elements of the Representation Theory of Associative Algebras, Vol. 1. If one considers more generally tilting complexes as opposed to just modules, then the difference between the global dimensions of the algebras involved can be made as large as you want. 2 added 85 characters in body There are lots of examples to be found in the theory of tilted algebras. A tilted algebra$B$is an algebra of the form$\operatorname{End}_A(T)$with$A$an hereditary finite dimensional algebra (of an algebraically closed field, say) and$T$a tilting module. Then$B$and$A$have equivalent derived categories, but$B$is usually not hereditary (that is, its global dimension is usually bigger than$1$).1$; one does have $\operatorname{gldim}B\leq2$, though, so it does not blow up too much).

A concrete example: let $A$ be the path algebra of the quiver $$\bullet \leftleftarrows \bullet \leftarrow \bullet$$ Number the vertices $1$, $2$ and $3$ from left to writeright, and let $T$ be the direct sum of the simple $S(1)$ and the indecomposable injective modules $I(1)$ and $I(3)$, which is a tilting module. Then $B=\operatorname{End}_A(T)$ is the algebra given by the same quiver, but bound by the relations given by the two paths of length two. In particular, $B$ is not hereditary.

You will find all this discussed at length in Assem, Simson and Slowroński's book Elements of the Representation Theory of Associative Algebras, Vol. 1.

If one considers more generally tilting complexes as opposed to just modules, then the difference between the global dimensions of the algebras involved can be made as large as you want.

1