Suppose $R$ is Noetherian ring with a dualizing complex. This paper shows that $R$ is a quotient ring of a finite dimensional Gorenstein ring. (Corollary 1.4)
My question is:
If $R$ is Noetherian local ring with a dualizing complex, then is $R$ a quotient ring of a Gorenstein local ring?
Any helps will be appreciated.
Yes, this is true. Think about it from the geometry side: saying every Noetherian ring $R$ with a dualizing complex is a quotient of a Gorenstein ring $S$ means that $R=S/I$ for some ideal $I$, or $\operatorname{Spec} R\to\operatorname{Spec} S$ is a closed immersion. If $R$ is local, then the image of the maximal ideal $\mathfrak{m}$ under this immersion is a closed point $\mathfrak{p}\subset S$, and so the natural map on stalks $S_{\mathfrak{p}} \to R_{\mathfrak{m}}=R$ is again a surjection. But as finite-dimensional and Gorenstein are preserved under taking stalks, we see that $S_\mathfrak{p}$ is finite-dimensional Gorenstein and thus the statement you want is true.