The $a$-number of Certain Hyperelliptic Curves

TL;DR

This paper derives formulas for the $a$-number of specific hyperelliptic curves, expanding understanding of their algebraic properties for infinitely many cases, including curves defined by $y^2= x^m+1$ and $y^2= x^m+x$.

Contribution

It provides explicit formulas for the $a$-number of hyperelliptic curves of the form $y^2= x^m+1$ and $y^2= x^m+x$, for infinitely many values of $m$, which was previously unknown.

Findings

Formulas for the $a$-number of $y^2= x^m+1$ curves.

Formulas for the $a$-number of $y^2= x^m+x$ curves.

Extension of $a$-number computations to infinitely many cases.

Abstract

In this paper, we compute a formula for the $a$-number of certain hyperelliptic curves given by the equation $y^2= x^m+1$ for infinitely many values of $m$. The same question is studied for the curve corresponding to $y^2= x^m+x$.

Equations

$$X\to 1\otimes X+X\otimes 1.$$

$$y^2=x^m+1\text{mbox}or{y}^2=x^m+x$$

$$\mathcal{C}(\omega ):=f_{p-1}dx.$$

$$\mathscr{C}(x^{j}dx)=\left\{\begin{array}[]{ccc}0&\mbox{if}&\hskip 11.38092ptp\nmid j+1\\ x^{s-1}dx&\mbox{if}&\hskip 11.38092ptj+1=ps.\\ \end{array}\right.$$

$$y^2=f(x)$$

$$\mathcal{B}=\{{\omega }_i=\frac{x^{i-1}dx}{y},i=1,\dots ,g\}.$$

$${\omega }_i=\frac{x^{i-1}dx}{y}=x^{i-1}{y}^{-p}{y}^{p-1}dx=y^{-p}{x}^{i-1}\sum _{j=0}^N{c}_j{x}^{j}dx,$$

$$y^{p-1}=f(x{)}^{(q-1)/2}=\sum _{j=0}^N{c}_j{x}^j\text{mbox}withN=\frac{p-1}{2}(d).$$

$${\omega }_i=$$

$$\mathcal{C}({\omega }_i)=\sum _{l=0}^{g-1}{(c_{(l+1)p-i})}^{1/p}.\frac{x^l}{y}dx.$$

$$\mathcal{C}(\omega )=M(\mathcal{X}{)}^{(1/p)}\omega ,$$

$$M(\mathcal{X})=\left(\begin{array}[]{cccc}c_{q-1}&c_{p-2}&\ldots&c_{p-g}\\ c_{2p-1}&c_{2p-2}&\ldots&c_{2p-g}\\ \vdots&\ldots&\ldots&\vdots\\ c_{gp-1}&c_{gp-2}&\ldots&c_{gp-g}\end{array}\right).$$

$$\begin{array}[]{ccc}\mathscr{C}(w_{i})&=&\frac{1}{y}\mathscr{C}(x^{i-1}y^{p-1}dx)\\ &=&\frac{1}{y}\mathscr{C}((x^{m}+1)^{\frac{p-1}{2}}x^{i-1}dx)\\ &=&\frac{1}{y}\mathscr{C}(\sum_{j=0}^{\frac{p-1}{2}}a_{j}x^{j+i-1}dx)\neq 0,\\ \end{array}$$

$$i+mj-1\equiv p-1$$

$$m(p-1-h)+i-1\equiv p-1$$

$$\frac{(k+1)(p-1)}{2}.$$

$$\frac{k(p-1)}{2}.$$

$$i-h-1\equiv 0$$

$$i\equiv h+1$$

$$\frac{p}{2}\le lp+h+1\le \frac{3p-1}{2}.$$

$$\Bigg{\{}\begin{array}[]{c}l\geq 0\\ l<3/2\\ \end{array}$$

$$i\equiv h+1$$

$$\frac{(2k-1)p}{2}\le lp+h+1\le \frac{(2k+1)p-1}{2}.$$

$$\Bigg{\{}\begin{array}[]{c}l\geq\dfrac{2k-1}{2}\\ l<\dfrac{2k+1}{2}\\ \end{array}$$

$$\text{mbox}rank(M_{(2k+1)p+1})=\text{mbox}rank(M_{(2k-1)p+1})+\frac{1}{2}(p+1).$$

$$a({\mathcal{X}}_{(2k+1)p+1})=g({\mathcal{X}}_{(2k+1)p+1})-rank(M_{(2k+1)p+1})$$

$$i-h-1\equiv 0$$

$$i\equiv h+1$$

$$1\le lp+h+1\le 2p.$$

$$\Bigg{\{}\begin{array}[]{c}l\geq 0\\ l<2\\ \end{array}$$

Full text

1. Introduction

Let $k$ be an algebraically closed field of characteristic $p>0$. Let $\mathcal{A}$ be an abelian variety defiend over $k$. Let $\alpha_{p}$ be the group scheme $\mbox{Spec}(k[X]/(X^{p}))$ with co-multiplication given by

[TABLE]

The group $\mbox{Hom}(\alpha_{p},A)$ can be considered as $k$-vector space since $\mbox{End}(\alpha_{p})=k$. The $a$-number $a(\mathcal{A})$ defined to be the dimension of the vector space $\mbox{Hom}(\alpha_{p},A)$.

Let $\mathcal{X}$ be a (non-singular, projective, geometrically irreducible, algebraic) curve defined over $k$. One can define the $a$-number $a(\mathcal{X})$ of $\mathcal{X}$ as the $a$-number of its Jacobian variety $\mathcal{J}_{\mathcal{X}}$. As a matter of fact, the $a$-number of a curve is a birational invariant which can defined as the dimension of the space of exact holomorphic differentials.

The $a$-number of Hermitian curves computed by Gross in [10], and for Fermat and Hurwitz curves computed by Maria [14]. A few results on the rank of the Carteir operator (especially $a$-number) of curves introduced by Kodama and Washio [11], González [8], Pries and Weir [15] and Yui [22].

In this work, we consider the hyperelliptic curve $\mathcal{X}$ given by the equation

[TABLE]

over $k$.

These families of hyperelliptic curves have been investigated for several reasons by many authors (see [12], [19], [18], [21]). Here we are going to determine the $a$-number $a(\mathcal{X})$ of $\mathcal{X}$ for infinitely many values of $m$. See Theorem 3.1, 3.2 and 4.1.

2. The Cartier operator

Let $k$ be an algebraically closed field of characteristic $p>0$. Let $\mathcal{X}$ be a curve defined over $k$. The Cartier operator is a $p$-linear operator acting on the sheaf $\Omega^{1}_{\mathcal{X}}$ of differential forms on $\mathcal{X}$ in positive characteristic.

Let $K=k(\mathcal{X})$ be the function field of a curve $\mathcal{X}$ of genus $g$ defined over $k$. A separating variable for $K$ is an element $x\in K\setminus K^{p}$.

Definition 2.1**.**

(The Cartier operator). Let $\omega\in\Omega_{K/k}$. There exists $f_{0},\cdots,f_{p-1}$ such that $\omega=(f^{p}_{0}+f^{p}_{1}x+\cdots+f^{p}_{p-1}x^{p-1})dx$. The Cartier operator $\mathscr{C}$ is defined by

[TABLE]

The definition does not depend on the choice of $x$ (see [[16], Proposition 1]).

We refer the reader to [[1], [2],[16], [20]] for the proofs of the following statements.

Proposition 2.2**.**

(Global Properties of $\mathscr{C}$). For all $\omega\in\Omega_{K/k}$ and all $f\in F$,

$\mathscr{C}(f^{p}\omega)=f\mathscr{C}(\omega)$;

• 2.

$\mathscr{C}(\omega)=0\Leftrightarrow\exists h\in K,\omega=dh$;

• 3.

$\mathscr{C}(\omega)=\omega\Leftrightarrow\exists h\in K,\omega=dh/h$.

Remark 2.3**.**

Moreover, one can easily show that

[TABLE]

If $div(\omega)$ is effective then differential $\omega$ is holomorphic. The set $H^{0}(\mathcal{X},\Omega^{1})$ of holomorphic differentials is a $g$-dimensional $k$-vector subspace of $\Omega^{1}$ such that $\mathscr{C}(H^{0}(\mathcal{X},\Omega^{1}))\subseteq H^{0}(\mathcal{X},\Omega^{1})$. If $\mathcal{X}$ is a curve, then the $a$-number of $\mathcal{X}$ equals the dimension of the kernel of the Cartier operator $H^{0}(\mathcal{X},\Omega^{1})$ (or equivalently, the dimension of the space of exact holomorphic differentials on $\mathcal{X}$) (see [11, 5.2.8]).

The Cartier operator and Hasse-Witt-matrix are dual to each other under the duality given by the Riemann-Roch theorem. Let $\mathcal{B}=\{\omega_{1},\cdots,\omega_{g}\}$ be a basis of the $k$-module of holomorphic differentials in $H^{0}(\mathcal{X},\Omega^{1})$. Then the representation matrix $M$ over $k$ of $\mathscr{C}$ with respect to this basis is called the Hasse-Witt matrix.

Let $k$ be a field of characteristic $p>2.$ Let $\mathcal{X}$ be a projective nonsingular hyperelliptic curve over $k$ of genus $g$. Then $\mathcal{X}$ can be defined by an affine equation of the form

[TABLE]

where $f(x)$ is a polynomial over $k$ of degree $d=2g+1$ or $d=2g+2$ without multiple roots.

The differential 1-forms of the first kind on $\mathcal{X}$ form a $k-$vector space $H^{0}(\mathcal{X},\Omega^{1})$ of dimension $g$ with basis

[TABLE]

The images under the operator $\mathscr{C}$ are determined in the following way (see [22]). Rewrite

[TABLE]

where the coefficients $c_{j}\in k$ are obtained from the expansion

[TABLE]

Then we get for $i=1,\ldots,g,$

[TABLE]

Note here that $0\leq l\leq\frac{N+i}{p}-1<g-\frac{1}{2}.$ On the other hand, we know from Remark 2.3 that if $\mathscr{C}(x^{r-1}dx)\neq 0$ then $r\equiv 0$ (mod $p$). Thus we have

[TABLE]

If we write $\omega=(\omega_{1},\ldots,\omega_{g})$ as a row vector we have

[TABLE]

where $M(\mathcal{X})$ is the $(g\times g)$ matrix with elements in $k$ given as

[TABLE]

3. The $a$-number of Hyperelliptic Curve $y^{2}=x^{m}+1$

In this section, we consider the hyperelliptic curve $\mathcal{X}$ given by the equation $y^{2}=x^{m}+1$ over $k$. This curve is of genus $g=(m-1)/2$ (resp. $g=(m-2)/2$) if $m$ is odd (resp. $m$ is even).

Let $\mathcal{B}=\{\omega_{i}=\frac{x^{i-1}dx}{y},\hskip 5.69046pti=1,\ldots,g\}$ be a basis for the differential 1-forms of the first kind on $\mathcal{X}$. Then the rank of the Cartier operator $\mathscr{C}$ on the curve $\mathcal{X}$ equals the number of $i$ with $i\leq g$ such that

[TABLE]

where $(x^{m}+1)^{\frac{p-1}{2}}=\sum_{j=0}^{\frac{p-1}{2}}a_{j}x^{jm}.$ From this we must have the equation of congruences mod $p$,

[TABLE]

for some $0\leq j\leq\frac{(p-1)}{2}$. Equivalently, the following equation

[TABLE]

has a solution $h$ for $0\leq h\leq\dfrac{p-1}{2}$.

For the rest of this section, $M_{m}:=M(\mathcal{X})$ is the matrix representing the $p$-th power of the Cartier operator $\mathscr{C}$ on the curve $\mathcal{X}$ with respect to the basis $\mathcal{B}$.

Theorem 3.1**.**

Let $\mathcal{X}$ be a hyperelliptic curve given by the equation $y^{2}=x^{m}+1$. Suppose that $m=sp+1$, then

If $s=2k+1$ and $k\geq 0$, then the $a$-number of the curve $\mathcal{X}$ equals

[TABLE]

• 2.

If $s=2k$ and $k\geq 1$, then the $a$-number of the curve $\mathcal{X}$ equals

[TABLE]

Proof.

• (1).

At the first, if $m=sp+1=(2k+1)p+1$ with $k\geq$, then we prove that $\mbox{rank}(M_{m})=\dfrac{k(p+1)}{2}$.

In this case, $i\leq g$ and Equation (3.2) mod $p$ reads

[TABLE]

In particular, if $k=0$ then $m=p+1$, where $i\leq g$ and Equation (3.3) be transformed into

[TABLE]

Take $l\in\mathbb{Z}_{0}^{+}$ so that $i=lp+h+1$, then $1\leq lp+h+1\leq\dfrac{p-1}{2}$. This implies that $h\geq 0$ and $h<-3/2$, a contradictions. Thus, $rank(M_{p+1})=0$.

If $k=1$ then $m=3p+1$, in this case we have $\dfrac{p}{2}\leq i\leq\dfrac{3p-1}{2}$. We need to find the solutions $h$ mode $p$ of the Equation (3.4). Then

[TABLE]

As $h+1\geq 0$ we obtain

[TABLE]

Thus, we have two choices for $l$, i.e, $l=0$ or $l=1$. From this we have $\frac{1}{2}(p+1)$ choices for $h$, and so we conclude $\mbox{rank}(M_{3p+1})=\frac{1}{2}(p+1)$.

For $k\geq 2$, and $m=sp+1$ we can say $\mbox{rank}(M_{(2k+1)p+1})$ equals

$\mbox{rank}(M_{(2k-1)p+1})$ plus the number of $i$ such that there is $h$ solution of the equation mod $p$

[TABLE]

with $\dfrac{(2k-1)p}{2}\leq i\leq\dfrac{(2k+1)p-1}{2}$. Then

[TABLE]

This implies that

[TABLE]

or equivalently we obtain $k=l$. In this case we have $\frac{1}{2}(p+1)$ choices for $h$. Therefore we get

[TABLE]

Now the our claim on the rank of $M_{(2k+1)p+1}$ follows by induction on $k$.

Then $a(\mathcal{X}_{(2k+1)p+1})=\dfrac{(k+1)(p-1)}{2}$ can be computed from

[TABLE]

.

• (2.)

At first we cliam that $rank(M_{sp+1})=\dfrac{k(p+1)}{2}$, with $m=2kp+1$ and $k\geq 1$. In this case, $i\leq g$ and Equation 3.2 mod $p$ reads

[TABLE]

In particular, if $k=1$ then $m=2p+1$, where $i\leq g$ and Equation 3.5 be transformed into

[TABLE]

Take $l\in\mathbb{Z}_{0}^{+}$ so that $i=lp+h+1$, then $1\leq lp+h+1\leq p$. Thus, we have one choices for $l$. From this we have $\frac{1}{2}(p+1)$ choices for $h$, and yielding $rank(M_{2p+1})=\frac{1}{2}(p+1)$.

If $k=2$, then $m=4p+1$, in this case we have $1\leq i\leq 2p$. We need to find the solutions $h$ mode $p$ of the above Equation 3.6. Then

[TABLE]

As $h+1\geq 0$

[TABLE]

Thus, we have two choices for $l$, i.e, $l=0$ or $l=1$. From this we have $(p+1)$ choices for $h$, and yielding $rank(M_{4p+1})=(p+1)$.

For $k\geq 3$, and $m=sp+1$ we can say $rank(M_{2kp+1})$ equals $rank(M_{2(k-1)p+1})$ plus the number of $i$ such that there is $h$ solution of the equation mod $p$

[TABLE]

with $1\leq i\leq 2kp$. Then

[TABLE]

Hence,

[TABLE]

In this case we have $\frac{1}{2}(p+1)$ choices for $h$. This implies that

[TABLE]

Now our claim on the rank of $M_{2kp+1}$ follows by induction on $k$.

Then $a(\mathcal{X}_{2kp+1})=\dfrac{(k)(p-1)}{2}$ can be computed from

[TABLE]

.

∎

Theorem 3.2**.**

Suppose that $m=sp-1$ then,

If $s=2k+1$ and $k\geq 0$, then the $a$-number of the curve $\mathcal{X}$ equals

[TABLE]

• 2.

If $s=2k$ and $k\geq 1$, then the $a$-number of the curve $\mathcal{X}$ equals

[TABLE]

Proof.

Proof of this theorem is similar to Theorem 3.1.

∎

4. The $a$-number of Hyperelliptic Curve $y^{2}=x^{m}+x$

In this section, we consider the hyperelliptic curve $\mathcal{X}$ given by the equation $y^{2}=x^{m}+x$ over $k$. This curve is of genus $g=(m-1)/2$ (resp. $g=(m-2)/2$) if $m$ is odd (resp. $m$ is even).

Let $\mathcal{B}=\{\omega_{i}=\frac{x^{i-1}dx}{y},\hskip 5.69046pti=1,\ldots,g\}$ be a basis for the differential 1-forms of the first kind on $\mathcal{X}$. Then the rank of the Cartier operator $\mathscr{C}$ on the curve $\mathcal{X}$ equals the number of $i$ with $i\leq g$ such that

[TABLE]

where $(x^{m-1}+1)^{\frac{p-1}{2}}=\sum_{j=0}^{\frac{p-1}{2}}a_{j}x^{j(m-1)}.$ From this we must have the equation of congruences mod $p$,

[TABLE]

for some $0\leq j\leq\frac{(p-1)}{2}$. Equivalently, the following equation

[TABLE]

has a solution $h$ for $0\leq t\leq h\leq\dfrac{p-1}{2}$.

Theorem 4.1**.**

If $m=sp$ for $s=2k+1$ and $k\geq 0$, then the $a$-number of the curve $\mathcal{X}$ equals

[TABLE]

Proof.

At first we cliam that $rank(M_{sp})=\dfrac{k(p+1)}{2}$, with $m=(2k+1)p$ and $k\geq 0$.

In this case, $i\leq g$ and Equation 4.2 mod $p$ reads

[TABLE]

Peculiarly, if $k=0$ then $m=p$, where $i\leq g$ and Equation 4.3 be transformed into

[TABLE]

Take $l\in\mathbb{Z}_{0}^{+}$ so that $i=lp-t$, then $1\leq lp-t\leq\dfrac{p}{2}$. From this $t\geq-1$ and $t\geq 0$, a contradictions. Thus, $rank(M_{p})=0$.

If $k=1$ then $m=3p$, in this case we have $\dfrac{p}{2}\leq i\leq\dfrac{3p}{2}$. We need to find the solutions $h$ mode $p$ of the above Equation 4.4. Then

[TABLE]

As $t\geq 0$

[TABLE]

Thus, we have two choices for $l$, i.e, $l=0$ or $l=1$. From this we have $\frac{1}{2}(p+1)$ choices for $t$, and yielding $rank(M_{3p})=\frac{1}{2}(p+1)$.

For $k\geq 2$, and $m=sp$ we can say $rank(M_{(2k+1)p})$ equals $rank(M_{(2k-1)p})$ plus the number of $i$ such that there is $t$ solution of the equation mod $p$

[TABLE]

with $\dfrac{(2k-1)p}{2}\leq i\leq\dfrac{(2k+1)p}{2}$. Then

[TABLE]

Hence,

[TABLE]

In this case we have $\frac{1}{2}(p+1)$ choices for $t$. This implies that

[TABLE]

Now our claim on the rank of $M_{(2k+1)p}$ follows by induction on $k$.

Then $a(\mathcal{X}_{(2k+1)p})=\dfrac{(k+1)(p-1)}{2}$ can be computed from

[TABLE]

∎

Acknowledgement

The third author was supported by FAPESP/SP-Brazil grant 2017/19190-5.