now turn to estimate the invariant $bg(R)$ in general. Suppose there exists a self-dual fractional ideal of $R$. Then we have an upper bound of $bg(R)$ as follows.

\[lem23\] Let $(R,\m)$ be a complete one-dimensional Cohen-Macaulay local ring. Assume $R$ contains an infinite coefficient field $k\cong R/\m$. Let $I\subseteq R$ be a fractional ideal of $R$. If $I$ is self-dual, then we have $bg(R)\leq l(R/I)$. In other words, the following inequality holds $$bg(R)\leq \inf\{\ell_R(R/I)\mid I\cong I^\dag \}.$$

In the case $I=R$, the self-duality of $I$ implies $R$ is Gorenstein. So we may assume $I\subseteq \m$. Take a non-zero divisor $t\in I$, and Put $B=k+I$. Then $B\subseteq R$ is a module-finite extension and $I$ is the maximal ideal of local ring $B$. Remark that $B$ is also complete. Since $B\subseteq R$ is birational, the $R$-isomorphism $I\to I^\dag$ is also a $B$-isomorphism, and $I^\dag$ is equal to the canonical dual of $I$ over $B$. By Theorem \[thA\], $bg(B)\leq 1$, that is, there is a Gorenstein ring $S$ and module-finite birational extension $S\subseteq B$. Then $S\subseteq R$ is also a module-finite birational extension. The calculation $$\ell_S(R/S)=\ell_S(R/B)+\ell_S(B/S)=\ell_R(\m/I)+1=\ell_R(R/I)$$ shows that $bg(R)\leq \ell_R(R/I)$.

As a corollary of this, we can see the finiteness of $bg(R)$ in 