When homogeneous coordinate ring of an Abelian variety is Gorenstein?

Let $A$ be an abelian variety of dimension at least $2$. If $A$ is embedded into a projective space by a very ample line bundle $\mathcal{L}$ under which assumptions on $\mathcal{L}$ the homogeneous coordinate ring $$R = \oplus_{i \geq 0} H^0(A, \mathcal{L}^i)$$ of $A$ is a graded Gorenstein ring?

• This ring is never Cohen-Macaulay and hence can not be Gorenstein. – Mohan Mar 27 '17 at 17:11
• How one can prove this? Any references? – Alex Mar 27 '17 at 19:43
• If it were CM, it is very easy to check that $H^i(L^k)=0$ for all $k$ and $0<i<\dim A$. In your case, since $\dim A>1$ and $H^1(\mathcal{O}_A)\neq 0$, you are done. – Mohan Mar 27 '17 at 20:04

The definition of Gorenstein ring is a local one, a ring $R$ is Gorenstein if for every prime ideal $R_p$ is a local Gorenstein ring. For an abelian variety, you will have a Gorenstein ring which is not arithmetically Gorenstein, i.e. if you look at the global equations they do not determine a Conhen-Macaulay ideal. On the other hand, for every localisation it is Gorenstein (and hence CM). Check the following thesis for the construction of the coordinate ring of an abelian surface with a polarisation of type (1,3). http://wrap.warwick.ac.uk/80934/