\m^\dag$, and hence $\m\cong \m^\dag$ by the Krull-Schmidt property. $\blacksquare$

All that remains is to show the direction (5) $\implies$ (1). Let $(R,\m)$ be a Noetherian local ring containing a coefficient field $k\cong R/\m$. Let $I\subset R$ be a fractional ideal such that $\ell_R(R/I)<\infty$. Put $k+I\coloneqq\{a+b\mid a\in k, b\in I\}\subseteq R$, which is a $k$-subalgebra of $R$. Then, since $\dim_k R/(k+I)\leq \ell_R(R/I)<\infty$, $R$ is finitely generated as a $k+I$-module and hence $k+I$ is Noetherian local ring with a maximal ideal $(k+I)\cap \m=I$. Thus, the ring extension $k+I\subseteq R$ is module-finite residually rational and birational.

\[lmlmlm\] Let $(R,\m)$ be a one-dimensional Cohen-Macaulay local ring. Assume $R$ has a canonical ideal $I\cong \omega$ such that $l(R/I)=2$. Put $S=k+I$. Then $S$ is Gorenstein, and the colength $\ell_S(R/S)$ is equal to $1$.

$S$ is local with a maximal ideal $\mathfrak{n}=I$. The extension $S\subseteq R$ is module-finite, residually rational and birational. Since $I$ is a canonical ideal, we have $I:I=R$. Equivalently, $\mathfrak{n}:\mathfrak{n}=R$. In particular, the colength $\ell_S(S/R)$ is equal to the Cohen-Macaulay type of $S$ (Lemma \[lem21\]). Since $R$ and $S$ have same residue field $k$, we can see the equalities $\ell_S(R/S)=l_S(\m/\mathfrak{n})=\ell_R(\m/I)$. On the other hand, we have $$\ell_R(\m/I)=\ell_R(R/I)-\ell_R(R/\m)=2-1=1.$$ It follows that $S$ has Cohen-Macaulay 