Is a complete Noetherian local ring a homomorphic image of a Gorenstein local ring of the same dimension?

Question

From the proof of Theorem 29.4 in Matsumura's book Commutative Ring Theory we can see that a complete Noetherian local ring is a homomorphic image of a regular local ring, but how can we prove that a complete Noetherian local ring $R$ is a quotient of a Gorenstein local ring $S$ of the same dimension as $R$?

Answer 1

Let $R$ be a local ring which is a homomorphic image of a regular local ring $S$. Let $I$ be the kernel $S \to R$, i.e., $R \cong S/I$. Since $S$ is Cohen-Macaulay, $I$ contains a regular sequence of length equal to its height. Let $J$ be the ideal generated by such regular sequence. Then $S/J$ is a complete intersection; hence it is Gorenstein. Furthermore by the isomorphism theorem the map $S \to R$ factors through $S/J$, i.e., $S/J \to R$ with kernel $I/J$. Observe that $S/J$ has the same dimension as $R$.

Here I'm using the fact for an ideal $I$ in a local Cohen-Macaulay ring $R$, $$ \operatorname{height} I + \dim R/I = \dim R. $$