#### [SOLVED] Quadratic algebras and Koszul algebras

By Mare

Let $$A$$ be a quadratic algebra and $$B$$ the Ext-algebra of $$A$$. In case $$A$$ is a Koszul algebra, we should have that the global dimension of $$A$$ plus one is equal to the Loewy length of $$B$$ (is there a reference for this?).

Namely we should have $$gldim(A)= \sup \{ i \geq 0 | Ext_A^i(A_0,A_0) \neq 0 \} = LL(B)-1$$, where LL stands for Loewy length and $$A_0$$ is the degree zero part of the graded algebra $$A$$. Im not sure in general about the first equality here (it should at least hold for $$A$$ finite dimensional), but the second equality should be correct since $$B$$ is generated in degree 0 and 1.

Thus $$gldim(A)+1=LL(B)$$.

Question 1: Is $$gldim(A)= \sup \{ i \geq 0 | Ext_A^i(A_0,A_0) \neq 0 \}$$ true in general or under some restrictions? Is there a reference?

Question 2: Is a quadratic algebra Koszul iff $$gldim(A)+1=LL(B)$$ holds?

Maybe on needs to assume further restrictions for question 1, but I think in some form it will be true. One should be able to apply this in two nice examples:

a) $$A=K[x_1,...,x_n]$$ the polynomial ring in $$n$$ variables. Here $$B$$ is the Grassmann algebra in $$n$$ variables which has Loewy length $$n+1$$ and this shows that $$A$$ has global dimension $$n$$.

b)$$A=kQ$$ the quiver algebra of an arbitrary quiver with finitely many points and at least one arrow (that may be infinite dimensional). Then $$B$$ is the algebra with the same quiver and radical square zero and thus Loewy length 2. Thus the formula would give here that $$A$$ has global dimension one and I think the proof of this is actually quite complicated without those tools.

A useful reference for answering your questions at least partially is Theorem 1.7 (especially part (5) of it) in the notes http://inmabb.criba.edu.ar/revuma/pdf/v48n2/v48n2a05.pdf .