Let $k$ be a field and $R$ an $\mathbb{N}$-graded $k$-algebra that is graded-commutative. Assume that $\dim_k R<\infty$ and that $R$ is Gorenstein (i.e. the injective dimension of $R$ over itself is finite). Then why are the following product maps non-degenerate?
$$R_i\times R_{n-i}\rightarrow R_n$$
where $R_n$ is the last non-zero degree of $R$.
EDIT: The proof on the fourth page of this paper works for standard graded rings (i.e. rings generated in degree 1). Is the claim still true for rings not generated in degree 1? And do we need $R_n$ to be one-dimensional?
All these can be found in the excellent classic `On ubiquity of Gorenstein rings' by Hyman Bass. So, let $R$ be a self injective local ring of finite length with $k$ the residue field. Then the functor $N\mapsto N^*=\operatorname{Hom}_R(N,R)$ is exact. This immediately implies $\ell(N^*)=e\ell(N)$ for all finite length modules where $\ell$ stands for length and $e=\ell(k^*)$. But, $R^*=R$ implies $e=1$. Now, I hope that it is clear $R$ is the injective hull of $k$.