a contradiction. Thus, it must be $\dim R\leq 1$.

When $\dim R\ge 2$, the maximal ideal $\m$ cannot be self-dual. However, we suggest the following generalization of the self-duality of the maximal ideal in higher dimensional case.

$(R,\m)$ be a non-Gorenstein Cohen-Macaulay local ring of dimension $d>0$ having an infinite residue field. Assume $R$ has a canonical ideal $I$ satisfying $e(R/I)=2$. Then there is a regular sequence $\underline{x}=x_1,\dots,x_{d-1}$ such that $R/(\underline{x})$ has self-dual maximal ideal.

Since $R/I$ is Cohen-Macaulay of dimension $d-1$, we can take a minimal reduction $\underline{y}=y_1,\dots,y_{d-1}$ of the maximal ideal $\m/I$ in $R/I$. Then the length $l(R/I)/(\underline{y}))$ is equal to $e(R/I)(\le2)$. Let $\underline{x}=x_1,\dots,x_{d-1}$ be a preimage of $\underline{y}$ in $R$. As $I$ is unmixed, we can take $\underline{x}$ as a regular sequence in $R$. The tensor product $I'=I\otimes R/(\underline{x})$ is naturally isomorphic a canonical ideal of $R'=R/(\underline{x})$. The quotient $R'/I'$ has length $l(R/(I+\underline{x}))=l(R/I)/(\underline{y})\le 2$. Therefore $R'$ has self-dual maximal ideal by Theorem \[thA\].

Let $R=k[[x^3,x^2y,xy^2,y^3]]$ be the third Veronese subring of $k[[x,y]]$. Then $I=(x^3,x^2y)R$ is a canonical ideal of $R$. The quotient ring $R/I$ is isomorphic to $k[[s,t]]/(s^2)$, and hence $e(R/I)=2$.

Go back to the subject on self-duality of the maximal ideal. Recall the notion of trace 