Differential theory of zero-dimensional schemes

TL;DR

This paper develops a differential theory for zero-dimensional schemes in projective space, providing formulas for Hilbert polynomials and bounds for regularity indices of Kähler differentials, with applications to scheme characterization.

Contribution

It introduces new formulas and bounds for Kähler differentials of zero-dimensional schemes, extending previous results and providing novel characterizations without primary decomposition.

Findings

Formulas for Hilbert polynomials of Kähler differentials

Bounds for regularity indices of differential modules

Characterization of weakly curvilinear schemes

Abstract

For a 0-dimensional scheme $\mathbb{X}$ in $\mathbb{P}^n$ over a perfect field $K$, we first embed the homogeneous coordinate ring $R$ into its truncated integral closure $\widetilde{R}$. Then we use the corresponding map from the module of K\"ahler differentials $\Omega^1_{R/K}$ to $\Omega^1_{\widetilde{R}/K}$ to find a formula for the Hilbert polynomial ${\rm HP}(\Omega^1_{R/K})$ and a sharp bound for the regularity index ${\rm ri}(\Omega^1_{R/K})$. Additionally, we extend this to formulas for the Hilbert polynomials ${\rm HP}(\Omega^m_{R/K})$ and bounds for the regularity indices of the higher modules of K\"ahler differentials. Next we derive a new characterization of a weakly curvilinear scheme $\mathbb{X}$ which can be checked without computing a primary decomposition of its homogeneous vanishing ideal. Moreover, we prove precise formulas for the Hilbert polynomial of $\Omega^m_{R/K}$ of a fat point scheme $\mathbb{X}$, extending and settling previous partial results and conjectures. Finally, we characterize uniformity conditions on $\mathbb{X}$ using the Hilbert functions of the K\"ahler differential modules of $\mathbb{X}$ and its subschemes.

Equations

$$\mathrm{HP}({\Omega }_{R/K}^1)=\mathrm{deg}(\mathbb{X})+{\dim }_K({\Omega }_{S/K}^1)$$

$$\mathrm{HP}({\Omega }_{R/K}^m)={\dim }_K({\Omega }_{S/K}^m)+{\dim }_K({\Omega }_{S/K}^{m-1})$$

$$\mathrm{HP}({\Omega }_{R/K}^1)>\mathrm{deg}(\mathbb{X}),\mathrm{HP}({\Omega }_{R/K}^2)=\mathrm{HP}({\Omega }_{R/K}^1)-\mathrm{deg}(\mathbb{X}),\mathrm{HP}({\Omega }_{R/K}^m)=0\mathrm{for}m\ge 3$$

$${\dim }_K({\Omega }_{S/K}^m)=\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\genfrac{}{}{0ex}{}{n+k-2}{n}\right)+\left(\genfrac{}{}{0ex}{}{m+k-2}{m}\right)\left(\genfrac{}{}{0ex}{}{n+k-2}{n-m-1}\right)$$

$$1={\mathrm{HF}}_{\mathbb{X}}(0)<{\mathrm{HF}}_{\mathbb{X}}(1)<\cdots <{\mathrm{HF}}_{\mathbb{X}}(r_{\mathbb{X}})={\mathrm{HF}}_{\mathbb{X}}(r_{\mathbb{X}}+1)=\cdots =\mathrm{deg}(\mathbb{X})$$

$$Q^h(R)=\{\frac{a}{b}\mid a,b\in R;bhomogeneous\; non-zerodivisor\}$$

$$x_0^{dm}{g}^m+a_{m-1}{x}_0^d(x_0^{(m-1)d}{g}^{m-1})+\cdots +a_1{x}_0^{(m-1)d}(x_0^{d}g)+a_0{x}_0^{dm}=0$$

$$I_{\mathbb{X}}/I_{\mathbb{X}}^2\stackrel{\alpha }{⟶}{R}^{n+1}(-1)\stackrel{\beta }{⟶}{\Omega }_{R/K}^1⟶0$$

$${\Omega }_{R/K}^1\cong {\Omega }_{P/K}^1/(I_{\mathbb{X}}\cdot {\Omega }_{P/K}^1+d{I}_{\mathbb{X}})$$

$${\mathrm{HF}}_{{\Omega }_{R/K}^1}(r_{\mathbb{X}}+1)>\cdots >{\mathrm{HF}}_{{\Omega }_{R/K}^1}(\mathrm{ri}({\Omega }_{R/K}^1))$$

$$0⟶\mathcal{G}(-1)⟶R^{n+1}(-1)\stackrel{\phi }{⟶}{\Omega }_{R/K}^1⟶0$$

$$(x_0{f}_0,\dots ,x_0{f}_n)\in \mathrm{Ker}(\phi )=[\mathcal{G}(-1){]}_{i+1}={\mathcal{G}}_i$$

$$(f_0,\dots ,f_n)=h_1{v}_1+\cdots +h_k{v}_k\in {\mathcal{G}}_{i-1}$$

$$T{\Omega }_{R/K}^1=\{w\in {\Omega }_{R/K}^1\mid rw=0\text{mbox}forsomehomogeneousnon-zerodivisorr\in R\}$$

$${\mathrm{HF}}_{{\Omega }_{R/K}^1}$$

$${\mathrm{HF}}_{T{\Omega }_{R/K}^1}$$

$${\mathrm{HF}}_{{\Omega }_{R_{{\mathbb{Y}}^{\prime }}/K}^1}:03914977\cdots ,\mathrm{ri}({\Omega }_{R_{{\mathbb{Y}}^{\prime }}/K}^1)=5=2r_{{\mathbb{Y}}^{\prime }}+1,$$

$${\mathrm{HF}}_{T{\Omega }_{R_{{\mathbb{Y}}^{\prime }}/K}^1}:0027200\cdots ,\mathrm{ri}(T{\Omega }_{R_{{\mathbb{Y}}^{\prime }}/K}^1)=5=2r_{{\mathbb{Y}}^{\prime }}+1.$$

$$\phi :{\Omega }_{\tilde{R}/K}^1\cong S[x_0]d{x}_0\oplus K[x_0]{\otimes }_K{\Omega }_{S/K}^1.$$

$${\Omega }_{R/K}^1⟶{\Omega }_{\tilde{R}/K}^1⟶{\Omega }_{Q^h(R)/K}^1\cong {\Omega }_{R_{x_0}/K}^1\cong ({\Omega }_{R/K}^1{)}_{x_0}$$

$$\Psi (F{x}_{0}dG)=\phi (F)x_{0}d(\phi (G))=f{x}_0^{r_{\mathbb{X}}+1}d(g{x}_0^{r_{\mathbb{X}}})=x_0^{2r_{\mathbb{X}}+1}\otimes fdg+r_{\mathbb{X}}fg{x}_0^{2r_{\mathbb{X}}}d{x}_0$$

$$\epsilon :{\Omega }_{R/K}^1⟶\mathfrak{m}given\; by\epsilon (d{x}_i)=x_i\mathrm{for}i=0,\dots ,n.$$

$$0⟶{\Lambda }^{n+1}{\Omega }_{R/K}^1\stackrel{{\epsilon }^{(n+1)}}{⟶}\cdots \stackrel{{\epsilon }^{(3)}}{⟶}{\Lambda }^2{\Omega }_{R/K}^1\stackrel{{\epsilon }^{(2)}}{⟶}{\Omega }_{R/K}^1\stackrel{\epsilon }{⟶}\mathfrak{m}⟶0$$

$$0⟶\mathrm{Ker}(\xi )⟶P^{n+1}\stackrel{\xi }{⟶}\mathfrak{M}⟶0$$

$$TF=\sum _{i=0}^{k-1}T{F}_i{X}_i+(\sum _{i=k}^n{F}_i{X}_i{T}^{\prime })X_k$$

$$\delta (TF)=\sum _{i=0}^{k-1}t{f}_{i}d{x}_i+\sum _{i=k}^n{f}_i{x}_i{t}^{\prime }d{x}_k+U_{R/K}$$

$$t\cdot \delta (F)$$

$$=\sum _{i=0}^{k-1}t{f}_{i}d{x}_i+\sum _{i=k}^n{t}^{\prime }{f}_i{x}_{i}d{x}_k+\sum _{i=k}^n{t}^{\prime }{f}_i(x_{k}d{x}_i-x_{i}d{x}_k)+U_{R/K}$$

$$=\delta (TF)$$

$$\vartheta (F)=(F_0,\dots ,F_n)=(G_0,\dots ,G_n)+v\mathrm{with}v\in V$$

$${\mathrm{HF}}_{T{\Omega }_{R/K}^1}:000100\cdots ,and\; thus\mathrm{ri}(T{\Omega }_{R/K}^1)=4,$$

$${\mathrm{HF}}_{U_{R/K}}:003444\cdots ,and\; thus\mathrm{ri}(U_{R/K})=3,$$

$${\mathrm{HF}}_{\mathrm{Ker}(\epsilon )}:003544\cdots ,and\; thus\mathrm{ri}(\mathrm{Ker}(\gamma ))=4.$$

Full text

1. Introduction

The study of 0-dimensional subschemes ${\mathbb{X}}$ of a projective space $\mathbb{P}^{n}$, in particular of finite set of points, has a long and rich history. Traditional tools used in this area are the homogeneous vanishing ideal $I_{{\mathbb{X}}}$ of ${\mathbb{X}}$ in $P=K[x_{0},\dots,x_{n}]$, the homogeneous coordinate ring $R=P/I_{{\mathbb{X}}}$, its Hilbert function $\mathop{\rm HF}\nolimits_{{\mathbb{X}}}(i)=\dim_{K}(R_{i})$, and the canonical module $\omega_{R}$ of $R$. In several previous works, the authors introduced and started using the Kähler differential module $\Omega^{1}_{R/K}$ of $R$ and even the entire Kähler differential algebra $\Omega^{\bullet}_{R/K}=\Lambda_{R}({\Omega^{1}_{R/K}})$ in order to advance this topic (see [4], [7], and [8]).

In the present paper we systematize this approach, extend it by several new techniques, solve and unify previous partial results and conjectures, and derive a number of useful applications to some computational tasks. Let us describe the individual contributions in more detail.

In Section 2 we recall some basic objects related to a 0-dimensional scheme ${\mathbb{X}}$ in the projective space $\mathbb{P}^{n}$ over a perfect field $K$, in particular its homogeneous vanishing ideal $I_{\mathbb{X}}\subseteq P=K[X_{0},\dots,X_{n}]$, its homogeneous coordinate ring $R=P/I_{\mathbb{X}}$, and its Hilbert function $\mathop{\rm HF}\nolimits_{\mathbb{X}}(i)=\dim_{K}(R_{i})$ for $i\in\mathbb{Z}$. We always assume that the support of ${\mathbb{X}}$ is contained in the affine space $D_{+}(X_{0})$ and denote its affine coordinate ring by $S=K[X_{1},\dots,X_{n}]/I_{\mathbb{X}}^{\rm deh}$. After identifying the homogeneous ring of quotients of $R$ with $S[x_{0},x_{0}^{-1}]$, we introduce a new tool, namely the embedding of $R$ into its truncated integral closure $\widetilde{R}\cong S[x_{0}]$.

These techniques come to fruition in Section 3, where we introduce and study the Kähler differential module ${\Omega^{1}_{R/K}}$ and its Hilbert function $\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}$. New results here are the description of $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})$ and a sharp bound $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})\leq 2\,r_{\mathbb{X}}+1$ for the (Hilbert) regularity index of ${\Omega^{1}_{R/K}}$. Here $r_{\mathbb{X}}$ is the regularity index of ${\mathbb{X}}$, i.e., the first degree from where on the Hilbert function agrees with the value of the Hilbert polynomial (see Prop. 3.4). After finding the Hilbert function and the Hilbert polynomial of $\tilde{R}$ (see Prop. 3.7), we construct the canonical map $\varPhi:\,{\Omega^{1}_{R/K}}\rightarrow{\Omega^{1}_{\widetilde{R}/K}}$ explicitly, show that $\mathop{\rm Ker}\nolimits(\varPhi)$ is exactly the torsion submodule of ${\Omega^{1}_{R/K}}$, and use it to prove our first main result, namely the formula

[TABLE]

for the Hilbert polynomial of ${\Omega^{1}_{R/K}}$.

Further tools are provided in Section 4 in order to study the torsion submodule $T{\Omega^{1}_{R/K}}$ and other important submodules of ${\Omega^{1}_{R/K}}$. The Euler derivation $\delta:\,R\rightarrow R$ given by $\delta(f)=i\,f$ for $f\in R_{i}$ gives rise to the Euler form ${\varepsilon}:\,{\Omega^{1}_{R/K}}\rightarrow R$ which satisfies ${\varepsilon}(dx_{i})=x_{i}$ for $i=0,\dots,n$. The Koszul complex over ${\varepsilon}$ is called the Euler-Koszul complex of $R$, and the image of ${\varepsilon}^{(2)}:\,\Lambda^{2}{\Omega^{1}_{R/K}}\rightarrow{\Omega^{1}_{R/K}}$ is called the Koszul submodule $U_{R/K}=\langle x_{i}dx_{j}-x_{j}dx_{i}\mid 0\leq i<j\leq n\rangle$ of ${\Omega^{1}_{R/K}}$. Both $T{\Omega^{1}_{R/K}}$ and $U_{R/K}$ are contained in $\mathop{\rm Ker}\nolimits({\varepsilon})$ (see Props. 4.2 and 4.4). The Koszul submodule is equal to $\mathop{\rm Ker}\nolimits({\varepsilon})$ if $\mathop{\rm char}\nolimits(K)$ is zero or $\geq 2r_{\mathbb{X}}+1$, and in the other cases the (small) difference between the two submodules is laid out in detail in Prop. 4.6.

Next up, we apply the preceding results in Section 5 to the higher Kähler differential modules ${\Omega^{m}_{R/K}}$, where $m\geq 1$. After recalling the presentation and computation of ${\Omega^{m}_{R/K}}$ (see Prop. 5.2), we use the embedding $R\hookrightarrow\tilde{R}$ to prove the formula

[TABLE]

for the Hilbert polynomial of ${\Omega^{m}_{R/K}}$ and the sharp bound $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}})\leq r_{\mathbb{X}}+m$ for its regularity index (see Prop. 5.4).

In final three sections we apply and extend these algebraic results to characterize and compute geometric properties of the scheme ${\mathbb{X}}$. In Section 6 we look at (weakly) curvilinear schemes which are defined by the property that the maximal ideals of their local rings are unigenerated. Notice that this generalizes slightly the condition that these local rings are of the form $K[z]/\langle p^{k}\rangle$ with an irreducible polynomial $p$. Special cases are, of course, smooth schemes for which it is known that they can be characterized by $\Omega^{1}_{S/K}=0$, or, equivalently, by $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=\deg({\mathbb{X}})$ (see Prop. 6.1). Note that this allows us to check smoothness of ${\mathbb{X}}$ without computing the primary decomposition of $I_{\mathbb{X}}$. In a similar vein, we characterize non-smooth weakly curvilinear schemes by $\Omega^{1}_{S/K}\neq 0$ and $\Omega^{m}_{S/K}=0$ for $m\geq 2$. Equivalently, the scheme ${\mathbb{X}}$ is weakly curvilinear, but not smooth if and only if

[TABLE]

(see Prop. 6.6). Thus we can check algorithmically whether ${\mathbb{X}}$ is weakly curvilinear without computing a primary decomposition. Using the dimensions of the residue class fields of the local rings of ${\mathbb{X}}$, we can also write down explicit formulas for the Hilbert polynomials of ${\Omega^{1}_{R/K}}$ and $\Omega^{2}_{R/K}$ in the weakly curvilinear case (see Cor. 6.8).

The next application is contained in Section 7 where we consider fat point schemes ${\mathbb{X}}$. They are defined by vanishing ideals of the form $I_{\mathbb{X}}=I_{p_{1}}^{m_{1}}\cap\cdots\cap I_{p_{t}}^{m_{t}}$ where the points $p_{i}$ in the support of ${\mathbb{X}}$ are assumed to be $K$-rational and $m_{i}\geq 1$. The case of a reduced scheme, i.e., the case $m_{1}=\cdots=m_{t}=1$, is easily characterized by $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=t$ and $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=0$ for $m\geq 2$ (see Remark 7.1). The general case reduces to the study of Kähler differential modules of rings of the form $S=A/\mathfrak{q}^{k}$, where $A=K[X_{1},\dots,X_{n}]$ and $\mathfrak{q}=\langle X_{1},\dots,X_{n}\rangle$. Here we obtain explicit formulas for the values of the Hilbert function of $\Omega^{m}_{S/K}$ which only depend on the value of $\delta=\dim_{K}(d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})$ (see Prop. 7.2).

If $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$, we succeed in determining this value $\delta$ using a number of subtle and detailed arguments (see Props. 7.4, 7.5, and 7.6). Thus we are able to derive the explicit formula

[TABLE]

for $S=K[X_{1},\dots,X_{n}]/\langle X_{1},\dots,X_{n}\rangle^{k}$ and $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$ (see Thm. 7.8). Under a more stringent assumption about the characteristic of $K$, we also provide another proof based on the exactness of the Euler-Koszul complex (see Prop. 7.9 and Remark 7.10).

Finally, we combine everything and get explicit formulas for the Hilbert polynomials of the modules ${\Omega^{m}_{R/K}}$ if $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>\max\{m_{1},\dots,m_{t}\}$. They generalize and vastly improve the partial results in [7] and [8].

To cap the paper off, we look at differential characterizations of uniformity properties of ${\mathbb{X}}$ in Section 8. One of them, the Cayley-Bacharach property of degree $d$, requires that every hypersurface of degree $d$ which contains a subscheme ${\mathbb{Y}}$ of ${\mathbb{X}}$ of colength one automatically contains ${\mathbb{X}}$. It can be characterized by the degrees of the minimal separators of ${\mathbb{X}}$ (see Remark 8.2). Our key result here is that the vanishing ideal $I_{{\mathbb{Y}}/{\mathbb{X}}}$ of a subscheme ${\mathbb{Y}}$ in ${\mathbb{X}}$ satisfies $(dI_{{\mathbb{Y}}/{\mathbb{X}}})_{\alpha_{{\mathbb{Y}}/{\mathbb{X}}}}\neq 0$ in its initial degree $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}$ provided that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>r_{\mathbb{X}}$ (see Prop. 8.3). This allows us to characterize schemes having the Cayley-Bacharach property of degree $d$ through the Hilbert function of their Kähler differential modules (see Cor. 8.4). More generally, we consider higher uniformity properties of ${\mathbb{X}}$, called $(i,j)$-uniformity, and characterize them through the Hilbert function of ${\Omega^{1}_{R/K}}$ as well (see Cor. 8.7).

Many examples in this paper illustrate our results. They were computed using the computer algebra system ApCoCoA (see [1]). For the notation and basic definitions we use, we refer to [11], [12], and [13]. Our main reference for results about Kähler differential modules is the fundamental work [14] of our late teacher, mentor, and supporter Ernst Kunz (1933-2021) to whom we dedicate this paper with deep gratitude.

2. Zero-Dimensional Schemes

In the following we let ${\mathbb{X}}$ be a 0-dimensional scheme in projective $n$-space $\mathbb{P}^{n}$ over a perfect field $K$. The homogeneous coordinate ring of $\mathbb{P}^{n}$ is the polynomial ring $P=K[X_{0},\dots,X_{n}]$, and the homogeneous coordinate ring of ${\mathbb{X}}$ is $R=P/I_{\mathbb{X}}$, where $I_{\mathbb{X}}$ denotes the homogeneous vanishing ideal of ${\mathbb{X}}$. Recall that $I_{\mathbb{X}}$ is a saturated homogeneous ideal, and that $R$ is a standard graded 1-dimensional Cohen-Macaulay ring. In particular, there exists a homogeneous non-zerodivisor in $R$. To simplify the presentation, we make the following assumption.

Assumption 2.1**.**

Suppose that no point of the support of ${\mathbb{X}}$ is contained in the hyperplane $\mathcal{Z}(X_{0})$.

Usually, if this assumption does not hold, it suffices to perform a homogeneous linear change of coordinates. This will not affect any results in this paper. In the case of a finite field $K$, it may be necessary to perform a small base field extension in order to find a suitable homogeneous linear change of coordinates. This will not affect any results either. As a consequence of this assumption, we get that ${\mathbb{X}}$ is contained in the affine space $\mathbb{A}^{n}\cong D_{+}(X_{0})$, and that the image $x_{0}$ of $X_{0}$ in $R$ is a non-zerodivisor.

The affine coordinate ring of ${\mathbb{X}}$, viewed as a subscheme of $\mathbb{A}^{n}$, is given by $S=R/\langle x_{0}-1\rangle\cong K[X_{1},\dots,X_{n}]/I_{\mathbb{X}}^{\rm deh}$, where $I_{\mathbb{X}}^{\rm deh}$ is the dehomogenization of $I_{\mathbb{X}}$ with respect to $X_{0}$. The ring $S$ is a 0-dimensional affine $K$-algebra and hence a finite dimensional $K$-vector space. The following functions are central to the study of 0-dimensional subschemes of projective spaces.

Definition 2.2**.**

In the above setting, the map $\mathop{\rm HF}\nolimits_{\mathbb{X}}:\;{\mathbb{Z}}\longrightarrow{\mathbb{Z}}$ given by $\mathop{\rm HF}\nolimits_{\mathbb{X}}(i)=\dim_{K}(R_{i})$ for all $i\in{\mathbb{Z}}$ is called the Hilbert function of ${\mathbb{X}}$.

Its first difference function $\Delta\mathop{\rm HF}\nolimits_{\mathbb{X}}:\;{\mathbb{Z}}\longrightarrow{\mathbb{Z}}$ which is given by $\Delta\mathop{\rm HF}\nolimits_{\mathbb{X}}(i)=\mathop{\rm HF}\nolimits_{\mathbb{X}}(i)-\mathop{\rm HF}\nolimits_{\mathbb{X}}(i-1)$ for $i\in{\mathbb{Z}}$ is called the Castelnuovo function of ${\mathbb{X}}$.

The Hilbert function contains a number of useful invariants of the embedding of ${\mathbb{X}}$ into $\mathbb{P}^{n}$. Let us collect some of its basic properties.

Remark 2.3**.**

The Hilbert function of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ satisfies

[TABLE]

where $r_{\mathbb{X}}$ is called the (Hilbert) regularity index of ${\mathbb{X}}$ and $\deg({\mathbb{X}})$ is the degree of ${\mathbb{X}}$, i.e., the value of the (constant) Hilbert polynomial of $R$.

The structure of the ring $R$ in degrees $\geq r_{\mathbb{X}}$ can be described as follows.

Proposition 2.4**.**

For $i\geq r_{\mathbb{X}}$, the map ${\varepsilon}_{i}:\;R_{i}\longrightarrow S$ given by $f\mapsto f^{\rm deh}$ is an isomorphism of $K$-vector spaces.

Proof.

Since $x_{0}$ is a non-zerodivisor of $R$, the definition of $r_{\mathbb{X}}$ shows that the multiplication map $\mu_{x_{0}}:\;R_{i}\longrightarrow R_{i+1}$ is an isomorphism for $i\geq r_{\mathbb{X}}$. Together with the observation that $\dim_{K}(R_{i})=\dim_{K}(S)=\deg({\mathbb{X}})$ for $i\geq r_{\mathbb{X}}$, it follows that it suffices to prove the claim for $i\gg 0$. Let $\{b_{1},\dots,b_{r}\}$ be a $K$-vector space basis of $S$. Using the fact that the map $\delta:\;R\longrightarrow R/\langle x_{0}-1\rangle\cong S$ is surjective and given by dehomogenization, we can find for every $j\in\{1,\dots,r\}$ a homogeneous element $f_{j}\in R$ such that $\delta(f_{j})=f_{j}^{\rm deh}=b_{j}$. By multiplying each $f_{j}$ with an appropriate power of $x_{0}$, we may assume that all elements $f_{j}$ have the same degree $i$. Thus the map ${\varepsilon}_{i}$ is surjective, hence bijective, and the claim follows. ∎

Subsequently, we will also use the following ring. The graded $K$-algebra

[TABLE]

is called the (full) homogeneous ring of quotients of $R$. We can simplify its description as follows.

Proposition 2.5**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above.

1. (a)

The canonical inclusion $R_{x_{0}}\rightarrow Q^{h}(R)$ is bijective. 2. (b)

The map ${\varphi}:\;Q^{h}(R)\longrightarrow S[x_{0},x_{0}^{-1}]$ given by ${\varphi}(f/x_{0}^{i})=f^{\rm deh}\,x_{0}^{d-i}$ for $f\in R_{d}$ and $i\geq 0$ is an isomorphism of graded $K$-algebras.

Proof.

To prove (a), we let $a\in R_{d}$ and $b\in R_{e}$ be homogeneous elements of $R$ of degrees $d,e\in\mathbb{N}$, and assume that $b$ is a non-zerodivisor. We need to show that there exist $i\geq 0$ and a homogeneous element $f\in R_{i}$ such that $a/b=f/x_{0}^{i}$. First we choose $i$ such that $i\geq r_{\mathbb{X}}+e-d$. Note that the multiplication map $\mu_{b}:\,R_{k}\longrightarrow R_{k+e}$ is bijective for every $k\geq r_{\mathbb{X}}$, because $b$ is a non-zerodivisor. Hence $d+i\geq r_{\mathbb{X}}+e$ implies that the element $a\,x_{0}^{i}\in R_{d+i}$ is of the form $a\,x_{0}^{i}=b\,f$ for some $f\in R_{j}$, where $j=d+i-e$. Thus we get $a/b=f/x_{0}^{i}$, as desired.

It remains to prove (b). Given $f\in R_{d}$ and $i\geq 0$, we let $j=r_{\mathbb{X}}+i$. Then we have $x_{0}^{j}\,(f/x_{0}^{i})\in R_{r_{\mathbb{X}}+d}$ and $x_{0}^{j}\,{\varphi}(f/x_{0}^{i})={\varphi}(x_{0}^{j}\,(f/x_{0}^{i}))=f^{\rm deh}x_{0}^{r_{\mathbb{X}}+d}$. Together with Proposition 2.4, it follows that ${\varphi}_{d-i}:\;Q^{h}(R)_{d-i}\longrightarrow Sx_{0}^{d-i}$ is an isomorphism of $K$-vector spaces. As ${\varphi}$ is clearly a homomorphism of graded rings, the claim follows. ∎

Using the identifications $Q^{h}(R)\cong R_{x_{0}}\cong S[x_{0},x_{0}^{-1}]$ provided by this proposition, we now examine the integral closure of $R$ in $Q^{h}(R)$.

Proposition 2.6**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above.

1. (a)

The map $\psi:\;R\longrightarrow S[x_{0}]$ given by $\psi(f)=f^{\rm deh}\,x_{0}^{d}$ for $f\in R_{d}$ is an injective graded $K[x_{0}]$-algebra homomorphism. 2. (b)

For $i\geq r_{\mathbb{X}}$, the map $\psi_{i}:\;R_{i}\longrightarrow Sx_{0}^{i}$ is an isomorphism of $K$-vector spaces. 3. (c)

The subring $\widetilde{R}$ of $Q^{h}(R)$ generated by all homogeneous elements of non-negative degree satisfies $\widetilde{R}\cong S[x_{0}]$ and is contained in the integral closure of $R$ in $Q^{h}(R)$. 4. (d)

The ring $\widetilde{R}$ is the integral closure of $R$ in $Q^{h}(R)$ if and only if $S$ is a reduced ring.

Proof.

Claims (a) and (b) follow immediately from Propositions 2.4 and 2.5. To prove (c), we let $f\in Q^{h}(R)$ be a homogeneous element of degree $d\geq 0$. Using the isomorphism ${\varphi}:\;Q^{h}(R)\longrightarrow S[x_{0},x_{0}^{-1}]$ given in Proposition 2.5, we write ${\varphi}(f)=x_{0}^{d}\,g$ with $g\in S$. By definition, the map ${\varphi}$ identifies $\widetilde{R}$ with $S[x_{0}]$. Since $S$ is a finite $K$-algebra, there exists a relation $g^{m}+a_{m-1}g^{m-1}+\cdots+a_{1}g_{1}+a_{0}=0$ with $a_{i}\in K$. Then we apply ${\varphi}^{-1}$ to both sides of

[TABLE]

and get $f^{m}+a_{m-1}x_{0}^{d}f^{m-1}+\cdots+a_{1}x_{0}^{(m-1)d}f+a_{0}x_{0}^{dm}=0$. Thus $a_{i}x_{0}^{j}\in R$ implies that $f$ is integral over $R$.

Finally, we prove (d). If $S$ is a reduced ring, then $\widetilde{R}$ is the integral closure of $R$ in $Q^{h}(R)$ by [2], Ch. V, §2, Prop. 9 and §8, Prop. 20. Conversely, if $S$ contains a non-zero nilpotent element $f\in S$ with $f^{k}=0$ for some $k\geq 1$, then we consider the homogeneous element $g={\varphi}^{-1}(fx_{0}^{-1})\in Q^{h}(R)_{-1}$. By definition, we have $g\notin\widetilde{R}$. Moreover, from ${\varphi}(g^{k})=f^{k}x_{0}^{-k}=0$, it follows that $g^{k}=0$, and therefore $g$ is integral over $R$. ∎

The ring $\widetilde{R}$ will prove very useful and deserves a name.

Definition 2.7**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above. Then the subring $\widetilde{R}$ of $Q^{h}(R)$ generated by all homogeneous elements of non-negative degree is called the truncated integral closure of $R$.

Frequently, we will use the identification $\widetilde{R}\cong S[x_{0}]$ given by $f\mapsto f^{\rm deh}\,x_{0}^{d}$ for a homogeneous element $f\in\widetilde{R}_{d}$ of degree $d\geq 0$.

3. The Kähler Differential Module

In the following we continue to use the assumptions and notation introduced above. In particular, let $K$ be a perfect field, and let ${\mathbb{X}}$ be a 0-dimensional subscheme of $\mathbb{P}^{n}$ with homogeneous coordinate ring $R=P/I_{\mathbb{X}}$ for $P=K[X_{0},\dots,X_{n}]$. For $i=0,\dots,n$, we denote the image of $X_{i}$ in $R$ by $x_{i}$, and we assume that $x_{0}$ is a non-zerodivisor in $R$. Furthermore, we let $\mathfrak{m}=\langle x_{0},\dots,x_{n}\rangle$ be the homogeneous maximal ideal of $R$. As the following $R$-module is the main object of study in this section, we recall its (well-known) definition.

Definition 3.1**.**

Let $J$ be the kernel of the multiplication map $\mu:R\otimes_{K}R\longrightarrow R$. Then ${\Omega^{1}_{R/K}}=J/J^{2}$ is an $R$-module via $r\cdot\sum_{i}a_{i}\otimes b_{i}=\sum_{i}ra_{i}b_{i}$ for $r,a_{i},b_{i}\in R$. It is called the module of Kähler differentials of $R/K$, or the Kähler differential module of $R/K$.

The map $d_{R/K}:\;R\longrightarrow{\Omega^{1}_{R/K}}$ defined by $d_{R/K}(f)=f\otimes 1-1\otimes f+J^{2}$ is called the universal derivation of $R/K$.

Notice that the map $d_{R/K}$ is indeed a derivation of $R/K$, i.e., it is $K$-linear and satisfies the product rule. If the algebra $R/K$ is clear from the context, we will usually simply write $df$ instead of $d_{R/K}(f)$. Let us recall some of the basic properties of the Kähler differential module of $R/K$.

Proposition 3.2**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above.

1. (a)

The $R$-module ${\Omega^{1}_{R/K}}$ is positively graded and generated by the homogeneous elements $\{dx_{0},\dots,dx_{n}\}$ of degree 1. 2. (b)

There is a homogeneous short exact sequence of graded $R$-modules

[TABLE]

where $\alpha$ is given by $\alpha(f+I_{\mathbb{X}}^{2})=({\textstyle\frac{\partial f}{\partial x_{0}}},\dots,{\textstyle\frac{\partial f}{\partial x_{n}}})$ and where $\beta$ is given by $\beta(e_{i})=dx_{i}$ for $i=0,\dots,n$. 3. (c)

The graded $R$-module ${\Omega^{1}_{R/K}}$ has a presentation

[TABLE]

where $\Omega^{1}_{P/K}=PdX_{0}\oplus\cdots\oplus PdX_{n}\cong P^{n+1}(-1)$ is a graded free $P$-module and $dI_{\mathbb{X}}=\langle{\textstyle\frac{\partial f}{\partial X_{0}}}dX_{0}+\cdots+{\textstyle\frac{\partial f}{\partial X_{n}}}dX_{n}\mid f\in I_{\mathbb{X}}\rangle$.

Proof.

Claims (a) and (b) follow from [14], Propositions 4.12 and 4.17 and claim (c) follows from ibid., Proposition 4.19. ∎

Every finitely generated graded $R$-module has a Hilbert function, a constant Hilbert polynomial, and a regularity index. This yields the following definition.

Definition 3.3**.**

In the above setting, let ${\Omega^{1}_{R/K}}$ be the Kähler differential module of $R$.

1. (a)

The map $\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}:\;{\mathbb{Z}}\longrightarrow{\mathbb{Z}}$ given by $\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}(i)=\dim_{K}({\Omega^{1}_{R/K}})_{i}$ for $i\in{\mathbb{Z}}$ is called the Hilbert function of ${\Omega^{1}_{R/K}}$. 2. (b)

The number $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}(i)$ for $i\gg 0$ is called the Hilbert polynomial of ${\Omega^{1}_{R/K}}$. 3. (c)

The number $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})=\min\{i\in{\mathbb{Z}}\mid\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}(j)=\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})\hbox{\ \rm for \ }j\geq i\}$ is called the regularity index of ${\Omega^{1}_{R/K}}$.

The following proposition shows that the Hilbert polynomial of ${\Omega^{1}_{R/K}}$ is well-defined and that there is a bound for its regularity index.

Proposition 3.4**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above.

1. (a)

For $i\leq 0$, we have $\mathop{\rm HF}\nolimits_{\Omega^{1}_{R/K}}(i)=0$. 2. (b)

For $i\geq r_{\mathbb{X}}+1$, the multiplication map $\mu_{x_{0}}:\;({\Omega^{1}_{R/K}})_{i}\longrightarrow({\Omega^{1}_{R/K}})_{i+1}$ is surjective. In particular, we have $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})\geq r_{\mathbb{X}}+1$. 3. (c)

If $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})>r_{\mathbb{X}}+1$ then we have

[TABLE] 4. (d)

For $i\geq 2r_{\mathbb{X}}+1$, the multiplication map $\mu_{x_{0}}:\;({\Omega^{1}_{R/K}})_{i}\longrightarrow({\Omega^{1}_{R/K}})_{i+1}$ is bijective. In particular, we have $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})\leq 2r_{\mathbb{X}}+1$.

Proof.

Claim (a) follows from the fact that ${\Omega^{1}_{R/K}}$ is generated by homogeneous elements of degree 1. To prove (b), we let $i\geq r_{\mathbb{X}}+1$ and $w\in({\Omega^{1}_{R/K}})_{i+1}$. Then we can write $w=f_{0}dx_{0}+\cdots+f_{n}dx_{n}$ with $f_{0},\dots,f_{n}\in R_{i}$. Since $i\geq r_{\mathbb{X}}+1$, Proposition 2.4 implies that there exist elements $g_{j}\in R_{i-1}$ such that $f_{j}=x_{0}g_{j}$ for $j=0,\dots,n$, and hence $w=x_{0}\,(g_{0}dx_{0}+\cdots+g_{n}dx_{n})$ is in the image of $\mu_{x_{0}}$.

Claim (c) is a consequence of (b) and the definition of $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})$. It remains to show (d). Let $w=f_{0}dx_{0}+\cdots+f_{n}dx_{n}$ be an element in the kernel of $\mu_{x_{0}}$, where $f_{j}\in R_{i-1}$ for $j=0,\dots,n$. By Proposition 3.2, we have an exact sequence of graded $R$-modules

[TABLE]

where $\mathcal{G}$ is generated by the tuples $(\frac{\partial f}{\partial x_{0}},\dots,\frac{\partial f}{\partial x_{n}})$ such that $f\in I_{\mathbb{X}}$ and where ${\varphi}(f_{0},...,f_{n})=f_{0}dx_{0}+\cdots+f_{n}dx_{n}$ for all $f_{0},...,f_{n}\in R$. Since $I_{\mathbb{X}}$ is generated in degrees $\leq r_{\mathbb{X}}+1$ by [6], Proposition 1.1, the graded $R$-module $\mathcal{G}$ is generated in degrees $\leq r_{\mathbb{X}}$. So, let $\{v_{1},...,v_{k}\}$ be a homogeneous system of generators of $\mathcal{G}$, where $\deg(v_{j})\leq r_{\mathbb{X}}$. Given an element $w=f_{0}dx_{0}+\cdots+f_{n}dx_{n}\in\mathop{\rm Ker}\nolimits(\mu_{x_{0}})$ with $f_{j}\in R_{i-1}$ for $j=0,\dots,n$, we have

[TABLE]

Thus we may write $(x_{0}f_{0},...,x_{0}f_{n})=g_{1}v_{1}+\cdots+g_{k}v_{k}$ with homogeneous elements $g_{j}\in R$ of degree $\deg(g_{j})=i-\deg(v_{j})\geq i-r_{\mathbb{X}}\geq r_{\mathbb{X}}+1$. Now Proposition 2.4 allows us to write $g_{j}=x_{0}h_{j}$ with $h_{j}\in R$ and we obtain

[TABLE]

This yields $w=f_{0}dx_{0}+\cdots+f_{n}dx_{n}=0$ in ${\Omega^{1}_{R/K}}$, and the proof is complete. ∎

The graded $R$-module

[TABLE]

is called the torsion submodule of ${\Omega^{1}_{R/K}}$. We can describe it as follows.

Proposition 3.5**.**

Let $R$ be the homogeneous coordinate ring of ${\mathbb{X}}$ as above.

1. (a)

We have $T{\Omega^{1}_{R/K}}=\{\,w\in{\Omega^{1}_{R/K}}\mid x_{0}^{i}w=0$ for some $i\geq 1\,\}$. 2. (b)

We have $\mathop{\rm HP}\nolimits(T{\Omega^{1}_{R/K}})=0$ and $\mathop{\rm ri}\nolimits(T{\Omega^{1}_{R/K}})\leq 2r_{\mathbb{X}}+1$.

Proof.

Let $w\in T{\Omega^{1}_{R/K}}$ be a non-zero homogeneous element with $fw=0$ for a homogeneous non-zerodivisor $f\in R_{i}$, $i\geq 0$. Then $f^{\rm deh}\in S$ is a non-zerodivisor. Since $S$ is an Artinian semilocal ring, it consists of only units and zero-divisors, and so $f^{\rm deh}\in S$ is a unit. Let $u\in S$ be such that $uf^{\rm deh}=1$ and $g=\psi_{r_{\mathbb{X}}}^{-1}(ux_{0}^{r_{\mathbb{X}}})$, where $\psi_{r_{\mathbb{X}}}:R_{r_{\mathbb{X}}}\rightarrow Sx_{0}^{r_{\mathbb{X}}}$ is the isomorphism of $K$-vector spaces in Proposition 2.6.b. We have $gf=x_{0}^{r_{\mathbb{X}}+i}$, and so $x_{0}^{r_{\mathbb{X}}+i}w=gfw=0$. This proves claim (a). Claim (b) follows from (a) and Proposition 3.4.d. ∎

Let us show by the following example that the above upper bound for the regularity index of $T{\Omega^{1}_{R/K}}$ is a sharp bound.

Example 3.6**.**

Let ${\mathbb{X}}$ and ${\mathbb{Y}}$ be two sets of five $K$-rational points in $\mathbb{P}^{2}$ such that ${\mathbb{X}}$ consists of three points on a line and two points on another line (two lines can intersect at one point of ${\mathbb{X}}$), and such that ${\mathbb{Y}}$ is contained in a non-singular conic. By $R_{\mathbb{Y}}$ we denote the homogeneous coordinate of ${\mathbb{Y}}$. Then the Hilbert functions of $R$ and $R_{\mathbb{Y}}$ agree, namely $\mathop{\rm HF}\nolimits_{\mathbb{X}}=\mathop{\rm HF}\nolimits_{\mathbb{Y}}:\ 1\ 3\ 5\ 5\cdots,$ and so $r_{\mathbb{X}}=r_{\mathbb{Y}}=2$. However, the Hilbert functions of ${\Omega^{1}_{R/K}}$ and $\Omega^{1}_{R_{\mathbb{Y}}/K}$ as well as of $T{\Omega^{1}_{R/K}}$ and $T\Omega^{1}_{R_{\mathbb{Y}}/K}$ are different:

[TABLE]

In particular, we get $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})=\mathop{\rm ri}\nolimits(T{\Omega^{1}_{R/K}})=\mathop{\rm ri}\nolimits(\Omega^{1}_{R_{\mathbb{Y}}/K})=\mathop{\rm ri}\nolimits(T\Omega^{1}_{R_{\mathbb{Y}}/K})=5=2r_{\mathbb{X}}+1$, and hence the upper bounds for the regularity index of ${\Omega^{1}_{R/K}}$ and $T{\Omega^{1}_{R/K}}$ given in Propositions 3.4 and 3.5 are sharp. Moreover, these upper bounds are also sharp for the non-reduced 0-dimensional scheme ${\mathbb{Y}}^{\prime}\subseteq\mathbb{P}^{2}$ whose support comprises five points $(1:0:1)$, $(1:1:2)$, $(1:2:2)$, $(1:3:1)$, and $(1:1:0)$ on a non-singular conic, in which only the last point $(1:1:0)$ is non-reduced with its ideal $I_{(1:1:0)}=\langle X_{1}-X_{0},X_{2}^{2}\rangle\subseteq K[X_{0},X_{1},X_{2}]$ and ${\rm char}(K)\neq 2,3$, because a calculation gives $\mathop{\rm HF}\nolimits_{{\mathbb{Y}}^{\prime}}:\ 1\ 3\ 6\ 6\cdots$, $r_{{\mathbb{Y}}^{\prime}}=2$, and

[TABLE]

Next we consider the module of Kähler differentials of the truncated integral closure $\widetilde{R}\cong S[x_{0}]$. Its Hilbert function, Hilbert polynomial, and regularity index can be described as follows.

Proposition 3.7**.**

Let $\widetilde{R}\cong S[x_{0}]$ be the truncated integral closure of $R$ in its homogeneous ring of quotients.

1. (a)

We have an isomorphism of graded $K[x_{0}]$-modules

[TABLE] 2. (b)

The Hilbert function $\mathop{\rm HF}\nolimits_{\Omega^{1}_{\widetilde{R}/K}}:\;{\mathbb{Z}}\longrightarrow{\mathbb{Z}}$ of ${\Omega^{1}_{\widetilde{R}/K}}$ is given by $\mathop{\rm HF}\nolimits_{\Omega^{1}_{\widetilde{R}/K}}(i)=0$ for $i<0$, by $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{\widetilde{R}/K}}}(0)=\dim_{K}({\Omega^{1}_{S/K}})$, and by $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{\widetilde{R}/K}}}(i)=\deg({\mathbb{X}})+\dim_{K}({\Omega^{1}_{S/K}})$ for $i>0$. 3. (c)

We have $\mathop{\rm HP}\nolimits({\Omega^{1}_{\widetilde{R}/K}})=\deg({\mathbb{X}})+\dim_{K}({\Omega^{1}_{S/K}})$ and $\mathop{\rm ri}\nolimits({\Omega^{1}_{\widetilde{R}/K}})=1$.

Proof.

Claim (a) follows from [14], Formula 4.11.a, and from $\Omega^{1}_{K[x_{0}]/K}\cong K[x_{0}]dx_{0}$. Part (b) is an immediate consequence of (a), and (c) follows from (b). ∎

Next we use the injective homomorphism of graded $K$-algebras ${\varphi}:\;R\to\widetilde{R}\cong S[x_{0}]$ given by $f\mapsto f^{\rm deh}x_{0}^{k}$ for $f\in R_{k}$ to compare the Kähler differential modules of $R$ and $\widetilde{R}$.

Proposition 3.8**.**

Let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above, and let $\widetilde{R}\cong S[x_{0}]$ be the truncated integral closure of $R$ in its homogeneous ring of quotients.

1. (a)

The homomorphism ${\varphi}:\;R\to\widetilde{R}\cong S[x_{0}]$ induces a homomorphism of graded $K[x_{0}]$-modules $\varPhi:\;{\Omega^{1}_{R/K}}\longrightarrow{\Omega^{1}_{\widetilde{R}/K}}\cong S[x_{0}]dx_{0}\oplus K[x_{0}]\otimes_{K}{\Omega^{1}_{S/K}}$ which is given by $\varPhi(fdx_{i})=f^{\rm deh}x_{i}x_{0}^{k}dx_{0}+x_{0}^{k+1}\otimes f^{\rm deh}dx_{i}$ for $i=1,\dots,n$, and by $\varPhi(fdx_{0})=f^{\rm deh}x_{0}^{k}dx_{0}$ for $f\in R_{k}$. 2. (b)

For $i\geq 2r_{\mathbb{X}}+1$, the map $\varPhi_{i}:\;({\Omega^{1}_{R/K}})_{i}\longrightarrow({\Omega^{1}_{\widetilde{R}/K}})_{i}$ is an isomorphism of $K$-vector spaces. 3. (c)

We have $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=\deg({\mathbb{X}})+\dim_{K}({\Omega^{1}_{S/K}})$ and $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})\leq 2r_{\mathbb{X}}+1$. 4. (d)

We have $T{\Omega^{1}_{R/K}}=\mathop{\rm Ker}\nolimits(\varPhi)$.

Proof.

To prove (a), it suffices to use ${\varphi}$ and calculate $\varPhi(fdx_{i})={\varphi}(f)d{\varphi}(x_{i})=f^{\rm deh}x_{0}^{k}d(x_{i}x_{0})=f^{\rm deh}x_{i}x_{0}^{k}dx_{0}+x_{0}^{k+1}\otimes f^{\rm deh}dx_{i}$ for $i=1,\dots,n$ and $\varPhi(fdx_{0})=f^{\rm deh}x_{0}^{k}dx_{0}$.

Next we show (b). Since the multiplication maps $\mu_{x_{0}}:\;({\Omega^{1}_{R/K}})_{i}\longrightarrow({\Omega^{1}_{R/K}})_{i+1}$ and $\mu_{x_{0}}:\;({\Omega^{1}_{\widetilde{R}/K}})_{i}\longrightarrow({\Omega^{1}_{\widetilde{R}/K}})_{i+1}$ are isomorphisms for $i\geq 2r_{\mathbb{X}}+1$, it suffices to prove that $\varPsi:=\varPhi_{2r_{\mathbb{X}}+1}:\;({\Omega^{1}_{R/K}})_{2r_{\mathbb{X}}+1}\longrightarrow({\Omega^{1}_{\widetilde{R}/K}})_{2r_{\mathbb{X}}+1}$ is an isomorphism. First we prove that $\varPsi$ is injective. The inclusions $R\subset\widetilde{R}\subset Q^{h}(R)=R_{x_{0}}$ induce canonical homomorphisms of graded $K[x_{0}]$-algebras

[TABLE]

whose composition is the canonical map to the localization of ${\Omega^{1}_{R/K}}$ in the element $x_{0}$. If $w\in(\Omega_{R})_{2r_{\mathbb{X}}+1}$ is in the kernel of $\varPsi$ then its image $\frac{w}{1}$ in $({\Omega^{1}_{R/K}})_{x_{0}}$ is zero. Hence there exists a number $j\geq 0$ such that $x_{0}^{j}\,w=0$, and Proposition 3.4.d yields $w=0$.

It remains to prove that $\varPsi$ is surjective. In view of Proposition 3.7.a, we have to consider two cases. For an element of the form $fx_{0}^{2r_{\mathbb{X}}}dx_{0}$, we let $F\in R_{r_{\mathbb{X}}}$ be a preimage of $fx_{0}^{r_{\mathbb{X}}}$ under the isomorphism $R_{r_{\mathbb{X}}}\cong Sx_{0}^{r_{\mathbb{X}}}$ in Proposition 2.6.b. Then we have $\varPsi(Fx_{0}^{r_{\mathbb{X}}}dx_{0})=F^{\rm deh}x_{0}^{2r_{\mathbb{X}}}dx_{0}=fx_{0}^{2r_{\mathbb{X}}}dx_{0}$. For an element of the form $x_{0}^{2r_{\mathbb{X}}+1}\otimes fdg$ with $f,g\in S$, we let $F,G\in R_{r_{\mathbb{X}}}$ be preimages of $fx_{0}^{r_{\mathbb{X}}}$ and $gx_{0}^{r_{\mathbb{X}}}$ under the isomorphism $R_{r_{\mathbb{X}}}\cong Sx_{0}^{r_{\mathbb{X}}}$, respectively. Then we calculate

[TABLE]

Since we know already that the second summand is in the image of $\varPsi$, also the first one is and the proof is complete.

Finally, claim (c) is an immediate consequence of (b) and Proposition 3.7.c, and claim (d) follows from Proposition 3.5.a and the fact that $\varPhi$ is identified with the canonical map from ${\Omega^{1}_{R/K}}$ to its localization in $x_{0}$. ∎

4. The Euler Form and the Koszul Submodule

Continuing to use the notation introduced in the preceding sections, let $K$ be a perfect field, let $P=K[X_{0},\dots,X_{n}]$, and let ${\mathbb{X}}$ be a 0-dimensional subscheme of $\mathbb{P}^{n}$ with homogeneous coordinate ring $R=P/I_{\mathbb{X}}$. For $i=0,\dots,n$, we denote the image of $X_{i}$ in $R$ by $x_{i}$, and we assume that $x_{0}$ is a non-zerodivisor in $R$.

Recall that the $K$-linear map $\delta:\;R\longrightarrow R$ given by $\delta(f)=i\cdot f$ for $i\geq 0$ and $f\in R_{i}$ is a derivation of $R$ (cf. [14], Example 1.5). It is called the Euler derivation of $R/K$.

Definition 4.1**.**

By the universal property of ${\Omega^{1}_{R/K}}$ (cf. [14], Theorem 1.19), the Euler derivation $\delta:R\longrightarrow R$ gives rise to an $R$-linear map ${\varepsilon}:\;{\Omega^{1}_{R/K}}\longrightarrow R$ such that $\delta={\varepsilon}\circ d_{R/K}$. This map will be called the Euler form on ${\Omega^{1}_{R/K}}$.

For $i=0,\dots,n$, we obtain ${\varepsilon}(dx_{i})=\delta(x_{i})=x_{i}$. Hence we have $\mathop{\rm Im}\nolimits({\varepsilon})=\mathfrak{m}$, where $\mathfrak{m}=\langle x_{0},...,x_{n}\rangle$ is the homogeneous maximal ideal of $R$, and we will view ${\varepsilon}$ as the surjective $R$-linear map

[TABLE]

The following proposition collects some properties of the kernel of the Euler form.

Proposition 4.2**.**

Let ${\varepsilon}:\;{\Omega^{1}_{R/K}}\longrightarrow\mathfrak{m}$ be the Euler form on ${\Omega^{1}_{R/K}}$.

1. (a)

We have $T{\Omega^{1}_{R/K}}\subseteq\mathop{\rm Ker}\nolimits({\varepsilon})$. 2. (b)

We have $\mathop{\rm Ann}\nolimits_{R}(dx_{0})=\{0\}$ and $\mathop{\rm Ker}\nolimits({\varepsilon})\cap Rdx_{0}=\{0\}$. 3. (c)

For $i\geq 2r_{\mathbb{X}}+1$, we have $\mathop{\rm HF}\nolimits_{\mathop{\rm Ker}\nolimits({\varepsilon})}(i)=\dim_{K}({\Omega^{1}_{S/K}})$.

Proof.

First we show (a). Let $w\in T{\Omega^{1}_{R/K}}$. By Proposition 3.5.a, there exists a number $i\geq 1$ such that $x_{0}^{i}\,w=0$. Since the set $\{dx_{0},\dots,dx_{n}\}$ generates ${\Omega^{1}_{R/K}}$, we can write $w=f_{0}dx_{0}+\cdots+f_{n}dx_{n}$ with $f_{0},\dots,f_{n}\in R$ and calculate $0={\varepsilon}(x_{0}^{i}w)={\varepsilon}(x_{0}^{i}f_{0}dx_{0}+\cdots+x_{0}^{i}f_{n}dx_{n})=x_{0}^{i}f_{0}x_{0}+\cdots+x_{0}^{i}f_{n}x_{n}$. Now we use the fact that $x_{0}$ is a non-zerodivisor for $R$ and conclude that ${\varepsilon}(w)=f_{0}x_{0}+\cdots+f_{n}x_{n}=0$.

To prove (b), we let $f\in R$ such that $fdx_{0}=0$. Then $0={\varepsilon}(fdx_{0})=f\,{\varepsilon}(dx_{0})=f\,x_{0}$ implies $f=0$. This shows $\mathop{\rm Ann}\nolimits_{R}(dx_{0})=\{0\}$. Next, let $f\in R$ such that $fdx_{0}\in\mathop{\rm Ker}\nolimits({\varepsilon})$. Then ${\varepsilon}(fdx_{0})=f\,x_{0}=0$ implies $f=0$, and hence $fdx_{0}=0$.

Finally, claim (c) follows from Proposition 3.8.c and $\dim_{K}(\mathfrak{m}_{i})=\dim_{K}(R_{i})=\deg({\mathbb{X}})$ for $i\geq r_{\mathbb{X}}+1$. ∎

Given a linear form on a module, one can construct an associated Koszul complex.

Definition 4.3**.**

As above, let $R$ be the homogeneous coordinate ring of a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$, and let ${\varepsilon}:\,{\Omega^{1}_{R/K}}\longrightarrow\mathfrak{m}$ be the Euler form.

1. (a)

The complex of $R$-modules

[TABLE]

is called the Euler-Koszul complex of $R$ (or of ${\mathbb{X}}$). Here ${\varepsilon}^{(i)}$ is given by ${\varepsilon}^{(i)}(w_{1}\wedge\cdots\wedge w_{i})=\sum_{j=1}^{i}(-1)^{j-1}{\varepsilon}(w_{j})\,w_{1}\wedge\cdots\wedge\widehat{w_{j}}\wedge\cdots\wedge w_{i}$ for $i\geq 2$ and $w_{1},\dots,w_{i}\in{\Omega^{1}_{R/K}}$. 2. (b)

The image of the $R$-linear map ${\varepsilon}^{(2)}:\,\Lambda^{2}{\Omega^{1}_{R/K}}\longrightarrow{\Omega^{1}_{R/K}}$, i.e., the $R$-submodule $U_{R/K}=\langle x_{i}dx_{j}-x_{j}dx_{i}\mid 0\leq i<j\leq n\rangle$ of ${\Omega^{1}_{R/K}}$ is called the Koszul submodule of ${\Omega^{1}_{R/K}}$.

It is clear that the Koszul submodule of ${\Omega^{1}_{R/K}}$ is contained in the kernel of the Euler form, and usually both submodules are equal. A more precise description of their relationship is given as follows.

Proposition 4.4**.**

Let ${\varepsilon}:\,{\Omega^{1}_{R/K}}\longrightarrow\mathfrak{m}$ be the Euler form and $U_{R/K}\subseteq{\Omega^{1}_{R/K}}$ the Koszul submodule of ${\Omega^{1}_{R/K}}$.

1. (a)

We have $U_{R/K}\subseteq\mathop{\rm Ker}\nolimits({\varepsilon})$, and this is an equality if we have $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)\geq 2r_{\mathbb{X}}+1$. 2. (b)

Assume that $\mathop{\rm char}\nolimits(K)=p>0$. If $i\geq 1$ satisfies $p\nmid i$ or if $i\geq 2r_{\mathbb{X}}+1$ then we have $(U_{R/K})_{i}=\mathop{\rm Ker}\nolimits({\varepsilon})_{i}$. 3. (c)

$\mathop{\rm Ker}\nolimits({\varepsilon})=\{w\in{\Omega^{1}_{R/K}}\mid x_{0}w\in U_{R/K}\}$.

Proof.

First of all, note that ${\varepsilon}(x_{i}dx_{j}-x_{j}dx_{i})=x_{i}x_{j}-x_{j}x_{i}=0$ for $0\leq i<j\leq n$. Thus we have $U_{R/K}\subseteq\mathop{\rm Ker}\nolimits({\varepsilon})$. Moreover, notice that both $U_{R/K}$ and $\mathop{\rm Ker}\nolimits({\varepsilon})$ are graded $R$-submodules of ${\Omega^{1}_{R/K}}$.

Now let $i\geq 1$. For a homogeneous element $w\in({\Omega^{1}_{R/K}})_{i}$, it is shown in the proof of [7], Proposition 2.4, that $({\varepsilon}^{(2)}\circ d+d\circ{\varepsilon})(w)=i\,w$. Therefore every element $w\in\mathop{\rm Ker}\nolimits({\varepsilon})_{i}$ satisfies ${\varepsilon}^{(2)}(dw)=i\,w$. Hence if $i$ is a unit in $K$ then we have $w={\varepsilon}^{(2)}(\frac{1}{i}\,dw)\in U_{R/K}$. This proves the case $\mathop{\rm char}\nolimits(K)=0$ in (a) and the first case $p\nmid i$ in (b).

Next we consider the second case in (a) and the second case in (b) simultaneously. By what we have already seen, we may assume that $\mathop{\rm char}\nolimits(K)=p>0$ and that $p$ divides the degree $i\geq 2r_{\mathbb{X}}+1$ of an element $w\in\mathop{\rm Ker}\nolimits({\varepsilon})_{i}\subseteq({\Omega^{1}_{R/K}})_{i}$. We use the fact that $\mu_{x_{0}}:\;({\Omega^{1}_{R/K}})_{i-1}\longrightarrow({\Omega^{1}_{R/K}})_{i}$ is surjective by Proposition 3.4.b and write $w=x_{0}\tilde{w}$ with $\tilde{w}\in({\Omega^{1}_{R/K}})_{i-1}$. Here $0={\varepsilon}(w)=x_{0}{\varepsilon}(\tilde{w})$ implies $\tilde{w}\in\mathop{\rm Ker}\nolimits({\varepsilon})_{i-1}$. Since $p\nmid(i-1)$, we have $\tilde{w}=\frac{1}{i-1}\,{\varepsilon}^{(2)}(d\tilde{w})\in U_{R/K}$, and thus $w=x_{0}\tilde{w}\in U_{R/K}$.

It remains to prove (c). By (b), the only case where the containment $(U_{R/K})_{i}\subseteq\mathop{\rm Ker}\nolimits({\varepsilon})_{i}$ can be strict for some $i\geq 1$ is when $\mathop{\rm char}\nolimits(K)=p>0$ and $p\mid i$. Then $p\nmid(i+1)$ implies $x_{0}w\in(U_{R/K})_{i+1}$. This proves the inclusion $\subseteq$ of the claim. Conversely, if $x_{0}w\in U_{R/K}\subseteq\mathop{\rm Ker}\nolimits({\varepsilon})$, then $0={\varepsilon}(x_{0}w)=x_{0}{\varepsilon}(w)$ shows ${\varepsilon}(w)=0$. ∎

Let us also try to describe those elements in $\mathop{\rm Ker}\nolimits({\varepsilon})$ which are not in the Koszul submodule of ${\Omega^{1}_{R/K}}$ as explicitly as possible. The following construction will be useful.

Remark 4.5**.**

Let $P=K[X_{0},\dots,X_{n}]$ and $\mathfrak{M}=\langle X_{0},\dots,X_{n}\rangle$ its homogeneous maximal ideal. Then every polynomial $F\in\mathfrak{M}$ has a unique decomposition of the form $F=F_{0}X_{0}+\cdots+F_{n}X_{n}$ where $F_{i}\in K[X_{i},\dots,X_{n}]$. In other words, we have $F_{n}\in K[X_{n}]$, $F_{n-1}\in K[X_{n-1},X_{n}]$, etc. We call it the triangular decomposition of $F$.

If we denote the tuple $(F_{0},\dots,F_{n})$ by $\vartheta(F)$, then the map $\vartheta:\,\mathfrak{M}\longrightarrow P^{n+1}$ is $K$-linear and yields a section of the epimorphism $\xi:\,P^{n+1}\longrightarrow\mathfrak{M}$ given by $\xi(e_{i})=X_{i}$ for $i=0,\dots,n$. In other words, the map $\vartheta$ splits the short exact sequence

[TABLE]

Recall that $\mathop{\rm Ker}\nolimits(\xi)=\langle X_{i}e_{j}-X_{j}e_{i}\mid 0\leq i<j\leq n\rangle$, as $(X_{0},\dots,X_{n})$ is a $P$-regular sequence.

With the help of the triangular decomposition, we obtain the following description of the kernel of the Euler form.

Proposition 4.6**.**

**(The Kernel of the Euler Form)

**Let $R=P/I_{\mathbb{X}}$ be the homogeneous coordinate ring of a 0-dimensional subscheme ${\mathbb{X}}$ of $\mathbb{P}^{n}$, and let ${\varepsilon}:\,{\Omega^{1}_{R/K}}\longrightarrow\mathfrak{m}$ be the Euler form.

1. (a)

Let $\gamma:\,P^{n+1}\longrightarrow{\Omega^{1}_{R/K}}$ be the epimorphism defined by $\gamma(e_{i})=dx_{i}$ for $i=0,\dots,n$, and let $\eta:\,{\Omega^{1}_{R/K}}\longrightarrow{\Omega^{1}_{R/K}}/U_{R/K}$ be the canonical epimorphism. Then the composition $\delta=\eta\circ\gamma\circ\vartheta:\,\mathfrak{M}\longrightarrow{\Omega^{1}_{R/K}}/U_{R/K}$ is a $P$-linear map. 2. (b)

We have $\delta(I_{\mathbb{X}})=\mathop{\rm Ker}\nolimits({\varepsilon})/U_{R/K}$. 3. (c)

Given a homogeneous system of generators $\{G_{1},\dots,G_{r}\}$ of $I_{\mathbb{X}}$, we have $\mathop{\rm Ker}\nolimits({\varepsilon})=U_{R/K}+\langle(\gamma\circ\vartheta)(G_{1}),\dots,(\gamma\circ\vartheta)(G_{r})\rangle$. 4. (d)

Let $m_{\mathbb{X}}=\max\{i\geq 0\mid(I_{\mathbb{X}}/\mathfrak{M}I_{\mathbb{X}})_{i}\neq 0\}$ be the maximal degree of a homogeneous minimal generator of $I_{\mathbb{X}}$. Then we have $(U_{R/K})_{i}=\mathop{\rm Ker}\nolimits({\varepsilon})_{i}$ for all $i>m_{\mathbb{X}}$.

Proof.

First we prove (a). As a composition of $K$-linear maps, the map $\delta$ is clearly $K$-linear. Now let $F\in\mathfrak{M}$, and let $F=F_{0}X_{0}+\cdots+F_{n}X_{n}$ be the triangular decomposition of $F$, where $F_{i}\in K[X_{i},\dots,X_{n}]$.

Given a term $T\in K[X_{0},\dots,X_{n}]$, we let $k$ be the least index of an indeterminate dividing $T$, and we write $T=X_{k}\cdot T^{\prime}$ with $T^{\prime}\in\mathbb{T}^{n+1}$. Then we have

[TABLE]

and this is clearly the triangular decomposition of $T\,F$. Hence we get

[TABLE]

where $f_{i}$ denotes the residue class of $F_{i}$ in $R$ for $i=0,\dots,n$ and $t$ (resp. $t^{\prime}$) the residue class of $T$ (resp. $T^{\prime}$). On the other hand, we calculate

[TABLE]

Consequently, the map $\delta$ is $P$-linear.

To show (b), we let $F\in I_{\mathbb{X}}$ and write the triangular decomposition of $F$ as above as $F=F_{0}X_{0}+\cdots+F_{n}X_{n}$. Then we get $\delta(F)=f_{0}dx_{0}+\cdots+f_{n}dx_{n}+U_{R/K}$ and for every element $u\in U_{R/K}$ we obtain ${\varepsilon}(f_{0}dx_{0}+\cdots+f_{n}dx_{n}+u)=f_{0}x_{0}+\cdots+f_{n}x_{n}=\allowbreak F+I_{\mathbb{X}}=0$.

For the reverse inclusion, let $w\in\mathop{\rm Ker}\nolimits({\varepsilon})$, and write $w=g_{0}dx_{0}+\cdots+g_{n}dx_{n}$. Represent $g_{i}\in R$ by polynomials $G_{i}\in P$, and let $F=G_{0}X_{0}+\cdots+G_{n}X_{n}$. Since ${\varepsilon}(w)=g_{0}x_{0}+\cdots+g_{n}x_{n}=0$, we have $F\in I_{\mathbb{X}}$. Now let $F=F_{0}X_{0}+\cdots+F_{n}X_{n}$ be the triangular decomposition of $F$. Then $(F_{0}-G_{0},\dots,F_{n}-G_{n})$ is a syzygy of $(X_{0},\dots,X_{n})$, and thus contained in $V:=\langle X_{i}e_{j}-X_{j}e_{i}\mid 0\leq i<j\leq n\rangle$. This implies

[TABLE]

and therefore $\delta(F)=g_{0}dx_{0}+\cdots+g_{n}dx_{n}+u+U_{R/K}=w+U_{R/K}$, where $u=\gamma(v)\in U_{R/K}$.

Since claim (c) follows immediately from (a) and (b), it remains to prove (d). If we have $(U_{R/K})_{m_{\mathbb{X}}}=\mathop{\rm Ker}\nolimits({\varepsilon})_{m_{\mathbb{X}}}$ then (c) shows that the claim holds. So, the only remaining case is $\mathop{\rm char}\nolimits(K)=p>0$ and $p\mid m_{\mathbb{X}}$. In this case we have $(U_{R/K})_{m_{\mathbb{X}}+1}=\mathop{\rm Ker}\nolimits({\varepsilon})_{m_{\mathbb{X}}+1}$ by Proposition 4.4.a,b. Hence we see that $f\cdot(\gamma\circ\vartheta)(G_{i})\in U_{R/K}$ for all $i\in\{1,\dots,r\}$ and all homogeneous elements $f\in R$ such that $\deg(f)+\deg(G_{j})=m_{\mathbb{X}}+1$. The claim follows from (c) and this observation. ∎

The next example shows that the torsion submodule of ${\Omega^{1}_{R/K}}$ is, in general, not contained in its Koszul submodule.

Example 4.7**.**

Let $K=\mathbb{F}_{3}$, let $P=K[x_{0},x_{1},x_{3}]$, and let ${\mathbb{X}}$ be the 0-dimensional complete intersection scheme in $\mathbb{P}^{2}$ defined by the homogeneous ideal $I_{\mathbb{X}}=\langle G_{1},G_{2}\rangle$, where $G_{1}=X_{1}^{2}+X_{2}^{2}$ and $G_{2}=X_{0}X_{1}^{2}+X_{1}^{3}+X_{2}^{3}$.

It is straightforward to calculate $\mathop{\rm HF}\nolimits_{\mathbb{X}}:\ 1\ 3\ 5\ 6\ 6\cdots$ which shows $r_{\mathbb{X}}=3$, as well as $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}:\ 0\ 3\ 8\ 11\ 10\ 10\cdots$ which implies $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})=4$. Moreover, we compute

[TABLE]

Consequently, the Koszul submodule $U_{R/K}$ is a proper submodule of $\mathop{\rm Ker}\nolimits({\varepsilon})$.

The triangular decomposition of $G_{2}\in I_{\mathbb{X}}$ is $G_{2}=(X_{0}X_{1}+X_{1}^{2})\,X_{1}+(X_{2}^{2})\,X_{2}$. Therefore we may check that

[TABLE]

Furthermore, it turns out that $T{\Omega^{1}_{R/K}}=K\cdot(\gamma\circ\vartheta)(G_{2})$, and hence the torsion submodule of ${\Omega^{1}_{R/K}}$ is in general not contained in its Koszul submodule. Notice that for reduced schemes we have $T{\Omega^{1}_{R/K}}=\mathop{\rm Ker}\nolimits({\varepsilon})$ (see [4], Proposition 2.1). However, even in this case $U_{R/K}\subsetneq\mathop{\rm Ker}\nolimits({\varepsilon})$ may occur, for instance, when ${\mathbb{X}}$ is a set of five $\mathbb{F}_{3}$-rational points on a non-singular conic in $\mathbb{P}^{2}$.

5. Kähler Differential $m$-Forms

Continuing to use the above notation and assumptions, we want to introduce and study the higher exterior powers of ${\Omega^{1}_{R/K}}$ next.

Definition 5.1**.**

Let $m\geq 0$, and let $R$ be the homogeneous coordinate ring of a zero-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$.

1. (a)

The $m$-th exterior power ${\Omega^{m}_{R/K}}:=\Lambda^{m}_{R}{\Omega^{1}_{R/K}}$ is called the module of Kähler differential $m$-forms of $R$. 2. (b)

The graded $R$-algebra $\Omega^{\bullet}_{R/K}:=\bigoplus_{m\geq 0}{\Omega^{m}_{R/K}}$ is called the Kähler differential algebra of $R/K$.

Notice that $\Omega^{0}_{R/K}=R$ and that we can define $P$-modules $\Omega^{m}_{P/K}$ analogously. For every $m\geq 0$, the module of Kähler differential $m$-forms is a finitely generated graded $R$-module. To calculate a presentation of this $R$-modules, we can use the following results.

Proposition 5.2**.**

Let $R$ be the homogeneous coordinate ring of a zero-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$.

1. (a)

For $m\geq n+2$, we have ${\Omega^{m}_{R/K}}=\langle 0\rangle$. 2. (b)

For $m=0,\dots,n+1$, the $P$-module $\Omega^{m}_{P/K}$ is a free $P$-module of rank $\binom{n+1}{m}$ with basis $\{dX_{i_{1}}\wedge\cdots\wedge dX_{i_{m}}\mid 0\leq i_{1}<\cdots<i_{m}\leq n\}$. 3. (c)

For $m\geq 2$, the module of Kähler differential $m$-forms satisfies

[TABLE]

Here $dI_{\mathbb{X}}$ is the $P$-submodule of $\Omega^{1}_{P/K}$ generated by $\{d_{P/K}f\mid f\in I_{\mathbb{X}}\}$. 4. (d)

Let $I_{\mathbb{X}}=\langle f_{1},\dots,f_{k}\rangle$ for some $f_{1},\dots,f_{k}\in P$. Then we have

[TABLE]

Proof.

Claim (a) is clearly true, claim (b) follows from [14], Example 2.5.d, and claim (c) follows from ibid., Proposition 4.12.

To show (d), we note that $I_{{\mathbb{X}}}\Omega^{1}_{P/K}$ is contained in $dI_{\mathbb{X}}$ because, for $f\in I_{\mathbb{X}}$ and $g\in P$, we have $f\,dg=d(fg)-g\,df\in dI_{\mathbb{X}}$. This implies the inclusion $\supseteq$. Conversely, let $g\in I_{\mathbb{X}}$ and write $g=h_{1}f_{1}+\cdots+h_{k}f_{k}$ with $h_{1},\dots,h_{k}\in P$. Then $dg=h_{1}df_{1}+\cdots+h_{k}df_{k}+f_{1}dh_{1}+\cdots+f_{k}dh_{k}$ is contained in the right-hand side, and the desired equality is proved. ∎

Since the modules of Kähler differential $m$-forms are finitely generated graded $R$-modules, we can define their Hilbert function, Hilbert polynomial and regularity index as usual.

Definition 5.3**.**

Let $R$ be the homogeneous coordinate ring of a zero-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$, and let $m\geq 1$.

1. (a)

The map $\mathop{\rm HF}\nolimits_{\Omega^{m}_{R/K}}:\;{\mathbb{Z}}\longrightarrow{\mathbb{Z}}$ given by $\mathop{\rm HF}\nolimits_{\Omega^{m}_{R/K}}(i)=\dim_{K}({\Omega^{m}_{R/K}})_{i}$ for every $i\in{\mathbb{Z}}$ is called the Hilbert function of ${\Omega^{m}_{R/K}}$. 2. (b)

The integer $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=\mathop{\rm HF}\nolimits_{{\Omega^{m}_{R/K}}}(i)$ for $i\gg 0$ is called the Hilbert polynomial of ${\Omega^{m}_{R/K}}$. 3. (c)

The number $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}}):=\min\{i\in{\mathbb{Z}}\mid\mathop{\rm HF}\nolimits_{{\Omega^{m}_{R/K}}}(j)=\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})\hbox{\ \rm for\ }j\geq i\}$ is called the regularity index of ${\Omega^{m}_{R/K}}$.

Our goal in this section is to determine the Hilbert polynomial of the module ${\Omega^{m}_{R/K}}$ and to find a sharp bound for its regularity index. As in the preceding section, our method to reach these goals is to compare it to the module of Kähler differential $m$-forms of the truncated integral closure $\widetilde{R}$ of $R$ in $Q^{h}(R)$. The modules ${\Omega^{m}_{\widetilde{R}/K}}$ can be computed and compared to ${\Omega^{m}_{R/K}}$ as follows.

Proposition 5.4**.**

Let $R$ be the homogeneous coordinate ring of a zero-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$, let $\widetilde{R}$ be the truncated integral closure of $R$ in $Q^{h}(R)$, let $S=R/\langle x_{0}-1\rangle$ be the affine coordinate ring of ${\mathbb{X}}$ in $\mathbb{A}^{n}\cong D_{+}(x_{0})$, and let $m\geq 1$.

1. (a)

The isomorphism $\widetilde{R}\cong S[x_{0}]$ induces an isomorphism

[TABLE] 2. (b)

Let $\varPhi:\;{\Omega^{1}_{R/K}}\longrightarrow{\Omega^{1}_{\widetilde{R}/K}}$ be the canonical $R$-linear map induced by $R\subseteq\widetilde{R}$. Then the $K$-linear map $(\Lambda^{m}_{R}\varPhi)_{i}:\;({\Omega^{m}_{R/K}})_{i}\longrightarrow({\Omega^{m}_{\widetilde{R}/K}})_{i}$ is an isomorphism for every $i\geq 2r_{\mathbb{X}}+m$. 3. (c)

We have $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=\mathop{\rm HP}\nolimits({\Omega^{m}_{\widetilde{R}/K}})=\dim_{K}(\Omega^{m}_{S/K})+\dim_{K}(\Omega^{m-1}_{S/K})$. 4. (d)

We have $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}})\leq 2r_{\mathbb{X}}+m$. 5. (e)

If $\mathop{\rm char}\nolimits(K)=0$ or if $\mathop{\rm char}\nolimits(K)=p>0$ and $p\nmid(2r_{\mathbb{X}}+n+1)$, then we even have $\mathop{\rm ri}\nolimits(\Omega^{n+1}_{R/K})\leq 2r_{\mathbb{X}}+n$.

Proof.

Claim (a) follows from

[TABLE]

To prove (b), we first show that $\widetilde{\varPhi}:=(\Lambda^{m}_{R}\varPhi)_{i}$ is surjective for $i\geq 2r_{\mathbb{X}}+m$. Notice that $K[x_{0}]\otimes_{K}\Omega^{m}_{S/K}\cong S[x_{0}]\otimes_{S}\Omega^{m}_{S/K}$. Using (a), it suffices to show that

[TABLE]

where $f,g\in S$ and $\{i_{1},...,i_{m-1}\},\{j_{1},...,j_{m}\}$ are subsets of $\{1,...,n\}$ of lengths $m-1$ and $m$. Let $F,G\in R_{r_{\mathbb{X}}}$ be preimages of $fx_{0}^{r_{\mathbb{X}}}$ and $gx_{0}^{r_{\mathbb{X}}}$ under the isomorphism $R_{r_{\mathbb{X}}}\cong Sx_{0}^{r_{\mathbb{X}}}$ in Proposition 2.6.b, respectively. Then we have

[TABLE]

Also

[TABLE]

where $\widehat{dx_{j_{k}}}$ means that $dx_{j_{k}}$ is omitted. In the last equality, the first summand is in the image of $\widetilde{\varPhi}$, and so is the second one. Therefore $\widetilde{\varPhi}$ is surjective. Next, we show that $\dim_{K}({\Omega^{m}_{R/K}})_{i}=\dim_{K}({\Omega^{m}_{\widetilde{R}/K}})_{i}$ for $i\geq 2r_{\mathbb{X}}+m$, and hence $\widetilde{\varPhi}$ is bijective. Set $W:=\mathop{\rm Im}\nolimits(\varPhi)\subseteq{\Omega^{1}_{\widetilde{R}/K}}$. Then $\varPhi:{\Omega^{1}_{R/K}}\rightarrow W$ is an epimorphism of graded $R$-modules. By [16], Satz 85.8, we have the exact sequence of $K$-vector spaces

[TABLE]

It follows from Proposition 3.8.d that $T{\Omega^{1}_{R/K}}=\mathop{\rm Ker}\nolimits(\varPhi)$. By Proposition 3.5, $\mathop{\rm Ker}\nolimits(\varPhi)_{k}=\{0\}$ for all $k\geq 2r_{\mathbb{X}}+1$, and so $(\mathop{\rm Ker}\nolimits(\varPhi)\wedge_{R}\Omega^{m-1}_{R/K})_{i}=\{0\}$. It follows that $\dim_{K}(\Omega^{m}_{R/K})_{i}=\dim_{K}(\Lambda_{R}^{m}(W))_{i}$. On the other hand, it is clearly true that $(\Lambda_{R}^{m}(W))_{i}\subseteq({\Omega^{m}_{\widetilde{R}/K}})_{i}$. Hence the surjectivity of $\widetilde{\varPhi}$ implies that $(\Lambda_{R}^{m}(W))_{i}=({\Omega^{m}_{\widetilde{R}/K}})_{i}$, and subsequently we get $\dim_{K}({\Omega^{m}_{R/K}})_{i}=\dim_{K}({\Omega^{m}_{\widetilde{R}/K}})_{i}$, as desired.

Claims (c) and (d) follow from (a) and (b). It remains to prove (e). Consider the Euler-Koszul complex of $R$

[TABLE]

This sequence is an exact sequence of graded $R$-modules in characteristic zero by [7], Proposition 2.4. If $\mathop{\rm char}\nolimits(K)=p>0$ and $p\nmid(2r_{\mathbb{X}}+n+1)$ then the proof of [7], Proposition 2.4, shows that the map ${\varepsilon}^{(n+1)}:\;(\Omega^{n+1}_{R/K})_{d}\longrightarrow(\Omega^{n}_{R/K})_{d}$ is also injective in degree $d=2r_{\mathbb{X}}+n+1$. In both cases we see that the claim follows from (d). ∎

The above upper bound for the regularity index of ${\Omega^{m}_{R/K}}$ is sharp, as the following example shows.

Example 5.5**.**

Let ${\mathbb{X}}=\{p_{1},...,p_{n+1}\}$ be a set of $n+1$ distinct $K$-rational points in $\mathbb{P}^{n}$ which are in general position. We claim that the regularity index of ${\Omega^{m}_{R/K}}$ satisfies $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}})=m+2$ for $1\leq m\leq n$ and $\mathop{\rm ri}\nolimits(\Omega^{n+1}_{R/K})=n+2$.

First of all, it is clearly true that $\mathop{\rm HF}\nolimits_{\mathbb{X}}:\ 1\ n+1\ n+1\cdots$ and $r_{\mathbb{X}}=1$. After a linear change of coordinates, we may assume that $p_{1}=(1:0:\dots:0)$, $\dots$, $p_{n+1}=(0:\dots:0:1)$. It follows that $I_{\mathbb{X}}=\langle X_{i}X_{j}\mid 0\leq i<j\leq n\rangle$.

Now let ${\mathbb{Y}}$ be the 0-dimensional subscheme of $\mathbb{P}^{n}$ defined by the saturated homogeneous ideal $I_{\mathbb{Y}}=\bigcap_{j=1}^{n+1}I_{p_{j}}^{2}$. Then we have $\mathop{\rm HF}\nolimits_{\mathbb{Y}}:1\;\binom{n+1}{n}\;\binom{n+2}{n}\;(n+1)^{2}\;(n+1)^{2}\cdots$ and $r_{\mathbb{Y}}=3$. From [14], Proposition 4.17, we obtain an exact sequence of graded $R$-modules

[TABLE]

and we deduce that $\mathop{\rm ri}\nolimits({\Omega^{1}_{R/K}})=3$. Moreover, $\Omega^{n+1}_{R/K}\cong(P/\langle X_{0},...,X_{n}\rangle)(-n-1)$, implies $\mathop{\rm HF}\nolimits_{\Omega^{n+1}_{R/K}}:\ 0\dots 0\ 1\ 0\ 0\dots$ and $\mathop{\rm ri}\nolimits(\Omega^{n+1}_{R/K})=n+2$.

Altogether, by Proposition 6.1, it suffices to show $\mathop{\rm HF}\nolimits_{\Omega^{m}_{R/K}}(m+1)\neq 0$ for all $m\in\{2,\dots,n\}$. Observe that the element $w=dx_{0}\wedge dx_{1}\wedge\cdots\wedge dx_{n}$ in $(\Omega^{n+1}_{R/K})_{n+1}$ is non-zero. Since $w=d(x_{0}\,dx_{1}\wedge\cdots\wedge dx_{n})\neq 0$, also the element $\tilde{w}=x_{0}\,dx_{1}\wedge\cdots\wedge dx_{n}$ in $\Omega^{n}_{R/K}$ is non-zero. This implies $\mathop{\rm HF}\nolimits_{\Omega^{n}_{R/K}}(n+1)\neq 0$.

As $\tilde{w}=(x_{0}\,dx_{1}\wedge\cdots\wedge dx_{m})\wedge(dx_{m+1}\wedge\cdots\wedge dx_{n})$, we obtain non-zero elements $x_{0}\,dx_{1}\wedge\cdots\wedge dx_{m}$ in ${\Omega^{m}_{R/K}}$ for all $m\in\{2,\dots,n\}$, and the claim is proved.

Remark 5.6**.**

Let ${\mathbb{X}}\subseteq\mathbb{P}^{n}$ be a set of $t$ distinct $K$-rational points. If $t\leq n$ then $\Omega^{n+1}_{R/K}=0$. When $t>n$ and $r:=\mathop{\rm ri}\nolimits(\Omega^{n+1}_{R/K})>0$, we get a lower bound for the regularity index of ${\Omega^{m}_{R/K}}$ given by $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}})\geq r+m-n$ for $2\leq m\leq n$. In particular, if $r=2r_{\mathbb{X}}+n$ then $\mathop{\rm ri}\nolimits({\Omega^{m}_{R/K}})=2r_{\mathbb{X}}+m$ for $2\leq m\leq n$.

6. Kähler Differentials of Curvilinear Schemes

As before, we let ${\mathbb{X}}$ be a 0-dimensional scheme in $\mathbb{P}^{n}$ over a perfect field $K$, we let $R=P/I_{\mathbb{X}}$ be the homogeneous coordinate ring of ${\mathbb{X}}$, we assume that $x_{0}\in R_{1}$ is a non-zero divisor of $R$, and we let $S=R/\langle x_{0}-1\rangle$ be the affine coordinate ring of ${\mathbb{X}}$ in $\mathbb{A}^{n}\cong D_{+}(x_{0})$. In this section we want to connect local properties of ${\mathbb{X}}$ to the structure of the modules of Kähler differential $m$-forms of $R/K$ for various $m\geq 1$.

Our first proposition characterizes smooth schemes.

Proposition 6.1**.**

In the above setting, the following conditions are equivalent.

1. (a)

The scheme ${\mathbb{X}}$ is smooth. 2. (b)

We have ${\Omega^{1}_{S/K}}=\langle 0\rangle$. 3. (c)

We have $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=\deg({\mathbb{X}})$ and $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=0$ for $m\geq 2$.

Proof.

The equivalence of (a) and (b) follows from [14], Corollary 6.9, and the equivalence of (b) and (c) follows from Proposition 3.8. ∎

Recall that a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{A}^{n}$ has a finite support and that $S$ is a 0-dimensional affine $K$-algebra. Let $\mathfrak{q}_{1},\dots,\mathfrak{q}_{t}$ be the primary components of $\langle 0\rangle$ in $S$. Then the Chinese Remainder Theorem yields an isomorphism $S\cong S/\mathfrak{q}_{1}\times\cdots\times S/\mathfrak{q}_{t}$ where the rings $\mathcal{O}_{i}=S/\mathfrak{q}_{i}$ are 0-dimensional local rings.

Definition 6.2**.**

In the above setting, let $(T,\mathfrak{n})$ be a 0-dimensional local ring with residue class field $L=T/\mathfrak{n}$.

1. (a)

The ring $T$ is called (weakly) curvilinear if its maximal ideal $\mathfrak{n}$ is generated by one element. 2. (b)

The scheme ${\mathbb{X}}$ is called (weakly) curvilinear if its local rings $\mathcal{O}_{1},\dots,\mathcal{O}_{t}$ are all (weakly) curvilinear.

The condition to be weakly curvilinear can be characterized as follows.

Proposition 6.3**.**

Let $(T,\mathfrak{n})$ be a 0-dimensional local Noetherian ring with residue class field $L=T/\mathfrak{n}$, and let $\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)$ be the associated graded ring of $T$. Then the following conditions are equivalent.

1. (a)

The ring $T$ is weakly curvilinear. 2. (b)

The associated graded ring of $T$ is of the form $\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)\cong L[z]/\langle z^{m}\rangle$ for some $m\geq 1$.

Proof.

First we prove that (a) implies (b). Notice that the ring $T$ is Artinian, and let $\mathfrak{n}=\langle a\rangle$ for some $a\in T$. Then every proper ideal of $T$ is a power of $\mathfrak{n}$, and therefore principal. In particular, there is a smallest number $m\geq 1$ such that $\langle 0\rangle=\mathfrak{n}^{m}=\langle a^{m}\rangle$. Thus the $L$-algebra epimorphism $L[z]\rightarrow\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)$ defined by $z\mapsto a+\mathfrak{n}^{2}$ yields an isomorphism $\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)\cong L[z]/\langle z^{m}\rangle$, as desired.

Conversely, if $\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)\cong L[z]/\langle z^{m}\rangle$, then $\dim_{L}(\mathfrak{n}/\mathfrak{n}^{2})=1$, and hence $\mathfrak{n}$ is principal by Nakayama’s Lemma. So, the ring $T$ is weakly curvilinear. ∎

Let us also compare the above definition of a weakly curvilinear 0-dimensional scheme to the one given in [13].

Remark 6.4**.**

In [13], Definition 4.3.1, a 0-dimensional local affine $K$-algebra $T$ is called weakly curvilinear if it is of the form $T\cong K[z]/\langle p(z)^{m}\rangle$, where $p(z)$ is an irreducible polynomial in $K[z]$ and $m\geq 1$, and it is called curvilinear if $T\cong K[z]/\langle z^{m}\rangle$ for some $m\geq 1$.

For a ring $T$ which is weakly curvilinear in this sense, its maximal ideal $\langle p(\bar{z})\rangle$ is a principal ideal. Therefore Definition 6.2 generalizes the definition in [13]. Since the definition of a weakly curvilinear 0-dimensional scheme uses the local rings of its affine coordinate ring, also this definition generalizes the one in [13].

Moreover, note that in the case of an algebraically closed field $K$, the maximal ideal of a local ring $\mathcal{O}_{i}$ of ${\mathbb{X}}$ is unigenerated if and only if it is the residue class ideal of an ideal of the form $\langle z-a\rangle$ in $K[z]$, where $a\in K$. Thus the ring $\mathcal{O}_{i}$ is isomorphic to a ring of the form $K[z]/\langle z^{m}\rangle$ with $m\geq 1$, in agreement with the classical definition of a curvilinear 0-dimensional scheme. Such schemes appear naturally in algebraic geometry, as for instance the well-known paper [5] demonstrates.

The next example shows that Definition 6.2 is a proper generalization of Definition 4.3.1 in [13].

Example 6.5**.**

Consider the 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{2}$ over the field $\mathbb{Q}$ defined by the ideal $I_{\mathbb{X}}=\langle X_{1}^{2}+X_{0}^{2},(X_{2}^{2}-2X_{0})^{2}\rangle\subseteq\mathbb{Q}[X_{0},X_{1},X_{2}]$. Its affine coordinate ring $S=\mathbb{Q}[X_{1},X_{2}]/\langle X_{1}^{2}+1,(X_{2}^{2}-2)^{2}\rangle$ is a 0-dimensional local ring with maximal ideal $\mathfrak{n}=\langle x_{2}^{2}-2\rangle$ and residue class field $L=S/\mathfrak{n}\cong\mathbb{Q}(i,\sqrt{2})$.

Since the ideal $\mathfrak{n}$ is principal, the ring $S$ is weakly curvilinear in the sense of Definition 6.2. However, the ring $S$ cannot be presented as $S\cong\mathbb{Q}[z]/\langle p(z)^{m}\rangle$ with an irreducible polynomial $p(z)\in\mathbb{Q}[z]$ and $m\geq 1$. Furthermore, note that $S\ncong\mathop{\rm gr}\nolimits_{\mathfrak{n}}(S)$.

The main topic of this section is to use the Hilbert polynomials of the modules of Kähler differential $m$-forms to characterize weakly curvilinear 0-dimensional schemes.

Proposition 6.6**.**

For a 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{n}$ as above, the following conditions are equivalent.

1. (a)

The scheme ${\mathbb{X}}$ is weakly curvilinear, but not smooth. 2. (b)

We have ${\Omega^{1}_{S/K}}\neq\langle 0\rangle$ and $\Omega^{m}_{S/K}=\langle 0\rangle$ for all $m\geq 2$. 3. (c)

We have $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})>\deg({\mathbb{X}})$ and $\mathop{\rm HP}\nolimits(\Omega^{2}_{R/K})=\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})-\deg({\mathbb{X}})$ and $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=0$ for $m>2$.

Proof.

As above, we let $S=R/\langle x_{0}-1\rangle$ be the affine coordinate ring of $R$ and $S\cong\mathcal{O}_{1}\times\cdots\times\mathcal{O}_{t}$ its decomposition into local rings.

First we show that (a) implies (b). Since ${\mathbb{X}}$ is not smooth, we have ${\Omega^{1}_{S/K}}\neq\langle 0\rangle$ by Proposition 6.1. Using the isomorphism $\Omega^{2}_{S/K}\cong\Omega^{2}_{\mathcal{O}_{1}/K}\times\cdots\times\Omega^{2}_{\mathcal{O}_{t}/K}$ (see [14], Proposition 4.7), we see that it suffices to show $\Omega^{2}_{\mathcal{O}_{i}/K}=\langle 0\rangle$ for $i=1,\dots,t$. As ${\mathbb{X}}$ is weakly curvilinear, the maximal ideal of $\mathcal{O}_{i}$ is unigenerated. Then the differential of this element generates the $\mathcal{O}_{i}$-module $\Omega^{1}_{\mathcal{O}_{i}/K}$. It follows that the second and all higher exterior powers of this module are zero.

Conversely, let us show that (b) implies (a). Since ${\Omega^{1}_{S/K}}\neq\langle 0\rangle$, the scheme ${\mathbb{X}}$ is not smooth. For $i=1,\dots,t$, the hypothesis and the above isomorphism yield $\Omega^{2}_{\mathcal{O}_{i}/K}=0$.

Hence we may assume that $S$ is local. Let $\mathfrak{n}$ be the maximal ideal of $S$. Then we have $\mathfrak{n}/\mathfrak{n}^{2}\cong{\Omega^{1}_{S/K}}/\mathfrak{n}\,{\Omega^{1}_{S/K}}$ by [14], Corollary 6.5, and this $S/\mathfrak{n}$-vector space satisfies

[TABLE]

since $\Lambda^{2}_{S}{\Omega^{1}_{S/K}}=\Omega^{2}_{S/K}=\langle 0\rangle$. Consequently, the $S/\mathfrak{n}$-vector space dimension of $\mathfrak{n}/\mathfrak{n}^{2}$ is one, and hence $\mathfrak{n}$ is unigenerated.

The equivalence of (b) and (c) follows from Proposition 5.4. ∎

Based on this characterization, we now derive explicit formulas for the Hilbert polynomials of ${\Omega^{1}_{R/K}}$ and $\Omega^{2}_{R/K}$ in the case of a weakly curvilinear scheme ${\mathbb{X}}$. The following proposition provides the key ingredients.

Proposition 6.7**.**

Let $(T,\mathfrak{n})$ be a 0-dimensional local affine $K$-algebra which is weakly curvilinear. Write $\mathfrak{n}=\langle a\rangle$ with $a\in T$, and let $\nu$ be the index of nilpotency of $a$, i.e., let $\nu=\min\{i\geq 1\mid a^{i}=0\}$. Furthermore, let $L=T/\mathfrak{n}$ be the residue field of $T$ and $\kappa=\dim_{K}(L)$. Then the following claims hold.

1. (a)

We have $\dim_{K}(T)=\dim_{K}(\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T))_{\mathstrut}=\nu\,\kappa$. 2. (b)

We have $\dim_{K}(\Omega^{1}_{\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)/K})=\begin{cases}\nu\,\kappa&\textrm{\ if\ }\mathop{\rm char}\nolimits(K)\mid\nu\\ (\nu-1)\,\kappa&\textrm{\ if\ }\mathop{\rm char}\nolimits(K)\nmid\nu\end{cases}$. 3. (c)

The Noether different $\vartheta_{N}(T/K)$ of the algebra $T/K$ satisfies

[TABLE] 4. (d)

We have $\dim_{K}(\Omega^{1}_{T/K})=\dim_{K}(\Omega^{1}_{\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)/K})$.

Proof.

To show (a), it suffices to note that $\langle 0\rangle=\mathfrak{n}^{\nu}\subseteq\mathfrak{n}^{\nu-1}\subseteq\cdots\subseteq\mathfrak{n}\subseteq T$ yields

[TABLE]

Next we prove (b). Letting $\bar{a}=a+\mathfrak{n}^{2}$, we have $\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)=L[\bar{a}]\cong L[z]/\langle z^{\nu}\rangle$, and therefore $\Omega^{1}_{\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)/K}\cong L[z]dz/(L[z]\cdot z^{\nu}\,dz+L[z]\cdot\nu\,z^{\nu-1}\,dz)$. So, if we set $\nu^{\prime}=\nu$ if $\mathop{\rm char}\nolimits(K)\mid\nu$ and $\nu^{\prime}=\nu-1$ if $\mathop{\rm char}\nolimits(K)\nmid\nu$ then we get $\Omega^{1}_{\mathop{\rm gr}\nolimits_{\mathfrak{n}}(T)/K}\cong L[z]/\langle z^{\nu^{\prime}}\rangle\cong L^{\nu^{\prime}}$, and the claim follows.

The first equality in (c) follows from [14], Proposition 10.18. The remaining claims were shown in [15], §18, Beispiel 3. For the convenience of the reader, we recall the main steps of the proof in our notation. Let $b_{1},\dots,b_{\kappa}\in T$ be elements such that their residue classes $\bar{b}_{1},\dots,\bar{b}_{\kappa}\in L$ form a $K$-basis of $L$. Without loss of generality we may assume that the canonical trace map $\mathop{\rm Tr}\nolimits_{L/K}:\,L\longrightarrow K$ satisfies $\mathop{\rm Tr}\nolimits_{L/K}(\bar{b}_{1})=1$ and $\mathop{\rm Tr}\nolimits_{L/K}(\bar{b}_{i})=0$ for $i=2,\dots,\kappa$. Then the elements in $B=\{a^{i}\,b_{j}\mid i\in\{0,\dots,\nu-1\},\,j\in\{1,\dots,\kappa\}\}$ form a $K$-basis of $T$. Since we have $\mathfrak{n}\cdot\langle a^{i}\rangle=\langle a^{i+1}\rangle$ for all $i\geq 0$, the canonical trace map $\mathop{\rm Tr}\nolimits_{T/K}:\,T\longrightarrow K$ satisfies $\mathop{\rm Tr}\nolimits_{T/K}(c)=\nu\cdot\mathop{\rm Tr}\nolimits_{L/K}(\bar{c})$ for every $c\in T$.

In the case $\mathop{\rm char}\nolimits(K)\mid\nu$, it follows that $\mathop{\rm Tr}\nolimits_{T/K}=0$. As $T$ is a Gorenstein ring with a trace map $\sigma$ and $\mathop{\rm Tr}\nolimits_{T/K}=g\cdot\sigma$ for a generating element $g\in T$ of $\vartheta_{N}(T/K)$ by [14], Corollary F.12, it follows that $\vartheta_{N}(T/K)=\langle 0\rangle$, as claimed.

Now assume that $\mathop{\rm char}\nolimits(K)\nmid\nu$. Let $\sigma:\;T\longrightarrow K$ be the projection to $a^{\nu-1}\,b_{1}$ along the rest of the basis $B$. It is straightforward to check that $\sigma$ is a trace map of $T/K$ and that $\mathop{\rm Tr}\nolimits_{T/K}=\nu\,a^{\nu-1}\,\sigma$. Therefore the element $a^{\nu-1}$ generates $\vartheta_{N}(T/K)$, as claimed.

It remains to prove (d). From $\Omega^{1}_{T/K}=T\,da$ we get $\dim_{K}(\Omega^{1}_{T/K})=\dim_{K}(T)-\dim_{K}(\vartheta_{N}(T/K))$. Thus the claim is a consequence of (b) and (c). ∎

As a consequence of this proposition and Proposition 6.6, we have the following formulas for the Hilbert polynomials of ${\Omega^{m}_{R/K}}$ when ${\mathbb{X}}$ is weakly curvilinear.

Corollary 6.8**.**

Let ${\mathbb{X}}$ be a 0-dimensional weakly curvilinear scheme in $\mathbb{P}^{n}$, let $S=\mathcal{O}_{1}\times\cdots\times\mathcal{O}_{t}$ be the decomposition of its affine coordinate ring into local rings, let $L_{i}$ be the residue class field of $\mathcal{O}_{i}$, let $\kappa_{i}=\dim_{K}(L_{i})$, and let $\nu_{i}=\dim_{K}(\mathcal{O}_{i})/\kappa_{i}$ for $i=1,\dots,t$.

1. (a)

We have $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=2\deg({\mathbb{X}})-\textstyle{\sum\limits_{\mathop{\rm char}\nolimits(K)\nmid\,\nu_{i}}}\kappa_{i}$. 2. (b)

We have $\mathop{\rm HP}\nolimits(\Omega^{2}_{R/K})=\deg({\mathbb{X}})-\textstyle{\sum\limits_{\mathop{\rm char}\nolimits(K)\nmid\,\nu_{i}}}\kappa_{i}$. 3. (c)

For $m>2$, we have $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=0$.

7. Kähler Differentials of Fat Point Schemes

For fat point schemes, the above results about the Hilbert functions and Hilbert polynomials of the modules of Kähler differentials can be made even more explicit. Let $K$ be a perfect field and $P=K[X_{0},\dots,X_{n}]$. Let $t\geq 1$, let $p_{1},\dots,p_{t}$ be distinct $K$-rational points in $\mathbb{P}^{n}$, and let $I_{p_{i}}\subseteq P$ be the homogeneous vanishing ideal of $p_{i}$ for $i=1,\dots,t$. Given positive integers $m_{1},\dots,m_{t}$, recall that the 0-dimensional scheme ${\mathbb{X}}$ defined by the saturated homogeneous ideal

[TABLE]

is called the fat point scheme with support $\mathop{\rm Supp}\nolimits({\mathbb{X}})=\{p_{1},\dots,p_{t}\}$ and multiplicities $m_{1},\dots,m_{t}$. Frequently, the scheme ${\mathbb{X}}$ is written as ${\mathbb{X}}=m_{1}p_{1}+\cdots+m_{t}p_{t}$. Let us collect some initial observations about this setting.

Remark 7.1**.**

For a fat point scheme ${\mathbb{X}}=m_{1}p_{1}+\cdots+m_{t}p_{t}$ in $\mathbb{P}^{n}$, we know that:

1. (a)

The degree of ${\mathbb{X}}$ is given by $\deg({\mathbb{X}})=\sum_{i=1}^{t}\tbinom{n+m_{i}-1}{n}$. 2. (b)

The scheme ${\mathbb{X}}$ is a reduced scheme if and only if $\mathop{\rm HP}\nolimits({\Omega^{1}_{R/K}})=t$ and $\mathop{\rm HP}\nolimits({\Omega^{m}_{R/K}})=0$ for $m\geq 2$.

It is natural to ask what the Hilbert polynomial of $\Omega^{m}_{R/K}$ is for $1\leq m\leq n+1$. More precisely, does the Hilbert polynomial of $\Omega^{m}_{R/K}$ depend only on $m$, $n$, and the multiplicties $m_{1},\dots,m_{t}$?

In view of Proposition 5.4, in order to determine the Hilbert polynomial of $\Omega^{m}_{R/K}$ for a fat point scheme ${\mathbb{X}}$, it suffices to work out the dimension of the $K$-vector spaces $\Omega^{m}_{S/K}$ for $m=1,\dots,n$. According to [14], Proposition 4.7, we have

[TABLE]

and so $\dim_{K}(\Omega^{m}_{S/K})=\sum_{i=1}^{t}\dim_{K}(\Omega^{m}_{\mathcal{O}_{i}/K})$. This leads us to compute $\dim_{K}(\Omega^{m}_{\mathcal{O}_{i}/K})$ at a fat point of ${\mathbb{X}}$.

Using a homogeneous linear change of coordinates, we assume that ${\mathbb{X}}=kp$ with $p=(1:0:...:0)$ and $k\geq 1$. Then we have $I_{p}=\langle X_{1},\dots,X_{n}\rangle\subseteq P$. Letting $A:=K[X_{1},\dots,X_{n}]$ and $\mathfrak{q}:=\langle X_{1},\dots,X_{n}\rangle$, the local ring of ${\mathbb{X}}$ at $p$ is $S=A/\mathfrak{q}^{k}$.

In the reduced case $k=1$, we have $\Omega^{m}_{S/K}=\langle 0\rangle$ for all $m\geq 2$. Thus we consider the case $k\geq 2$ now. We equip $A$ with the standard grading and note that $\mathfrak{q}^{k}$ is a homogeneous ideal. Therefore $S$ is a graded 0-dimensional affine $K$-algebra. Its Hilbert function is given by

[TABLE]

and thus $\dim_{K}(S)=\sum_{i\geq 0}\mathop{\rm HF}\nolimits_{S}(i)=\binom{n+k-1}{n}$.

Our next task is to describe the Hilbert functions of the graded $S$-modules $\Omega^{m}_{S/K}$ for all $m\geq 1$ explicitly. Notice that $\Omega^{m}_{S/K}=0$ for $m>n$.

Proposition 7.2**.**

In the above setting, let $1\leq m\leq n$. Then the Hilbert function of $\Omega^{m}_{S/K}$ is given by

[TABLE]

where $\delta=\dim_{K}(d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})_{m+k-1}$.

Proof.

By [14], Proposition 4.12, we have a homogeneous short exact sequence of graded $A$-modules

[TABLE]

where $\Omega^{m-1}_{A/K}=A$ if $m=1$. Here $\Omega^{m}_{A/K}=\bigoplus_{1\leq i_{1}<\cdots<i_{m}\leq n}AdX_{i_{1}}\wedge\cdots\wedge dX_{i_{m}}$ is a graded free $A$-module of rank $\binom{n}{m}$ with basis elements of degree $m$. Moreover, we have $\mathfrak{q}^{k}=\langle\,t_{1},\dots,t_{N}\,\rangle$ where $\mathbb{T}^{n}_{k}=\{t_{1},\dots,t_{N}\}$ and $N=\binom{n+k-1}{n-1}$. By Proposition 5.2, we have

[TABLE]

In particular, the graded $A$-module $\mathfrak{q}^{k}\,\Omega^{m}_{A/K}+d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K}$ is generated in degrees $m+k-1$ and $m+k$. So, for $i<m+k-1$, we have

[TABLE]

Furthermore, for $i\geq m+k$, we have $A_{i-m}=(\mathfrak{q}^{k})_{i-m}$ and

[TABLE]

Consequently, we get $\mathop{\rm HF}\nolimits_{\Omega^{m}_{S/K}}(i)=0$. In the case $i=m+k-1$, we have

[TABLE]

because $(\mathfrak{q}^{k}\Omega^{m}_{A/K})_{m+k-1}=0$. ∎

To describe the Hilbert function of $\Omega^{m}_{S/K}$ completely, it remains to compute $\delta=\dim_{K}(d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})_{m+k-1}$. For that, we simplify the notation as follows.

Remark 7.3**.**

For every $m\geq 1$, the map $\varrho:\;\Omega^{m}_{A/K}\longrightarrow A(-m)^{\binom{n}{m}}$ given by $\varrho(fdX_{i_{1}}\wedge\cdots\wedge dX_{i_{m}})=fe_{i_{1},\dots,i_{m}}$ for $f\in A$ and $1\leq i_{1}<\cdots<i_{m}\leq n$ is an isomorphism of graded $A$-modules. Note that the degrees of the standard basis vectors satisfy $\deg(e_{i_{1},\dots,i_{m}})=m$ here.

To compute the desired number $\delta$, we have to consider the graded $A$-submodule $U=\varrho(d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})$ of $A(-m)^{\binom{n}{m}}$. By Proposition 7.2, we have $U_{i}=\{0\}$ for $i<m+k-1$ and $U_{i}=(A(-m)^{\binom{n}{m}})_{i}$ for $i>m+k-1$. Therefore the only interesting homogeneous component of $U$ is $U_{m+k-1}$, and its $K$-dimension is the number $\delta$ we are looking for. The vector space $U_{m+k-1}$ will be called the initial defining vector space for $\Omega^{1}_{S/K}$.

In order to describe a $K$-basis of $U_{m+k-1}$, we use the module term ordering $\sigma={\tt DegRevLex-Pos}$ on $A(-m)^{\binom{n}{m}}$ (see [11], Definition 1.4.16) and determine a system of generators of $U_{m+k-1}$ having distinct leading terms. Here the standard basis vectors $e_{i_{1},\dots,i_{m}}$ are ordered by using the lexicographic ordering on the tuples $(i_{1},\dots,i_{m})$. The next three propositions set the stage for our main result.

Proposition 7.4**.**

In the setting of the preceding remark, assume that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$. Let $\{t_{1},\dots,t_{N}\}$ be the set of terms of degree $m+k-1$ in $A$, where $N=\binom{n+k-1}{n-1}$, and where the terms are ordered decreasingly with respect to DegRevLex.

1. (a)

For $i=1,\dots,N$, we have $\varrho(dt_{i})={\textstyle\frac{\partial t_{i}}{\partial X_{1}}}e_{1}+\cdots+{\textstyle\frac{\partial t_{i}}{\partial X_{n}}}e_{n}$. We denote the vector on the right-hand side by $\mathop{\rm grad}\nolimits(t_{i})$. 2. (b)

Let $i\in\{1,\dots,N\}$ and write $t_{i}=X_{1}^{\alpha_{1}}\cdots X_{n}^{\alpha_{n}}\in\mathbb{T}^{n}_{k}$, where $\alpha_{j}\geq 0$. Then we have

[TABLE] 3. (c)

For $i\in\{1,\dots,N\}$, we can write $\mathop{\rm LT}\nolimits_{\sigma}(\mathop{\rm grad}\nolimits(t_{i}))=X_{1}^{\beta_{1}}X_{2}^{\beta_{2}}\cdots X_{\ell}^{\beta_{\ell}}\,e_{\ell}$ with $\ell\in\{1,\dots,n\}$ and $\beta_{j}\geq 0$. Then we have $t_{i}=X_{1}^{\beta_{1}}X_{2}^{\beta_{2}}\cdots X_{\ell}^{\beta_{\ell}+1}$.

Consequently, $\mathop{\rm LT}\nolimits_{\sigma}(\mathop{\rm grad}\nolimits(t_{i}))$ and $t_{i}$ uniquely determine each other and all leading terms $\mathop{\rm LT}\nolimits_{\sigma}(\mathop{\rm grad}\nolimits(t_{i}))$ are pairwise distinct. 4. (d)

In the case $m=1$, the set $B_{1}=\{\mathop{\rm LT}\nolimits_{\sigma}(\mathop{\rm grad}\nolimits(t_{1})),\dots,\mathop{\rm LT}\nolimits_{\sigma}(\mathop{\rm grad}\nolimits(t_{N}))\}$ has $N=\binom{n+k-1}{n-1}$ distinct elements.

Proof.

Claim (a) follows from $dt_{i}={\textstyle\frac{\partial t_{i}}{\partial X_{1}}}X_{1}+\cdots+{\textstyle\frac{\partial t_{i}}{\partial X_{n}}}X_{n}$ and claim (b) is a consequence of $\mathop{\rm grad}\nolimits(t_{i})=\sum_{j=1}^{n}\alpha_{j}X_{1}^{\alpha_{1}}\cdots X_{j}^{\alpha_{j}-1}\cdots X_{n}^{\alpha_{n}}e_{j}$. Claim (c) follows from (b) and claim (d) follows from (c). ∎

In the case $m\geq 2$, the determination of the leading terms in $U_{m+k-1}$ is slightly more involved.

Proposition 7.5**.**

**(Generators of the Initial Defining Vector Space)

**In the setting described above, let $m\geq 2$, and assume that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$. For $i\in\{1,\dots,N\}$ and $1\leq j_{1}<\cdots<j_{m-1}\leq n$, we let $J=\{j_{1},\dots,j_{m-1}\}$, and we define

[TABLE]

1. (a)

The set $G$ of all vectors $v_{i,J}$ generates the $K$-vector space $U_{m+k-1}$. 2. (b)

For a set $J=\{j_{1},\dots,j_{m-1}\}$ as above and $\ell\notin J$, choose $\nu_{\ell}\in\{0,\dots,m-1\}$ such that $1\leq j_{1}<\cdots<j_{\nu_{\ell}}<\ell<j_{\nu_{\ell}+1}<\cdots<j_{m-1}\leq n$, and let $\bar{e}_{\ell,J}=(-1)^{\nu_{\ell}}\,e_{j_{1},\dots,j_{\nu_{\ell}},\ell,j_{\nu_{\ell}+1},\dots,j_{m-1}}$. Then we have

[TABLE]

Proof.

Claim (a) follows from the fact that the elements $dt_{i}\wedge dX_{j_{1}}\wedge\cdots\wedge dX_{j_{m-1}}$ generate the vector space $(\mathfrak{q}^{k}\Omega^{m}_{A/K}+d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})_{m+k-1}$ since $(\mathfrak{q}^{k}\Omega^{m}_{A/K})_{m+k-1}=0$.

To show (b), we calculate

[TABLE]

and apply the map $\varrho$. ∎

After describing a system of generators of the initial defining vector space, we now turn to finding its leading term vector space with respect to $\sigma$. Then we are able to deduce its dimension via $\dim_{K}(U_{m+k-1})=\dim_{K}(\mathop{\rm LT}\nolimits_{\sigma}(U_{m+k-1}))$.

Proposition 7.6**.**

**(Leading Terms of the Initial Defining Vector Space)

**In the setting described above, assume that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$.

1. (a)

Let $i\in\{1,\dots,N\}$ and $J=\{j_{1},\dots,j_{m-1}\}$, where $1\leq j_{1}<\cdots<j_{m-1}\leq n$. Then we have

[TABLE]

where $\ell\in\{1,\dots,n\}$ is the largest index such that $\ell\notin J$ and $X_{\ell}\mid t_{i}$, and where $e_{\ell,J}=e_{j_{1},\dots,j_{\nu},\ell,j_{\nu+1},\dots,j_{m-1}}$. If no such $\ell$ exists, we have $v_{i,J}=0$. 2. (b)

Let $i,i^{\prime}\in\{1,\dots,N\}$ be such that $i\leq i^{\prime}$, and assume that $J=\{j_{1},\dots,j_{m-1}\}$ and $J^{\prime}=\{j^{\prime}_{1},\dots,j^{\prime}_{m-1}\}$ satisfy $1\leq j_{1}<\cdots<j_{m-1}\leq n$ as well as $1\leq j^{\prime}_{1}<\cdots<j^{\prime}_{m-1}\leq n$. Suppose that we have $v_{i,J}\neq v_{i^{\prime},J^{\prime}}$ and $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})=\mathop{\rm LT}\nolimits_{\sigma}(v_{i^{\prime},J^{\prime}})$. Then the following properties hold.

• (1)

$i<i^{\prime}$**

• (2)

There exists indices $\ell\in\{1,\dots,n\}\setminus J$ and $\ell^{\prime}\in\{1,\dots,n\}\setminus J^{\prime}$ such that $t_{i}=X_{1}^{\alpha_{1}}\cdots X_{n}^{\alpha_{n}}$ and $t_{i^{\prime}}=X_{1}^{\alpha^{\prime}_{1}}\cdots X_{n}^{\alpha^{\prime}_{n}}$ satisfy $\alpha_{\ell}=\alpha^{\prime}_{\ell}+1$ and $\alpha^{\prime}_{\ell^{\prime}}=\alpha_{\ell^{\prime}}+1$.

• (3)

$\ell<\ell^{\prime}$**

• (4)

The set $\hat{J}=J\cap J^{\prime}$ satisfies $\hat{J}=J\setminus\{\ell^{\prime}\}=J^{\prime}\setminus\{\ell\}$.

• (5)

We have decompositions $t_{i}=\hat{t}\cdot\tilde{t}\cdot X_{\ell}^{\alpha_{\ell}}X_{\ell^{\prime}}^{\alpha_{\ell^{\prime}}}$ and $t_{i^{\prime}}=\hat{t}\cdot\tilde{t}\cdot X_{\ell}^{\alpha^{\prime}_{\ell}}X_{\ell^{\prime}}^{\alpha^{\prime}_{\ell^{\prime}}}$ where $\hat{t}\in K[X_{j}\mid j\in\hat{J}]$ and where $\tilde{t}=X_{\nu_{1}}^{\alpha_{\nu_{1}}}\cdots X_{\nu_{r}}^{\alpha_{\nu_{r}}}$ with $\nu_{1},\dots,\nu_{r}\notin\hat{J}\cup\{\ell,\ell^{\prime}\}$. 3. (c)

In the setting of (b), we have

[TABLE] 4. (d)

The leading terms of the elements of $G$ generate $\mathop{\rm LT}\nolimits_{\sigma}(U_{m+k-1})$. 5. (e)

The set of distinct $\sigma$-leading terms of elements of $G$ is

[TABLE] 6. (f)

We have $\#B_{m}=\tbinom{n}{m}\tbinom{n+k-2}{n-1}-\tbinom{m+k-2}{m}\tbinom{n+k-2}{n-m-1}$.

Proof.

First we prove (a). If $\ell\in J$ for all indices $\ell\in\{1,\dots,n\}$ such that $X_{\ell}\mid t_{i}$ then Proposition 7.5.b shows $v_{i,J}=0$. So, let us assume that the largest index $\ell$ with $X_{\ell}\mid t_{i}$ and $\ell\notin J$ exists. In view of Proposition 7.5.b, we can write $\mathop{\rm LM}\nolimits_{\sigma}(v_{i,J})={\textstyle\frac{\partial t_{i}}{\partial X_{\ell}}}\,\bar{e}_{\ell,J}$, and we only have to note that ${\textstyle\frac{\partial t_{i}}{\partial X_{\ell}}}$ differs from ${\textstyle\frac{t_{i}}{X_{\ell}}}$ by a unit of $K$.

Next we show the properties claimed in (b) one by one. To verify (1), we remark that $i=i^{\prime}$ implies that the power products in $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})$ and $\mathop{\rm LT}\nolimits_{\sigma}(v_{i^{\prime},J^{\prime}})$ are equal. But then also the positions of these two leading terms have to be equal, i.e., we have $J=J^{\prime}$. Altogether, the equality $(i,J)=(i^{\prime},J^{\prime})$ contradicts $v_{i,J}\neq v_{i^{\prime},J^{\prime}}$.

To check (2), we use (a) and find $\ell,\ell^{\prime}\in\{1,\dots,n\}$ such that the power product in $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})=\mathop{\rm LT}\nolimits_{\sigma}(v_{i^{\prime},J^{\prime}})$ is ${\textstyle\frac{t_{i}}{X_{\ell}}}={\textstyle\frac{t_{i^{\prime}}}{X_{\ell^{\prime}}}}$. As we just saw, the two power products cannot be equal, whence $\ell\neq\ell^{\prime}$. Now $X_{\ell^{\prime}}\,t_{i}=X_{\ell}\,t_{i^{\prime}}$ yields the desired equalities for the exponents of the indeterminates.

In order to see why (3) holds, we have to recall that the terms $t_{1},\dots,t_{N}$ were ordered decreasingly with respect to $\tt DegRevLex$. Hence $i<i^{\prime}$ yields $t_{i}>_{\tt DegRevLex}t_{i^{\prime}}$. Since the two terms differ only in the exponents of $X_{\ell}$ and $X_{\ell^{\prime}}$, it follows that $\ell<\ell^{\prime}$, as the exponent of $t_{i^{\prime}}$ is larger for the indeterminate $X_{\ell^{\prime}}$.

Claim (4) follows immediately from (2). Thus it remains to verify (5). Let $\tilde{t}_{i}$ be the power product of indeterminates of $t_{i}$ which are not in $\hat{J}\cup\{\ell,\ell^{\prime}\}$, and define $\tilde{t}_{i^{\prime}}$ analogously. Recall that $\ell=\max\{\lambda\notin J\mid\alpha_{\lambda}>0\}$. Therefore all indices $\lambda\notin\hat{J}\cup\{\ell,\ell^{\prime}\}$ of indeterminates dividing $\tilde{t}_{i}$ satisfy $\lambda<\ell$. Similarly, we have $\ell^{\prime}=\max\{\lambda\notin\hat{J}\cup\{\ell\}\mid\alpha_{\lambda}>0\}$. This maximum is larger than $\ell$, and the maximality of $\ell$ implies that no such indices $\lambda$ exist between $\ell$ and $\ell^{\prime}$. Hence all indices $\lambda\notin\hat{J}\cup\{\ell,\ell^{\prime}\}$ of indeterminates dividing $\tilde{t}_{i^{\prime}}$ satisfy $\lambda<\ell$, and we obtain $\tilde{t}:=\tilde{t}_{i}=\tilde{t}_{i^{\prime}}$. Since the exponents of $t_{i}$ and $t_{i^{\prime}}$ differ only at $X_{\ell}$ and $X_{\ell^{\prime}}$, combining all indeterminates with indices in $\hat{J}$ into $\hat{t}$ yields the claimed decompositions of $t_{i}$ and $t_{i^{\prime}}$.

To show (c), we have to consider the image of $dt_{i}\wedge e_{j_{1}}\wedge\cdots\wedge e_{j_{m-1}}$ under $\varrho$. Using the decomposition of $t_{i}$ given in (b.5), and considering the fact that $e_{\lambda}\wedge e_{j_{1}}\wedge\cdots\wedge e_{j_{m-1}}=0$ for $\lambda\in J=\hat{J}\cup\{\ell^{\prime}\}$, only the terms involving the partial derivatives ${\textstyle\frac{\partial t_{i}}{\partial X_{\lambda}}}$ with $X_{\lambda}\mid\tilde{t}$ or $\lambda=\ell$ survive the wedge product. In view of the formula for $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})$ shown in (a), this yields the claim for $v_{i,J}$. For $v_{i^{\prime},J^{\prime}}$, it follows analogously.

Next we prove (d). If in a linear combination $\sum_{i}c_{i}v_{i,J}$ with $c_{i}\in K$ the largest leading terms do not cancel, the result is a vector whose leading term is one of the $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})$, and the claim holds. Now suppose that the largest term appearing in the linear combination cancels out. We use induction on the largest term appearing in the result of the linear combination and show that it can be reduced using $G$. Then it follows that the leading terms of the elements of $G$ generate $\mathop{\rm LT}\nolimits_{\sigma}(U_{n+k-1})$.

To start the induction, we note that the smallest term in degree $m+k-1$ is $X_{n}^{k-1}e_{n-m+1,\dots,n}$. It is the leading term of $\varrho(dX_{n}^{k}\,e_{n-m+1}\wedge\cdots\wedge e_{n-1})$ and thus a leading term in $\mathop{\rm LT}\nolimits_{\sigma}(G)$. Subtracting the appropriate multiple of the corresponding element of $G$ has to yield zero, because no smaller term exists.

For the induction step, we note that any linear combination of the vectors $v_{i,J}$ whose largest term cancels out can be written as a linear combination of fundamental syzygies of the form

[TABLE]

Thus it suffices to consider a fundamental syzygy whose support contains the largest term for which the claim has not yet been shown. As $X_{\ell}^{\alpha_{\ell}}X_{\ell^{\prime}}^{\alpha_{\ell^{\prime}}}$ is larger than $X_{\ell}^{\alpha_{\ell}-1}X_{\ell^{\prime}}^{\alpha_{\ell^{\prime}}+1}$, it follows that the leading term of this syzygy is $\hat{t}\,X_{\ell^{\prime}}^{\alpha_{\ell^{\prime}}}{\textstyle\frac{\tilde{t}}{X_{\nu_{r}}}}\,X_{\ell}^{\alpha_{\ell}}e_{\nu_{r},J}$. This term is the leading term of $v_{i^{\prime\prime},J^{\prime\prime}}$ where $t_{i^{\prime\prime}}=t_{i}\cdot X_{\ell^{\prime}}/X_{\nu_{r}}$ and $J^{\prime\prime}=\{\nu_{r}\}\cup\hat{J}$, because we have $t_{i^{\prime\prime}}=\hat{t}\,X_{\nu_{1}}^{\alpha_{\nu_{1}}}\cdots X_{\nu_{r-1}}^{\alpha_{\nu_{r-1}}}\cdot X_{\nu_{r}}^{\alpha_{\nu_{r}}-1}\,X_{\ell}^{\alpha_{\ell}}\,X_{\ell^{\prime}}^{\ell^{\prime}+1}$ and $\{\nu_{r},\ell^{\prime}\}\cup\hat{J}=\{\nu_{r}\}\cup J$. Consequently, if we subtract the appropriate multiple of $v_{i^{\prime\prime},J^{\prime\prime}}$, we get zero or a smaller leading term, and the claim is a consequence of the induction hypothesis.

For the proof of (e) we first show that every element of $B_{m}$ is indeed a leading term of an element of $G$. Given $t\,e_{j_{1},\dots,j_{m}}\in B_{m}$ with $t\in\mathbb{T}^{n}_{k}$, we let $\tilde{t}=t\,X_{j_{m}}$ and $J=\{j_{1},\dots,j_{m-1}\}$. Then the degree of $\tilde{t}$ is $k$ and the hypothesis about the characteristic of $K$ implies that $\alpha_{j_{m}}+1\leq\deg(\tilde{t})=k$ is a unit in $K$. Choosing $i\in\{1,\dots,N\}$ such that $t_{i}=\tilde{t}$, we get from (a) that $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})=t\,e_{j_{1},\dots,j_{m}}$.

Conversely, part (a) shows that every leading term of an element of $G$ is in $B_{m}$. We also observe that the elements of $B_{m}$ are clearly pairwise distinct.

Finally, to prove (f), we count $\#B_{m}$. Given a fixed index $\ell\in\{1,\dots,n\}$, in order to form a term $t=X_{\ell}\cdot\hat{t}$ with $\hat{t}\in\mathbb{T}^{\ell}_{k-2}$, we have $\binom{\ell+k-3}{\ell-1}$ choices. From all $\binom{n}{m}$ choices for $(j_{1},\dots,j_{m})$ with $j_{1}<\cdots<j_{m}$ we have to subtract all $\binom{\ell-1}{m}$ choices where $\{j_{1},\dots,j_{m}\}\subseteq\{1,\dots,\ell-1\}$. Altogether, we get

[TABLE]

and the proof is complete. ∎

Notice that in the key step of the proof of part (d) we can even be more explicit, as the following remark shows.

Remark 7.7**.**

Let $v_{i,J}$ and $v_{i^{\prime},J^{\prime}}$ be two vectors such that $\mathop{\rm LT}\nolimits_{\sigma}(v_{i,J})=\mathop{\rm LT}\nolimits_{\sigma}(v_{i^{\prime},J^{\prime}})$, where $i,i^{\prime}\in\{1,\dots,N\}$ and $J,J^{\prime}$ satisfy the conditions of part (b) of the proposition. Then we can write the fundamental syzygy of $v_{i,J}$ and $v_{i^{\prime},J^{\prime}}$ in the form

[TABLE]

where $v_{\mu_{\kappa},J_{\kappa}}$ is the image of $\mathop{\rm grad}\nolimits(X_{\ell^{\prime}}\,t_{i}/X_{\nu_{\kappa}})\wedge e_{\nu_{\kappa},\hat{J}}$ under $\varrho$. This can be shown by a straightforward, but lengthy calculation and provides another proof of (d).

Now we are ready to prove the main result of this section which allows us to determine the Hilbert function of the Kähler differential modules of a fat point.

Theorem 7.8**.**

Let $k\geq 1$, let $K$ be a field of characteristic $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>k$, let $\mathfrak{q}=\langle X_{1},\dots,X_{n}\rangle$ be the homogeneous maximal ideal of $A=K[x_{1},\dots,x_{n}]$, and let $S=A/\mathfrak{q}^{k}$.

1. (a)

For every $m\geq 1$, we have

[TABLE] 2. (b)

For every $m\geq 1$, we have

[TABLE]

Proof.

To prove (a), we identify $(d\mathfrak{q}^{k}\wedge\Omega^{m-1}_{A/K})_{m+k-1}$ with the $K$-vector subspace $U_{m+k-1}$ of $(A(-m)^{\binom{n}{m}})_{m+k-1}$ using the map $\varrho$ of Remark 7.3.

In the case $m=1$, we apply Proposition 7.4 and get

[TABLE]

Thus the claim follows from

[TABLE]

Now let $m\geq 2$. In view of Proposition 7.6, it suffices to note that the elements of $B_{m}$ form a $K$-basis of $\mathop{\rm LT}\nolimits_{\sigma}(U_{m+k-1})$ and to apply the well-known equality $\dim_{K}(U_{m+k-1})=\dim_{K}(\mathop{\rm LT}\nolimits_{\sigma}(U_{m+k-1}))$ (see for instance [12], Thm. 5.1.18).

To prove (b) in the case $m=1$, we use Proposition 7.2 and get

[TABLE]

Therefore we have

[TABLE]

Finally, we show (b) in the case $m\geq 2$. By (a) and Proposition 7.2, we have

[TABLE]

Consequently, we get

[TABLE]

Thus the proof of the theorem is complete. ∎

Under somewhat stricter assumptions on the characteristic of the ground field, we can provide another proof of this theorem which is based on the Euler-Koszul complex of the affine coordinate ring $S=R/\langle x_{0}-1\rangle=K[X_{1},\dots,X_{n}]/\mathfrak{q}^{k}$ of ${\mathbb{X}}$, where $\mathfrak{q}=\langle X_{1},\dots,X_{n}\rangle$. This complex is constructed as follows. In analogy to Definition 4.1, the algebra $S/K$ has the Euler form ${\varepsilon}_{S}:\Omega^{1}_{S/K}\rightarrow\mathfrak{q}/\mathfrak{q}^{k}$, which is given by ${\varepsilon}_{S}(d_{S/K}f)=\deg(f)\cdot f$ for every homogeneous element $f\in S$. After forming the Koszul complex of ${\varepsilon}_{S}$, we get the following result.

Proposition 7.9**.**

In the above setting, let ${\varepsilon}_{S}:\Omega^{1}_{S/K}\rightarrow\mathfrak{q}/\mathfrak{q}^{k}$ be the Euler form of $S/K$.

1. (a)

The sequence of $S$-linear maps

[TABLE]

is a complex, called the Euler-Koszul complex of $S/K$. Here ${\varepsilon}_{S}^{(i)}$ is given by ${\varepsilon}_{S}^{(i)}(w_{1}\wedge\cdots\wedge w_{i})=\sum_{j=1}^{i}(-1)^{j-1}{\varepsilon}_{S}(w_{j})\,w_{i}\wedge\cdots\wedge\widehat{w_{j}}\wedge\cdots\wedge w_{i}$ for $i\geq 2$ and $w_{1},\dots,w_{i}\in\Omega^{1}_{S/K}$. 2. (b)

Let $i\geq 1$, and assume that $\mathop{\rm char}\nolimits(K)=0$ or that $\mathop{\rm char}\nolimits(K)>0$ is not a divisor of $i$. Then the sequence of $K$-vector spaces

[TABLE]

is exact.

Proof.

Claim (a) follows from the observation that this sequence is nothing but the Koszul complex associated to the $S$-linear map ${\varepsilon}_{S}$.

To prove (b) we have to show that $\mathop{\rm Ker}\nolimits({\varepsilon}_{S}^{(m)})\subseteq\mathop{\rm Im}\nolimits({\varepsilon}_{S}^{(m+1)})$ for $m\geq 1$. In view of Proposition 7.2, it is enough to check this condition in the case $i\in\{1,\dots,m+k-1\}$. It is straightforward verify that $({\varepsilon}_{S}^{(m+1)}\circ d+d\circ{\varepsilon}_{S}^{(m)})(w)=iw$ for every $w\in(\Omega^{m}_{S/K})_{i}$. Given $w\in\mathop{\rm Ker}\nolimits({\varepsilon}^{(m)})$, we use the fact that $i\in K\setminus\{0\}$ and let $w^{\prime}=\frac{1}{i}dw\in(\Omega^{m+1}_{S/K})_{i}$. Then $w^{\prime}$ satisfies ${\varepsilon}_{S}^{(m+1)}(w^{\prime})=\frac{1}{i}({\varepsilon}^{(m+1)}\circ d+d\circ{\varepsilon}^{(m)})(w)=w$. Hence we have $w\in\mathop{\rm Im}\nolimits({\varepsilon}^{m+1})$, and the proof is complete. ∎

Based on this proposition, we can give an alternative proof of Theorem 7.8.b under a more stringent assumption about $\mathop{\rm char}\nolimits(K)$.

Remark 7.10**.**

In the setting of Theorem 7.8, assume that $\mathop{\rm char}\nolimits(K)=0$ or that $\mathop{\rm char}\nolimits(K)>m+k-1$. Then the proposition yields

[TABLE]

where the last equality can be shown by induction. From this formula the claim of Theorem 7.8.b follows as in the proof of that theorem.

The condition on the characteristic of $K$ in Theorem 7.8 is necessary, as the following example shows.

Example 7.11**.**

Consider the fat point scheme ${\mathbb{X}}=2p$ in $\mathbb{P}^{2}$ over the field $K=\mathbb{F}_{2}$, where $p=(1:0:0)$. Then we have $k=2=\mathop{\rm char}\nolimits(K)$ and

[TABLE]

This shows that $\dim_{K}(d\mathfrak{q}^{2})_{2}=1\neq 3=\tbinom{n}{1}\tbinom{n}{n-1}-\tbinom{k-1}{1}\tbinom{n}{n-2}$ and $\dim_{K}(\Omega^{1}_{S/K})=5\neq 3=\tbinom{n}{1}\tbinom{n}{n}+\tbinom{k-1}{1}\tbinom{n}{n-2}$. In other words, the formulas given in Theorem 7.8 do not hold true for this case.

Moreover, we see that $\mathop{\rm HF}\nolimits_{S}(2)-\mathop{\rm HF}\nolimits_{\Omega^{1}_{S/K}}(2)+\mathop{\rm HF}\nolimits_{\Omega^{2}_{S/K}}(2)\neq 0$. Hence the sequence of $K$-vector spaces given in Proposition 7.9 is not exact for $i=2$.

Our final result of this section provides an explicit formula for the Hilbert polynomials of all Kähler differential modules provided the characteristic of the base field is not too small.

Theorem 7.12**.**

Let ${\mathbb{X}}=m_{1}p_{1}+\cdots+m_{t}p_{t}$ be a fat point scheme in $\mathbb{P}^{n}$, let $m\in\{1,\dots,n+1\}$, and assume $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>\max\{m_{1},\dots,m_{t}\}$. Then we have

[TABLE]

where $\delta_{i,m}=\tbinom{m+m_{i}-2}{m}\tbinom{n+m_{i}-2}{n-m-1}+\tbinom{m+m_{i}-3}{m-1}\tbinom{n+m_{i}-2}{n-m}$.

Proof.

By Proposition 5.4 and [14], Proposition 4.7, we have

[TABLE]

For $m=1$, Theorem 7.8 yields

[TABLE]

For $2\leq m\leq n+1$, Theorem 7.8 shows

[TABLE]

In particular, we have $\dim_{K}(\Omega^{n+1}_{\mathcal{O}_{i}/K})+\dim_{K}(\Omega^{n}_{\mathcal{O}_{i}/K})=\dim_{K}(\Omega^{n}_{\mathcal{O}_{i}/K})=\tbinom{n+m_{i}-2}{n}$. Therefore we obtain the desired formula for $\mathop{\rm HP}\nolimits(\Omega^{m}_{R/K})$. ∎

This theorem generalizes and unifies a number of special cases which were considered previously. Let us mention one of them.

Remark 7.13**.**

In the case $m=n+1$, Theorem 7.12 yields

[TABLE]

where ${\mathbb{Y}}$ is the fat point scheme ${\mathbb{Y}}=(m_{1}-1)\,p_{1}+\cdots+(m_{t}-1)\,p_{t}$ in $\mathbb{P}^{n}$. This provides a positive answer to a question posed in [7]. See also [8] for a different proof of this result.

8. Kähler Differentials of Uniform Schemes

In this section we examine how uniformity properties of 0-dimensional schemes in $\mathbb{P}^{n}$, in particular the Cayley-Bacharach property, are reflected in the structure of their Kähler differential modules. Thus we continue to let $K$ be a perfect field, let ${\mathbb{X}}$ be a 0-dimensional subscheme of $\mathbb{P}^{n}$ having $K$-rational support, and we let $R=P/I_{\mathbb{X}}$ be the homogeneous coordinate ring of ${\mathbb{X}}$. Recall that the Cayley-Bacharach property of ${\mathbb{X}}$ is defined as follows.

Definition 8.1**.**

Let $d\in\mathbb{N}$, and let $r_{\mathbb{X}}$ be the regularity index of ${\mathbb{X}}$.

1. (a)

The scheme ${\mathbb{X}}$ is said to have the Cayley-Bacharach property of degree $d$ (in short, ${\mathbb{X}}$ has CBP($d$)) if every hypersurface of degree $d$ which contains a subscheme ${\mathbb{Y}}$ of ${\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$ automatically contains ${\mathbb{X}}$. 2. (b)

If ${\mathbb{X}}$ has the Cayley-Bacharach property of degree $r_{\mathbb{X}}-1$, then ${\mathbb{X}}$ is called a Cayley-Bacharach scheme.

Algebraically, the Cayley-Bacharach property can be characterized as follows.

Remark 8.2**.**

Let ${\mathbb{Y}}$ be a subscheme of ${\mathbb{X}}$ of degree $\deg({\mathbb{X}})-1$.

1. (a)

The image of the vanishing ideal $I_{\mathbb{Y}}\subseteq P$ of ${\mathbb{Y}}$ in $R=P/I_{\mathbb{X}}$ is denoted by $I_{{\mathbb{Y}}/{\mathbb{X}}}=I_{\mathbb{Y}}/I_{\mathbb{X}}$. Its initial degree $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}=\min\{k\in\mathbb{N}\mid(I_{{\mathbb{Y}}/{\mathbb{X}}})_{k}\neq 0\}$ is called the separator degree of ${\mathbb{Y}}$ in ${\mathbb{X}}$. Then the Hilbert function of ${\mathbb{Y}}$ satisfies

[TABLE]

and we have $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}\leq r_{\mathbb{X}}$. 2. (b)

A non-zero homogeneous element $f^{\ast}_{\mathbb{Y}}\in I_{{\mathbb{Y}}/{\mathbb{X}}}$ of degree $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}$ is called a minimal separator of ${\mathbb{Y}}$ in ${\mathbb{X}}$. This case we have $(I_{{\mathbb{Y}}/{\mathbb{X}}})_{\alpha_{{\mathbb{Y}}/{\mathbb{X}}}+i}=K\,x_{0}^{i}\,f^{\ast}_{\mathbb{Y}}$ for every $i\geq 0$. 3. (c)

Using this terminology, the following conditions are equivalent for every $d\geq 1$ (see also [10], Proposition 2.1):

• (1)

${\mathbb{X}}$ has ${\rm CBP}(d)$.

• (2)

Every subscheme ${\mathbb{Y}}$ of ${\mathbb{X}}$ of degree $\deg({\mathbb{X}})-1$ satisfies $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}\geq d+1$.

• (3)

No element of $(I_{{\mathbb{Y}}/{\mathbb{X}}})_{r_{\mathbb{X}}}\setminus\{0\}$ is divisible by $x_{0}^{r_{\mathbb{X}}-d}$.

Notice that the number $r_{{\mathbb{X}}}-1$ is the largest degree $d\geq 0$ such that ${\mathbb{X}}$ can have CBP($d$). Thus ${\mathbb{X}}$ is a Cayley-Bacharach scheme if and only if it has the CBP with respect to the largest possible degree. The Kähler differential module of a subscheme ${\mathbb{Y}}$ of ${\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$ can be described as follows.

Proposition 8.3**.**

Let ${\mathbb{Y}}$ be a proper subscheme of ${\mathbb{X}}$, and let $R_{{\mathbb{Y}}}=R/I_{{\mathbb{Y}}/{\mathbb{X}}}$ be the homogeneous coordinate ring of ${\mathbb{Y}}$.

1. (a)

We have $\Omega^{1}_{R_{{\mathbb{Y}}}/K}\cong\Omega^{1}_{R/K}/dI_{{\mathbb{Y}}/{\mathbb{X}}}$. Moreover, for a system of non-zero homogeneous generators $\{g_{1},\dots,g_{r}\}$ of $I_{{\mathbb{Y}}/{\mathbb{X}}}$, we have

[TABLE] 2. (b)

If $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$ then we have $dI_{{\mathbb{Y}}/{\mathbb{X}}}=\langle df_{{\mathbb{Y}}}^{\ast},\,f_{{\mathbb{Y}}}^{\ast}\,dx_{0}\rangle$, where $f_{{\mathbb{Y}}}^{\ast}$ is a minimal separator of ${\mathbb{Y}}$ in ${\mathbb{X}}$. 3. (c)

If $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>r_{{\mathbb{X}}}$ then we have $(dI_{{\mathbb{Y}}/{\mathbb{X}}})_{\alpha_{{\mathbb{Y}}/{\mathbb{X}}}}\neq 0$.

Proof.

Claim (a) is a consequence of Proposition 5.2.d. Claim (b) follows from $(I_{{\mathbb{Y}}/{\mathbb{X}}})_{k}=K\,x_{0}^{k-\alpha_{{\mathbb{Y}}/{\mathbb{X}}}}\,f_{{\mathbb{Y}}}^{\ast}$ for $k\geq\alpha_{{\mathbb{Y}}/{\mathbb{X}}}$.

It remains to prove (c). Let $g\in I_{{\mathbb{Y}}/{\mathbb{X}}}$ be a non-zero homogeneous generator of least degree $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}$. Since $g$ is homogeneous, its image $g^{\rm deh}$ is non-zero in the affine coordinate ring $S=R/\langle x_{0}-1\rangle$ of ${\mathbb{X}}$. By (a), it suffices to show $dg\neq 0$.

Let $\varPhi:\;{\Omega^{1}_{R/K}}\longrightarrow{\Omega^{1}_{\widetilde{R}/K}}\cong S[x_{0}]dx_{0}\oplus K[x_{0}]\otimes_{K}{\Omega^{1}_{S/K}}$ be the canonical $K[x_{0}]$-linear map constructed in Proposition 3.8. The composition of $\varPhi$ with the projection to the first summand $\Theta:{\Omega^{1}_{R/K}}\stackrel{{\scriptstyle\varPhi}}{{\rightarrow}}{\Omega^{1}_{\widetilde{R}/K}}\rightarrow S[x_{0}]dx_{0}$ satisfies $\Theta(fdx_{i})=f^{\rm deh}x_{i}x_{0}^{k}dx_{0}$ for $i=1,\dots,n$ and $\Theta(fdx_{0})=f^{\rm deh}x_{0}^{k}dx_{0}$ for $f\in R_{k}$. Using $dg=\frac{\partial g}{\partial x_{0}}dx_{0}+\frac{\partial g}{\partial x_{1}}dx_{1}+\cdots+\frac{\partial g}{\partial x_{n}}dx_{n}$, we obtain

[TABLE]

where the last equality follows from Euler’s relation. Since $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}\neq 0$ in $K$ and $g^{\rm deh}\neq 0$ in $S$, we get $\Theta(dg)\neq 0$. Therefore we have $dg\neq 0$, as desired. ∎

This proposition yields the following characterization of the Cayley-Bacharach property of ${\mathbb{X}}$ using modules of Kähler differentials.

Corollary 8.4**.**

Suppose that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>r_{\mathbb{X}}$. Then the scheme ${\mathbb{X}}$ has CBP($d$) if and only if $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(d)=\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{\mathbb{Y}}/K}}(d)$ for every subscheme ${\mathbb{Y}}\subseteq{\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$.

Proof.

First we assume that ${\mathbb{X}}$ has CBP($d$). Then $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}\geq d+1$ by Remark 8.2.c, and hence the homogeneous generators of $dI_{{\mathbb{Y}}/{\mathbb{X}}}$ have degree $\geq d+1$. This implies $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(d)=\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{\mathbb{Y}}/K}}(d)$.

Conversely, assume that ${\mathbb{X}}$ does not have CBP($d$) and that ${\mathbb{Y}}$ is a subscheme of ${\mathbb{X}}$ with $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$ and $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}\leq d$. Then Proposition 8.3.c yields $(dI_{{\mathbb{Y}}/{\mathbb{X}}})_{\alpha_{{\mathbb{Y}}/{\mathbb{X}}}}\neq 0$, and hence $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(d)>\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{\mathbb{Y}}/K}}(d)$. ∎

Recall our assumption that the scheme ${\mathbb{X}}$ has $K$-rational support. In this way we make sure that there exist subschemes ${\mathbb{Y}}$ of ${\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-1$, because at each point in the support of ${\mathbb{X}}$ the local ring contains a socle element that generates an ideal which is a 1-dimensional $K$-vector space.

In Sections 2 and 3 of [9], the notion of a Cayley-Bacharach scheme was generalized to arbitrary 0-dimensional schemes ${\mathbb{X}}$ in $\mathbb{P}^{n}$. The following example shows that Corollary 8.4 fails to hold when the scheme ${\mathbb{X}}$ does not have $K$-rational support.

Example 8.5**.**

Consider the 0-dimensional scheme ${\mathbb{X}}$ in $\mathbb{P}^{2}$ over $\mathbb{Q}$ defined by $I_{\mathbb{X}}=\langle(X_{1}-X_{0})^{2},X_{2}^{3}+2X_{0}^{2}X_{2}+X_{0}^{3}\rangle$. Then $\deg({\mathbb{X}})=6$ and ${\mathbb{X}}$ does not have $K$-rational support.

The scheme ${\mathbb{X}}$ does not have subschemes of degree $\deg({\mathbb{X}})-1$. Instead, it has a unique maximal subscheme ${\mathbb{Y}}$ of degree $3$, namely the scheme defined by $I_{{\mathbb{Y}}}=\langle X_{1}-X_{0},X_{2}^{3}+2X_{0}^{2}X_{2}+X_{0}^{3}\rangle$. We compute $\mathop{\rm HF}\nolimits_{\mathbb{X}}:\ 1\ 3\ 5\ 6\ 6\cdots$, $r_{\mathbb{X}}=3$, and $\mathop{\rm HF}\nolimits_{\mathbb{Y}}:\ 1\ 2\ 3\ 3\cdots$. Thus the scheme ${\mathbb{X}}$ is a Cayley-Bacharach scheme in the sense of [9], Definition 3.10.

However, further calculations show that we have $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}:0\ 3\ 8\ 12\ 12\ 10\ 9\ 9\cdots$ and $\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{{\mathbb{Y}}}/K}}:0\ 2\ 4\ 5\ 4\ 3\ 3\cdots$. In particular, we have $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(2)\neq\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{{\mathbb{Y}}}/K}}(2)$, and hence Corollary 8.4 does not hold.

The Cayley-Bacharach property can be interpreted as a weak uniformity condition on the scheme ${\mathbb{X}}$. It is a special case of the following more general property.

Definition 8.6**.**

Let $1\leq i<\deg({\mathbb{X}})$ and $1\leq j\leq r_{\mathbb{X}}-1$. We say that the scheme ${\mathbb{X}}$ is $(i,j)$-uniform, if every subscheme ${\mathbb{Y}}\subseteq{\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-i$ satisfies $\mathop{\rm HF}\nolimits_{\mathbb{Y}}(j)=\mathop{\rm HF}\nolimits_{\mathbb{X}}(j)$.

In this terminology, the scheme ${\mathbb{X}}$ has CBP($d$) iff it is $(1,d)$-uniform. Another well-known case is $(\deg({\mathbb{X}})-n-1,1)$-uniformity which is also called linearly general position. If all possible $(i,j)$-uniformities are satisfied, the scheme ${\mathbb{X}}$ is commonly said to be in uniform position. Using Proposition 8.3.c, we can characterize $(i,j)$-uniformity in terms of Hilbert functions of Kähler differential modules as follows.

Corollary 8.7**.**

Suppose that $\mathop{\rm char}\nolimits(K)=0$ or $\mathop{\rm char}\nolimits(K)>r_{\mathbb{X}}$. Then the scheme ${\mathbb{X}}$ is $(i,j)$-uniform if and only if $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(j)=\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{\mathbb{Y}}/K}}(j)$ for every subscheme ${\mathbb{Y}}\subseteq{\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-i$.

Proof.

For a subscheme ${\mathbb{Y}}$ of ${\mathbb{X}}$ of degree $\deg({\mathbb{Y}})=\deg({\mathbb{X}})-i$, we have the exact sequence

[TABLE]

Clearly, the condition $\mathop{\rm HF}\nolimits_{\mathbb{Y}}(j)=\mathop{\rm HF}\nolimits_{\mathbb{X}}(j)$ is equivalent to $\alpha_{{\mathbb{Y}}/{\mathbb{X}}}>j$. By Proposition 8.3.c, this is in turn equivalent to $(dI_{{\mathbb{Y}}/{\mathbb{X}}})_{j}=\langle 0\rangle$, and hence to $\mathop{\rm HF}\nolimits_{{\Omega^{1}_{R/K}}}(j)=\mathop{\rm HF}\nolimits_{\Omega^{1}_{R_{\mathbb{Y}}/K}}(j)$. ∎