Some Results on Betti Series of Universal Modules of Differential Operators

TL;DR

This paper investigates the rationality of Betti series for universal modules of differential operators, establishing conditions under which the series is rational for certain algebraic structures.

Contribution

It proves the rationality of Betti series for universal modules of derivations over coordinate rings of affine irreducible curves with at most one singularity.

Findings

Betti series is rational for coordinate rings of affine irreducible curves with at most one singularity

Established conditions for the rationality of Betti series in this context

Provides new insights into the structure of universal modules of differential operators

Abstract

In this article, we discuss the rationality of the Betti Series of the universal module of nth order derivations of R_{m} where m is a maximal ideal of R. We proved that if R is a coordinate ring of an affine irreducible curve and if it has at most one singularity point, then the Betti series is rational.

Equations

$$0\to I\to R{\otimes }_{k}R\stackrel{\phi }{\to }R\to 0$$

$$d_n(r)=1\otimes r-r\otimes 1+I^{n+1}\text{ and }{d}_n(1)=0$$

$$\left\{d_n(x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}...x_s^{{\alpha }_s}):1\le {\alpha }_1+{\alpha }_2+...+{\alpha }_s\le n\right\}$$

$$\dots \to F_2\stackrel{{\partial }_2}{\to }{F}_1\stackrel{{\partial }_1}{\to }{F}_0\stackrel{\epsilon }{\to }{\Omega }_n(R)\to 0$$

$$B({\Omega }_n(R),t)=\sum _{i\ge 0}{\dim }_{R/m}Ex{t}^i({\Omega }_n(R),\frac{R}{m})t^i\text{ for all }n\ge 1.$$

$$0\to F_1\stackrel{\partial }{\to }{F}_0\to M\to 0$$

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

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Full text

††*Key Words: universal differential operator modules, minimal resolution * Mathematics subject classification 2010: 13N05

1 Introduction and Preliminaries

The following notations will be fixed throughout the paper: ring means commutative with identity and $R$ is a commutative $k$-algebra where $k$ is an algebraically closed field of characteristic zero.

An $n$th order $k$-derivation $D$ of $R$ into an $R$-module $F$ is an element of $Hom_{k}(R,F)$ such that for any $n+1$ elements $r_{0},r_{1},\ldots,r_{n}$ of $R$, the following identity holds:

$D(r_{0}r_{1}\ldots r_{n})=\sum\limits_{i=1}^{n}(-1)^{i-1}\sum\limits_{j_{1}<j_{2}<\ldots<j_{i}}r_{j_{1}}\ldots r_{j_{i}}D(r_{0}\ldots\hat{r}_{j_{1}}\ldots\hat{r}_{j_{i}}\ldots r_{n})$

where the hat over $r_{i}$’s means that it is missed. It can be easily seen that a first order derivation is the ordinary derivation of $R$ into an $R$-module $F$.

In [5], a universal object for $n$th order derivations constructed in the following way: Consider the exact sequence

[TABLE]

where $\varphi$ is defined as $\varphi(\sum\limits_{i=1}^{n}a_{i}\otimes b_{i})=\sum\limits_{i=1}^{n}a_{i}b_{i}$ for $a_{i},b_{i}\in R$ and $I$  is the kernel of $\varphi$ . It is known that $\ker\varphi$ is generated by the set

{$1\otimes r-r\otimes 1:r\in R$}

as an $R$-module. Then the mapping $d_{n}$ from $R$ into $I/I^{n+1}$ given by

[TABLE]

is called the universal derivation of order $n$ that is, any $n$th order derivation $D$ from $R$ into $F$ can be factored through $I/I^{n+1}$ where $F$ is an $R$-module. Here, the $R$-module $I/I^{n+1}$ is called the universal module of $n$th order derivations and is denoted by $\Omega_{n}(R)$.

Note that if $R$ is a finitely generated $k$-algebra, then $\Omega_{n}(R)$ is a finitely generated $R$-module. It is proved in [5, Prop. 2] that if $R=k[x_{1},...,x_{s}]$ is a polynomial algebra over $k$ with $s$ variables, then $\Omega_{n}(R)$ is a free $R$-module of rank $\binom{n+s}{s}-1$ with basis

[TABLE]

and in [5, Theo. 9] that

$\Omega_{n}(R)\otimes_{R}R_{S}\cong\Omega_{n}(R_{S})$

where $S$ is a multiplicatively closed subset of $R$.

A free resolution of $\Omega_{n}(R)$ where $R$ is a local $k$-algebra with maximal ideal $m$ is called a minimal resolution if the followings are satisfied:

[TABLE]

$F_{i}$’s are free $R$-modules of finite rank for all $i$ and $\partial_{n}(F_{n})\subseteq mF_{n-1}$ for all $n\geq 1$ (see [4] for definition).

Let $(R,m)$ be a local ring. The Betti series of $\Omega_{n}(R)$ is defined to be the series

[TABLE]

Lemma 1

Let $R$ be a local ring with maximal ideal $m$ and $M$ be a finitely generated $R$-module. Suppose that

[TABLE]

is a minimal resolution of $M.$ Then $Ext^{1}(M,R/m)$ is not zero.

Dealing with $n$th order case presents extra difficulties. Let $R=k[x,y,z]$ be a $k$-algebra with $z^{2}=x^{3}$ and $y^{2}=xz$. It is shown in [1, ex. 3.1.6 and 3.4.7] that $pd(\Omega_{1}(R))\leq 1$, but $pd(\Omega_{2}(R))$ is not finite.

In [6], the following proposition is proved:

Proposition 2

Let $R=k[x_{1},\ldots,x_{s}]$ and $S=k[y_{1},\ldots,y_{t}]$ be polynomial algebras and let $I$ be an ideal of $R$ generated by elements $\{f_{1},\ldots,f_{m}\}$. Assume that $R/I=k[x_{1},\ldots,x_{s}]/(f_{1},\ldots,f_{m})$ is an affine $k$-algebra of dimension $s-m$ and $pd(\Omega_{2}(R/I))\leq 1$. Then $pd(\Omega_{2}(R/I\otimes_{k}S))\leq 1$.

But, unfortunately, this result is not true even for $n=3$.    So, there are two natural questions arise from these results. Can we generalize the dimension of the ring $R$? Can we generalize the dimension of the universal module $\Omega_{n}$? In [2] and [3], it is studied the following question:

When is the Betti series of a universal module of second order derivations rational?

Our goal is to establish a result analogue of this question for $n$th order universal differential operator modules.

2 Main Results

Proposition 3

Let $k[x_{1},x_{2},\ldots,x_{s}]$ be a polynomial algebra and $m$ be a maximal ideal of $k[x_{1},x_{2},\ldots,x_{s}]$ containing an irreducible element $f$. If the elements $d_{n}(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{s}^{\alpha_{s}}f)$ belong to $m\Omega_{n}(k[x_{1},x_{2},\ldots,x_{s}])$ whenever $0\leq\alpha_{1}+\alpha_{2}+\ldots+\alpha_{s}\leq n-1$, then $\Omega_{n}(\frac{k[x_{1},x_{2},\ldots,x_{s}]}{(f)})_{\overset{-}{m}}$ admits a minimal resolution of $(\frac{k[x_{1},x_{2},\ldots,x_{s}]}{(f)})_{\overset{-}{m}}-$modules where $\overset{-}{m}=m/(f)$ is a maximal ideal of $\frac{k[x_{1},x_{2},\ldots,x_{s}]}{(f)}$.

Proof. Let $R=S/I=\frac{k[x_{1},x_{2},\ldots,x_{s}]}{(f)}$ and $\overset{-}{m}$ be a maximal ideal of $R$. Then by [5, Theo. 14 pg. 24] we have the following short exact sequence of $R$-modules

[TABLE]

where $N$ is a submodule of $\Omega_{n}(S)$ generated by the elements of the form

$\{d_{n}(g):g\in fk[x_{1},x_{2},\ldots,x_{s}]\}$.

By localizing (1) at $\overset{-}{m}$, we get the following exact sequence of $R_{\overset{-}{m}}-$modules:

[TABLE]

Step 1. A module generated by the set $\{d_{n}(g):g\in fk[x_{1},x_{2},\ldots,x_{s}]\}$ is a submodule of $m\Omega_{n}(k[x_{1},x_{2},\ldots,x_{s}])$.

Proof of Step 1. Since $d_{n}$ is $k$-linear, it suffices to show

$d_{n}(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{s}^{\alpha_{s}}f)\in m\Omega_{n}(k[x_{1},x_{2},\ldots,x_{s}])$.

By using the properties of $d_{n}$, we get

$d_{n}(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{s}^{\alpha_{s}}f)=\underset{\gamma}{\Sigma}a_{\gamma}(x_{1},x_{2},\ldots,x_{s})d_{n}(x_{1}^{\gamma_{1}}x_{2}^{\gamma_{2}}\ldots x_{s}^{\gamma_{s}}f)$ $+f(\underset{\beta}{\Sigma}a^{\prime}_{\beta}(x_{1},x_{2},\ldots,x_{s})d_{n}(x_{1}^{\beta_{1}}x_{2}^{\beta_{2}}\ldots x_{s}^{\beta_{s}})$ where $a_{\gamma}(x_{1},x_{2},\ldots,x_{s}),a^{\prime}_{\beta}(x_{1},x_{2},\ldots,x_{s})\in k[x_{1},x_{2},\ldots,x_{s}]$, $0\leq\gamma_{1}+\gamma_{2}+\ldots+\gamma_{s}\leq n-1$, $0<\beta_{1}+\beta_{2}+\ldots+\beta_{s}\leq n$. By assumption, we know

$d_{n}(x_{1}^{\gamma_{1}}x_{2}^{\gamma_{2}}\ldots x_{s}^{\gamma_{s}}f)\in m\Omega_{n}(k[x_{1},x_{2},\ldots,x_{s}])$

whenever $0\leq\gamma_{1}+\gamma_{2}+\ldots+\gamma_{s}\leq n-1$ and $f\in m$, then the result follows.

Step 2. $(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}\subseteq\overset{-}{m}(\frac{\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}$.

Proof of Step 2. By step 1, we know $N\subseteq m\Omega_{n}(S)$ and the rest is clear.

Step 3. $(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}$ is generated by $\binom{n+s-1}{s}$ elements.

Proof of Step 3. It is known that $(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})$ is generated by the set

$\{d_{n}(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{s}^{\alpha_{s}}f)+f\Omega_{n}(S):0\leq\alpha_{1}+\alpha_{2}+\ldots+\alpha_{s}\leq n-1\}$.

And, it has $\binom{n+s-1}{s}$ elements.

Step 4. $(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}$ is a free $R_{\overset{-}{m}}$-module.

Proof of Step 4. The Krull dimension of $R_{\overset{-}{m}}$ is $s-1$ and let $K$ be the field of fractions of $R_{\overset{-}{m}}$. Then by tensoring the exact sequence in (2) by $K$, we get

[TABLE]

We know that, $(\frac{\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}$ is a free $R_{\overset{-}{m}}$- module of rank $\binom{n+s}{s}-1$. By using the following isomorphism,

$K\otimes_{R_{\overset{-}{m}}}\Omega_{n}({R_{\overset{-}{m}}})\cong\Omega_{n}(K)$

we have

$dim$ $K\otimes_{R_{\overset{-}{m}}}(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}=$dimK$\otimes_{R_{\overset{-}{m}}}(\frac{\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}-dim\Omega_{n}(K)$ $=\binom{n+s}{s}-\binom{n+s-1}{s-1}=\binom{n+s-1}{s}$. Since the rank of $(\frac{N+f\Omega_{n}(S)}{f\Omega_{n}(S)})_{\overset{-}{m}}$ is equal to its number of minimal generators, it is a free $R_{\overset{-}{m}}$-module. Therefore, the short exact sequence given in (2) is a minimal resolution for $\Omega_{n}({R_{\overset{-}{m}}})$.

Let $R$ be a finitely generated regular $k$-algebra and $m$ be a maximal ideal of R then $\Omega_{n}({R_{m}})$ is a free $R_{m}$-module. Then it is clear that $B(\Omega_{n}({R_{m}},t))$ is rational.

Theorem 4

Let $k[x_{1},x_{2},\ldots,x_{s}]$ be a polynomial algebra and $m$ be a maximal ideal of $k[x_{1},x_{2},\ldots,x_{s}]$ containing an irreducible element $f$. Let $d_{n}(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{s}^{\alpha_{s}}f)\in m\Omega_{n}(k[x_{1},x_{2},\ldots,x_{s}])$ for $0\leq\alpha_{1}+\alpha_{2}+\ldots+\alpha_{s}\leq n-1$. Assume that $R=\frac{k[x_{1},x_{2},\ldots,x_{s}]}{(f)}$ is not a regular ring at $\overset{-}{m}=m/(f)$. Then $B(\Omega_{n}({R_{\overset{-}{m}}},t),)$ is a rational function.

Proof. By the previous proposition, the exact sequence of $R_{\overset{-}{m}}$-modules in (2) is a minimal resolution of $\Omega_{n}({R_{\overset{-}{m}}})$. And we get the result.

Question: It should be interesting to know whether the Betti Series is rational for the algebra $R=k[x,y,z]$ with $z^{2}=x^{3}$ and $y^{2}=xz$.

Acknowledgement

This work is a part of the first autor’s PhD thesis. The first author is grateful to TUBITAK for their financial support during her visit at the University of Sheffield.

Ali Erdogan

Hacettepe University

Department of Mathematics

Ankara,Turkey

e-mail:alier@hacettepe.edu.tr

Halise Melis Akcin

Hacettepe University

Department of Mathematics

Ankara,Turkey

e-mail:hmtekin@hacettepe.edu.tr