Exercise 3, § 8

Let $k$ be a field, $S$ the ring $k[T_1, \ldots, T_d]$, $\mathfrak{m}$ the ideal $(T_1, \ldots, T_d)$. We denote by $S^{*gr}$ the graded dual of $S$ (No. 6), and by $(\mathbf{T}^{-\alpha})_{\alpha \in \mathbf{N}^d}$ the dual basis of $(\mathbf{T}^{\alpha})_{\alpha \in \mathbf{N}^d}$ (denoted $(u_{\alpha})_{\alpha \in \mathbf{N}^d}$ in loc. cit.).

(a) To every finitely generated $S$-submodule $M$ of $S^{*gr}$ we associate the ideal $\mathrm{Ann}(M)$ of $S$. Prove that this defines a bijective correspondence between the finitely generated submodules of $S^{*gr}$ and the ideals $\mathfrak{a}$ of $S$ contained in $\mathfrak{m}$ and such that $S/\mathfrak{a}$ has finite length; the inverse bijection associates to the ideal $\mathfrak{a}$ the submodule $M$ of $S^{*gr}$ formed by the elements annihilated by $\mathfrak{a}$ (observe that $M$ is an $S/\mathfrak{a}$-module of Matlis).

(b) In order that $S/\mathfrak{a}$ be a Gorenstein ring, it is necessary and sufficient that the $S$-module $M$ be monogenic.

(c) Let $f$ be the element $\sum T_i^{-2}$ of $S^{*gr}$; prove that the ideal $\mathfrak{a}$ corresponding to the submodule $Af$ of $S^{*gr}$ is generated by the quadratic forms $T_i T_j$ for $i < j$ and $T_i^2 - T_1^2$ for $i > 1$. If $d \geqslant 3$ the ideal $\mathfrak{a}$ is not completely secant; the Gorenstein ring $A/\mathfrak{a}$ is not a complete intersection ring (cf. § 5, exerc. 5).