Gorenstein, complete intersection Let $S = k[X_1,...,X_n]$
Example 3.2.11(b) of Bruns-Herzog's book "Cohen-Macaulay Rings", gives a Gorenstein ring that is a complete intersection iff $n \leq 2$. They have proved it as a special case of the example. Is there a direct and independent proof? 
thanks for any help
 A: I assume you're referring to the explicit example at the end - let me state it for completeness: if $S = k[x_1, \ldots, x_n]$, $\varphi : S_2 \to k$ is $k$-linear, given by $\varphi(x_ix_j) = \delta_{ij}$, then for the graded ideal $I = \oplus_{j \ge 0} I_j$ (where $I_0 = 0 = I_1$, $I_2 = \ker \varphi$, and $I_j = S_j$ for $j > 2$), $R := S/I$ is Artinian Gorenstein, and a complete intersection iff $n \le 2$. If $n = 1$, then $R \cong k[x]/(x^2)$, which is a complete intersection, hence Gorenstein. Thus assume that $n \ge 2$.
To see that $R$ is Gorenstein in general, note that the top degree of $R$ is $2$ (i.e. $R_j = 0$ for all $j > 2$), so $R_2 \subseteq \text{Soc}(R)$, and since $I_0 = I_1 = 0$, it's not hard to check that the socle is just $R_2$. Thus $R$ is Gorenstein iff $\dim_k R_2 = 1$. There are ${n+1 \choose 2}$ forms in $S_2$, and ${n \choose 2} + (n-1)$ forms in $I_2$ (${n \choose 2}$ from $x_ix_j$ for $i \ne j$, and $(n-1)$ from $x_i^2 - x_{i+1}^2$ for $i = 1, \ldots, n-1$). Thus $\dim_k R_2 \le {n+1 \choose 2} - {n \choose 2} - (n-1) = 1$, and since $R_2 \ne 0$ (since $\varphi \ne 0$), equality must hold.
However, we just checked that $I$ was (minimally) generated by ${n \choose 2} + (n-1)$ elements, but $\text{ht}(I) = n$ since $R$ is Artinian. Thus $I$ is a complete intersection iff ${n \choose 2} + (n-1) \le n$, i.e. ${n \choose 2} \le 1$, or $n \le 2$.
