$n+1$. Then $C$ is a rational normal curve, in particular it is good determinantal. In the next proposition we produce a nondegenerate, arithmetically Cohen-Macaulay, reduced curve of degree $n+1$ in ${\mathbb{P}}^{n+1}$ that is not standard determinantal and whose general hyperplane section is good determinantal. The curve is necessarily reducible, because of what we just observed.

\[n+1curve\] Let $C_1\subseteq{\mathbb{P}}^{n+1}$ be a cone over $n$ generic points in ${\mathbb{P}}^n$. Let $C_2\subseteq{\mathbb{P}}^{n+1}$ be a generic line through a point in $C_1$. Let $C=C_1\cup C_2$. Then $C$ is not standard determinantal, and a general hyperplane section of $C$ is good determinantal.

Without loss of generality we can let the $n$ generic points in ${\mathbb{P}}^n$ be all the coordinate points except for $[1:0:\ldots:0]$. Then the saturated ideal of $C_1\subseteq{\mathbb{P}}^{n+1}$ is $$I_{C_1}=\bigcap_{i=1}^n(x_0,x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)=
(x_0)+\sum_{1\leq i<j\leq n}(x_ix_j).$$ We can also assume that $I_{C_2}=(x_2,\ldots,x_{n+1}).$ Then the saturated ideal of $C$ is $$I_C=I_{C_1}\cap I_{C_2}=x_0(x_2,\ldots,x_{n+1})
+\sum_{1\leq i<j\leq n}(x_ix_j).$$ Since $I_{C_1}+I_{C_2}=(x_0,x_2,\ldots,x_{n+1})=I_P$ where $P$ is the point $[0:1:0:\ldots:0]$, it follows that $C$ is arithmetically Cohen-Macaulay. This follows from computing cohomology from the short exact sequence $$0{\longrightarrow}{\mathcal{I}}_C{\longrightarrow}{\mathcal{I}}_{C_1}\oplus{\mathcal{I}}_{C_2}{\longrightarrow}{\mathcal{I}}_P{\longrightarrow}0.$$ In fact we obtain the long exact sequence $$0{\longrightarrow}I_C{\longrightarrow}I_{C_1}\oplus I_{C_2}{\longrightarrow}I_{C_1}+I_{C_2}=I_P\stackrel{0}{{\longrightarrow}} H^1_*({\mathcal{I}}_C){\longrightarrow}H^1_*({\mathcal{I}}_{C_1})\oplus H^1_*({\mathcal{I}}_{C_2})=0$$ where vanishing of the last module follows from the observation that $C_1$ and $C_2$ are 