aCM. The curve $C$ has degree $n+1$, and its general hyperplane section consists of $n+1$ generic points in ${\mathbb{P}}^n$ by construction. Therefore a general hyperplane section of $C$ is good determinantal by Proposition \[genpts\].

We now study the last morphism in a minimal free resolution of $I_C$, in order to show that $C$ is not standard determinantal. Let $I=x_0(x_2,\ldots,x_{n+1})$, $J=\sum_{1\leq i<j\leq
  n}(x_ix_j)$. Then clearly $I_C=I+J$. So we have the short exact sequence $$\label{mapp}
0{\longrightarrow}I\cap J{\longrightarrow}I\oplus J{\longrightarrow}I+J{\longrightarrow}0.$$ Let $$\label{mfrI}
0{\longrightarrow}{\mathbb{F}}_n{\longrightarrow}{\mathbb{F}}_{n-1}{\longrightarrow}\cdots{\longrightarrow}{\mathbb{F}}_1{\longrightarrow}I{\longrightarrow}0$$ be a minimal free resolution of $I$. Then ${\mathbb{F}}_n=R(-n-1)$ and ${\mathbb{F}}_{n-1}=R(-n)^n.$ The last morphism in (\[mfrI\]) is $$\label{lastI}
R(-n-1)\stackrel{(x_2,-x_3,x_4,\ldots,(-1)^{n+1}x_{n+1})}{{\longrightarrow}}
R(-n)^n.$$ Let $$\label{mfrJ}
0{\longrightarrow}{\mathbb{G}}_{n-1}\stackrel{M}{{\longrightarrow}}{\mathbb{G}}_{n-2}{\longrightarrow}\cdots{\longrightarrow}{\mathbb{G}}_1{\longrightarrow}J{\longrightarrow}0$$ be a minimal free resolution of $J$. The ideal $J$ is a lexsegment squarefree monomial ideal, hence morphisms in a minimal free resolution are explicitly computed in [@ar98], Theorem 2.1. It turns out that ${\mathbb{G}}_{n-1}=R(-n)^{n-1}$, ${\mathbb{G}}_{n-2}=R(-n+1)^{n(n-2)}$, and the matrix $M$ describing the last morphism in (\[mfrJ\]) has size $n(n-2)\times(n-1)$ and is of the form $$M=\left(\begin{array}{ccc}c_1 & \ldots & c_{n-1} 
\end{array}\right)$$ where each $c_i$ is a column with exactly $n-1$ nonzero entries (all the indeterminates but $x_i$). Finally, $I\cap J=x_0J$, so the minimal free resolution (\[mfrJ\]) twisted by $-1$ is a minimal free resolution of $I\cap J$ $$\label{mfrIJ}
0{\longrightarrow}{\mathbb{G}}_{n-1}(-1)\stackrel{M}{{\longrightarrow}}{\mathbb{G}}_{n-2}(-1){\longrightarrow}\cdots{\longrightarrow}{\mathbb{G}}_1(-1){\longrightarrow}I\cap 