An explicit triangular integral basis for any separable cubic extension of a function field

TL;DR

This paper provides an explicit construction of a triangular integral basis for all separable cubic extensions of rational function fields over finite fields, along with a formula for their discriminants.

Contribution

It introduces a universal explicit basis for separable cubic extensions of function fields and derives a discriminant formula in a standard Galois closure form.

Findings

Explicit triangular integral basis for all such extensions

Discriminant formula in standard Galois tower form

Applicable in any characteristic

Abstract

We determine an explicit triangular integral basis for any separable cubic extension of a rational function field over a finite field in any characteristic. We obtain a formula for the discriminant of every such extension in terms of a standard form in a tower for the Galois closure.

Equations

$$({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=N_{L/{\mathbb{F}}_q(x)}(({\mathfrak{D}}_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]})=\prod _{\mathfrak{p}\in \mathcal{R}}{\mathfrak{p}}^{{\sum }_{\mathfrak{P}|\mathfrak{p}}d(\mathfrak{P}|\mathfrak{p})f(\mathfrak{P}|\mathfrak{p})},$$

$$X^3+b{X}^2+cX+d=0,$$

$$\Delta (\omega )=b^2{c}^2-4c^3-4b^{3}d-27d^2+18bcd.$$

$$(\Delta (\omega ){)}_{{\mathbb{F}}_q[x]}=\text{Ind}(\omega {)}^2({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}.$$

$$X^3-3{\gamma }^2{\beta }_1^2{\beta }_2^{2}X-\alpha {\beta }_1^2{\beta }_2,$$

$$\Delta (\omega )=108{\gamma }^6{\beta }_1^6{\beta }_2^6-27{\alpha }^2{\beta }_1^4{\beta }_2^2=27{\beta }_1^4{\beta }_2^2(4{\gamma }^6{\beta }_1^2{\beta }_2^4-{\alpha }^2)=27{\beta }_1^4{\beta }_2^2(4{\gamma }^6{\beta }^2-{\alpha }^2).$$

$$({\beta }_1^4{\beta }_2^2(4{\gamma }^6{\beta }^2-{\alpha }^2){)}_{{\mathbb{F}}_q[x]}=(\Delta (\omega ){)}_{{\mathbb{F}}_q[x]}=\text{Ind}(\omega {)}^2({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}.$$

$$({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=(({\beta }_1{\beta }_2{)}^2{\eta }_1{)}_{{\mathbb{F}}_q[x]}\text{ and }\text{Ind}(w)=({\beta }_1{\eta }_2{)}_{{\mathbb{F}}_q[x]}.$$

$$z^2-z=b,$$

$$v_{\mathfrak{p}}\left(b\right)<0,\left(v_{\mathfrak{p}}\left(b\right),2\right)=1,$$

$$v_{\mathfrak{p}}\left(b\right)\ge 0.$$

$$({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=(({\beta }_1{\beta }_2{)}^2{)}_{{\mathbb{F}}_q[x]}\prod _{\begin{array}{c}(v_{\mathfrak{p}}(b),2)=1\\ and{v}_{\mathfrak{p}}(b)<0\end{array}}{\mathfrak{p}}^{{\ell }_{\mathfrak{p}}+1}\text{ and }\text{Ind}(\omega )=({\beta }_1{)}_{{\mathbb{F}}_q[x]}\prod _{\mathfrak{p}\mid (\alpha {)}_{{\mathbb{F}}_q[x]}}{\mathfrak{p}}^{v_{\mathfrak{p}}(\alpha )-\frac{1}{2}({\ell }_{\mathfrak{p}}+1)}.$$

$$({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=\prod _{\begin{array}{c}\mathfrak{p}|(\beta {)}_{{\mathbb{F}}_q[x]}\end{array}}{\mathfrak{p}}^2\prod _{\begin{array}{c}\mathfrak{p}|({\alpha }^2-4{\gamma }^3{\beta }^2{)}_{{\mathbb{F}}_q[x]}\\ (v_{\mathfrak{p}}({\alpha }^2-4{\gamma }^3{\beta }^2),2)=1\end{array}}\mathfrak{p}=(({\beta }_1{\beta }_2{)}^2{\eta }_1{)}_{{\mathbb{F}}_q[x]}.$$

$$\text{Ind}(\omega {)}^2$$

$$=({\beta }_1^4{\beta }_2^2{\eta }_1{\eta }_2^2({\beta }_1{\beta }_2{)}^{-2}{\eta }_1^{-1}{)}_{{\mathbb{F}}_q[x]}$$

$$=({\beta }_1{\eta }_2{)}_{{\mathbb{F}}_q[x]}^2.$$

$$d(\mathfrak{P}|\mathfrak{p})=-v_{\mathfrak{p}}\left(b\right)+1={\ell }_{\mathfrak{p}}+1,$$

$$({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=\prod _{\begin{array}{c}\mathfrak{p}|(\beta {)}_{{\mathbb{F}}_q[x]}\end{array}}{\mathfrak{p}}^2\prod _{\begin{array}{c}\mathfrak{p}|(b{)}_{{\mathbb{F}}_q[x]}\\ (v_{\mathfrak{p}}(b),2)=1\\ and{v}_{\mathfrak{p}}(b)<0\end{array}}{\mathfrak{p}}^{{\ell }_{\mathfrak{p}}+1}=(({\beta }_1{\beta }_2{)}^2{)}_{{\mathbb{F}}_q[x]}\prod _{\begin{array}{c}\mathfrak{p}|(b{)}_{{\mathbb{F}}_q[x]}\\ (v_{\mathfrak{p}}(b),2)=1\\ and{v}_{\mathfrak{p}}(b)<0\end{array}}{\mathfrak{p}}^{{\ell }_{\mathfrak{p}}+1}.$$

$$({\beta }_1^4{\beta }_2^2{\alpha }^2{)}_{{\mathbb{F}}_q[x]}=(\Delta (\omega ){)}_{{\mathbb{F}}_q[x]}=\text{Ind}(\omega {)}^2({\partial }_{L/{\mathbb{F}}_q(x)}{)}_{{\mathbb{F}}_q[x]}=\text{Ind}(\omega {)}^2({\beta }_1{\beta }_2{)}_{{\mathbb{F}}_q[x]}^2\prod _{\mathfrak{p}\mid (\alpha {)}_{{\mathbb{F}}_q[x]}}{\mathfrak{p}}^{{\ell }_{\mathfrak{p}}+1},$$

$$\text{Ind}(\omega )=({\beta }_1{)}_{{\mathbb{F}}_q[x]}\prod _{\mathfrak{p}\mid (\alpha {)}_{{\mathbb{F}}_q[x]}}{\mathfrak{p}}^{v_{\mathfrak{p}}(\alpha )-\frac{1}{2}({\ell }_{\mathfrak{p}}+1)}.$$

$$X^3+bX+b^2=0$$

$$v_{\mathfrak{q}}(b)\ge 0,\text{ or }{v}_{\mathfrak{q}}(b)<0\text{ and }(v_{\mathfrak{q}}(b),3)=1.$$

$$X^3+aX+a^2=0,$$

$$a_2=\frac{(j{a}_1^2+(w_2^3+a_1{w}_2){)}^2}{a_1^3}.$$

$$v_{\mathfrak{p}}(\alpha )=v_{\mathfrak{p}}(\alpha -\lambda +\lambda )=\min \{v_{\mathfrak{p}}(\alpha -\lambda ),v_{\mathfrak{p}}(\lambda )\}=v_{\mathfrak{p}}(\lambda )=2v_{\mathfrak{p}}(a_1)/3.$$

$$v_{\mathfrak{p}}({\alpha }^{-3}j{a}_1^2+w_0)>0.$$

$$v_{\mathfrak{q}}(j{a}_1^2+(w_2^3+a_1{w}_2))$$

$$v_{\mathfrak{q}}(a_2)=v_{\mathfrak{q}}\left(\frac{(j{a}_1^2+(w_2^3+a_1{w}_2){)}^2}{a_1^3}\right)\ge 4v_{\mathfrak{q}}(a_1)-v_{\mathfrak{q}}(a_1^3)=v_{\mathfrak{q}}(a_1)>0.$$

$$v_{\mathfrak{q}}(j{a}_1^2+(w_2^3+a_1{w}_2))\ge \min \{v_{\mathfrak{q}}(j{a}_1^2),v_{\mathfrak{q}}(w_2^3+a_1{w}_2))\}=0.$$

$$v_{\mathfrak{q}}(a_2)=v_{\mathfrak{q}}\left(\frac{(j{a}_1^2+(w_2^3+a_1{w}_2){)}^2}{a_1^3}\right)\ge 0-v_{\mathfrak{q}}(a_1^3)=0.$$

$$v_{\mathfrak{q}}(j{a}_1^2)<v_{\mathfrak{q}}(w_2^3+a_1{w}_2),$$

Full text

Let $\mathbb{F}_{q}(x)$ be a rational field in one variable, where $q=p^{n}$ and $p$ is a prime integer. In this paper, our main theorems determine explicitly a triangular integral basis in any characteristic for any separable extension: Theorem 2.1 for characteristic different from $3$ and Theorem 3.1 for characteristic $3$. The algorithm proposed only depends only on the factorisation of polynomials, Hasse’s algorithm to obtain a global standard form in the sense of Artin-Schreier theory (see [9, Example 5.8.9] or [3]) and the Chinese remainder theorem. In any characteristic, obtaining generators in a type of standard form in a tower generated by the Galois closure of a cubic appears to be key to explicitly determining this triangular integral basis. We note that this work generalises the results begun in [6, Theorem 5.17], and that our work in characteristic $p\geq 5$ is a consequence of our computations and results appearing in [4, 8]. In characteristic 3, the basis we construct bears some resemblance to that of [5, Theorem 9], although the construction is different.

In §1, we compute explicitly the discriminant for any separable cubic function field over $\mathbb{F}_{q}(x)$. In §2, we use those computations to determine that a triangular integral basis always exists in terms of our choices of generators, and we determine such a basis explicitly in each characteristic.

1 Discriminant of a cubic extension

Let $\mathfrak{D}_{L/\mathbb{F}_{q}(x)}$ be the different and $\partial_{L/\mathbb{F}_{q}(x)}$ the discriminant divisor of $L/\mathbb{F}_{q}(x)$ [9, Section 5.6]. We have by[9, Definition 5.6.8] that

[TABLE]

where $\mathcal{R}$ is the set of finite places of $\mathbb{F}_{q}(x)$ ramified in $L$, $d(\mathfrak{P}|\mathfrak{p})$ is the differential exponent of $\mathfrak{P}|\mathfrak{p}$, and $f(\mathfrak{P}|\mathfrak{p})$ is the inertia degree. Let $\mathcal{O}_{L,x}$ denote the integral closure of $\mathbb{F}_{q}[x]$ in $L$. If $\omega\in\mathcal{O}_{L,x}$ with minimal polynomial

[TABLE]

then the discriminant $\Delta(\omega)$ of $\omega$ is defined [2, Theorem 2.1] as

[TABLE]

The ideal $\text{Ind}(\omega)$ of $\mathbb{F}_{q}[x]$ is defined [4, p. 789] by

[TABLE]

We suppose that $L/\mathbb{F}_{q}(x)$ is an impure cubic extension. We split the calculations into the cases $3\nmid q$ and $3\mid q$, as the associated generating equations are different [7, Corollary 1.3].

1.1 $p\neq 3$

Let $y$ be a generator of $L/\mathbb{F}_{q}(x)$ which satisfies the equation $y^{3}-3y-a=0$, where $a\in\mathbb{F}_{q}(x)$, which exists by [7, Corollary 1.3]. We let $\mathfrak{p}$ a place of $K$ and $\mathfrak{P}$ a place of $L$ above $\mathfrak{p}$. We let $a=\alpha/(\gamma^{3}\beta)$, where $(\alpha,\beta\gamma)=1$, $\beta$ is cube-free, and $\beta=\beta_{1}\beta_{2}^{2}$, where $\beta_{1}$ and $\beta_{2}$ are square-free. Let $\omega=\gamma\beta_{1}\beta_{2}y$. The minimal polynomial of $\omega$ over $\mathbb{F}_{q}(x)$ is equal to

[TABLE]

and $\omega$ is integral over $\mathbb{F}_{q}[x]$. By definition, the discriminant of $\omega$ is equal to

[TABLE]

We have (up to constant multiples)

[TABLE]

In the following lemma, we determine $(\partial_{L/\mathbb{F}_{q}(x)})_{\mathbb{F}_{q}[x]}$ and $\text{Ind}(\omega)$ when $p\neq 3$, for which the cases $p\neq 2$ and $p=2$ are different.

Lemma 1.1**.**

1. (a)

If $p\neq 2$, let $(4\gamma^{6}\beta^{2}-\alpha^{2})=\eta_{1}\eta_{2}^{2}$, where $\eta_{1}$ is square-free. Then

[TABLE] 2. (b)

If $p=2$, let $r$ be a root of the quadratic resolvent of the polynomial $X^{3}-3X-a$. One can find an Artin-Schreier generator $z$ in global standard form for $\mathbb{F}_{q}(x,r)/\mathbb{F}_{q}(x)$ **[9, Example 5.8.8]**, i.e., an Artin-Schreier generator such that

[TABLE]

where $b=\frac{1}{a^{2}}+1+\eta^{2}-\eta$ for some $\eta\in\mathbb{F}_{q}(x)$ such that each finite place $\mathfrak{p}$ of $\mathbb{F}_{q}(x)$ which ramifies in $\mathbb{F}_{q}(x,z)=\mathbb{F}_{q}(x,r)$ satisfies

[TABLE]

and any unramified place $\mathfrak{p}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ satisfies

[TABLE]

We let $\ell_{\mathfrak{p}}=-v_{\mathfrak{p}}\left(b\right)$ for each ramified $\mathfrak{p}\mid(\alpha)_{\mathbb{F}_{q}[x]}$, and otherwise we let $\ell_{\mathfrak{p}}=-1$ whenever $\mathfrak{p}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ is unramified. Then

[TABLE]

Proof.

When $p\neq 2$ we have $d(\mathfrak{P}|\mathfrak{p})=2$ whenever $e(\mathfrak{P}|\mathfrak{p})=3$, which occurs when $v_{\mathfrak{p}}(a)<0$ and $(v_{\mathfrak{p}}(a),3)=1$. We also have $d(\mathfrak{P}|\mathfrak{p})=1$ if, and only if, $v_{\mathfrak{p}}(4\gamma^{6}\beta^{2}-\alpha^{2})$ is odd (see [7, Corollary 3.15]). In either of these cases, only one place $\mathfrak{P}|\mathfrak{p}$ above has $d(\mathfrak{P}|\mathfrak{p})\neq 0$, and for this place $\mathfrak{P}$, $f(\mathfrak{P}|\mathfrak{p})=1$. Thus we obtain from (1) that

[TABLE]

Thus we obtain from (2) that

[TABLE]

Therefore, $\text{Ind}(\omega)=(\beta_{1}\eta_{2})_{\mathbb{F}_{q}[x]}$. 2. 2.

When $p=2$, then again $d(\mathfrak{P}|\mathfrak{p})=2$ whenever $e(\mathfrak{P}|\mathfrak{p})=3$, which occurs when $v_{\mathfrak{p}}(a)<0$ and $(v_{\mathfrak{p}}(a),3)=1$. Let $b$ be as in the statement of the lemma. By Artin-Schreier theory [9, Example 5.8.8], only finite places $\mathfrak{p}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ and possibly the place at infinity can be ramified in $\mathbb{F}_{q}(x,r)/\mathbb{F}_{q}(x)$. More precisely, by [7, Lemma 3.23], we know that $e(\mathfrak{P}|\mathfrak{p})=2$ if and only if $v_{\mathfrak{p}}\left(b\right)<0,\ \left(v_{\mathfrak{p}}\left(b\right),2\right)=1$. Then for such $\mathfrak{p}$, we have (Ibid.) that

[TABLE]

and similarly for $\mathfrak{p}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ unramified by definition of $\ell_{\mathfrak{p}}$. Once again, in either ramified case, there is only one place $\mathfrak{P}|\mathfrak{p}$ with $d(\mathfrak{P}|\mathfrak{p})\neq 0$, and for this place, $f(\mathfrak{P}|\mathfrak{p})=1$ for each such place. Thus, we obtain from (1) that

[TABLE]

From (2), we then find

[TABLE]

and hence (note $\frac{1}{2}(\ell_{\mathfrak{p}}+1)\in\mathbb{Z}$ since $(\ell_{\mathfrak{p}},2)=1$) that

[TABLE]

∎

1.2 $p=3$

We now treat the case where $p=3$. Let $y_{1}$ be a generator of $L/\mathbb{F}_{q}(x)$ such that $y_{1}^{3}+a_{1}y_{1}+a_{1}^{2}=0$, which exists by [7, Corollary 1.3]. We first establish the existence of a generator in standard form at any finite place. This will allow us to use [7, Theorem 3.19/20] to determine $(\partial_{L/\mathbb{F}_{q}(x)})_{\mathbb{F}_{q}[x]}$, which will be useful for finding an integral basis of $\mathcal{O}_{L,x}$.

Lemma 1.2**.**

There exists $b\in\mathbb{F}_{q}(x)$ and generator $z$ of $L/\mathbb{F}_{q}(x)$ with minimal polynomial

[TABLE]

for which, for all finite places of $\mathbb{F}_{q}(x)$,

[TABLE]

Proof.

We prove the lemma by way of the following algorithm. Let $\mathfrak{p}$ be a finite place of $\mathbb{F}_{q}(x)$ for which $v_{\mathfrak{p}}(a_{1})<0$ and $3\mid v_{\mathfrak{p}}(a_{1})$. We will find an element $a\in\mathbb{F}_{q}(x)$ and a generator $y$ of $L/\mathbb{F}_{q}(x)$ with minimal polynomial

[TABLE]

and such that

• —

$v_{\mathfrak{p}}(a)\geq 0$ or $v_{\mathfrak{p}}(a)<0$ and $(v_{\mathfrak{p}}(a),3)=1$;

• —

all finite places $\mathfrak{q}\neq\mathfrak{p}$ such that $v_{\mathfrak{q}}(a_{1})<0$ also have $v_{\mathfrak{q}}(a)=v_{\mathfrak{q}}(a_{1})$;

• —

all finite places $\mathfrak{q}$ such that $v_{\mathfrak{q}}(a_{1})\geq 0$ also have $v_{\mathfrak{q}}(a)\geq 0$.

By [7, Lemma 1.11], any other generator $y_{2}$ of $L/\mathbb{F}_{q}(x)$ with a minimal equation of the same form $y_{2}^{3}+a_{2}y_{2}+a_{2}^{2}=0$ is such that $y_{2}=-\beta(\frac{j}{a_{1}}y_{1}+\frac{1}{a_{1}}w_{2})$ for some $w_{2}\in K$, and we have

[TABLE]

Let $\lambda\in\mathbb{F}_{q}(x)$ such that $v_{\mathfrak{p}}(\lambda)=2v_{\mathfrak{p}}(a_{1})/3$. By strong approximation [1, Chap. 2.15], there exists $\alpha\in\mathbb{F}_{q}(x)$ such that

• —

$v_{\mathfrak{p}}(\alpha-\lambda)\geq\left(2v_{\mathfrak{p}}(a_{1})/3\right)+1$ and

• —

$v_{\mathfrak{q}}(\alpha)\geq 0$ at all other finite places $\mathfrak{q}\neq\mathfrak{p}$ of $\mathbb{F}_{q}(x)$.

As $v_{\mathfrak{p}}(\alpha-\lambda)\geq\left(2v_{\mathfrak{p}}(a_{1})/3\right)+1=v_{\mathfrak{p}}(\lambda)+1>v_{\mathfrak{p}}(\lambda)$, the first condition implies by the non-Archimedean property

[TABLE]

We may therefore apply [7, Lemma 3.18], where $w_{0}\in K$ was chosen so that $w_{0}\neq-\alpha^{-3}ja_{1}^{2}$ and

[TABLE]

As $p=3$, the map $X\rightarrow X^{3}$ is an isomorphism of $k(\mathfrak{p})$, so we may find an element $w^{*}\in\mathbb{F}_{q}(x)$ such that ${w^{*}}^{3}=w_{0}\mod\mathfrak{p}$ (note also that $v_{\mathfrak{p}}(w^{*})=0$ by construction). Again by strong approximation, we choose an element $w_{1}\in\mathbb{F}_{q}(x)$ such that

• —

$v_{\mathfrak{p}}(w_{1}-w^{*})\geq 1$, so that $w_{1}=w^{*}\mod\mathfrak{p}$, which implies $w_{1}^{3}={w^{*}}^{3}=w_{0}\mod\mathfrak{p}$.

• —

$v_{\mathfrak{q}}(w_{1}-0)\geq v_{\mathfrak{q}}(a_{1})$ for any finite place $\mathfrak{q}$ such that $v_{\mathfrak{q}}(a_{1})>0$.

• —

$v_{\mathfrak{q}}(w_{1})\geq 0$ at all other finite places $\mathfrak{q}$.

As in [7, Lemma 3.18], we let $w_{2}=\alpha w_{1}$. By the same argument as in (Ibid.), we have $v_{\mathfrak{p}}(a_{2})>v_{\mathfrak{p}}(a_{1})$. We now observe how the process behaves at the other finite places of $\mathbb{F}_{q}(x)$:

• —

If $\mathfrak{q}$ is a finite place such that $v_{\mathfrak{q}}(a_{1})>0$, then $v_{\mathfrak{q}}(w_{2})=v_{\mathfrak{q}}(\alpha w_{1})=v_{\mathfrak{q}}(\alpha)+v_{\mathfrak{q}}(w_{1})\geq v_{\mathfrak{q}}(a_{1}).$ Then

[TABLE]

Thus

[TABLE]

• —

If $\mathfrak{q}$ is a finite place such that $v_{\mathfrak{q}}(a_{1})=0$, then $v_{\mathfrak{q}}(w_{1})\geq 0$, so that

[TABLE]

Hence

[TABLE]

• —

If $\mathfrak{q}\neq\mathfrak{p}$ is a finite place of $\mathbb{F}_{q}(x)$ such that $v_{\mathfrak{q}}(a_{1})<0$, then $v_{\mathfrak{q}}(w_{2})\geq 0$, so that

[TABLE]

whence

[TABLE]

So

[TABLE]

Summarising, we obtain

• —

$v_{\mathfrak{p}}(a_{2})>v_{\mathfrak{p}}(a_{1})$.

• —

If $\mathfrak{q}$ is a finite place such that $v_{\mathfrak{q}}(a_{1})>0$, then $v_{\mathfrak{q}}(a_{2})>0$.

• —

If $\mathfrak{q}$ is a finite place such that $v_{\mathfrak{q}}(a_{1})=0$, then $v_{\mathfrak{q}}(a_{2})\geq 0$.

• —

If $\mathfrak{q}\neq\mathfrak{p}$ is a finite place of $\mathbb{F}_{q}(x)$ such that $v_{\mathfrak{q}}(a_{1})<0$, then $v_{\mathfrak{q}}(a_{2})=v_{\mathfrak{q}}(a_{1})$.

We now repeat this process as $\mathfrak{p}$ to obtain $y$ and $a$ as desired. Once $a$ is obtained for $\mathfrak{p}$, we may choose another finite place $\mathfrak{p}^{\prime}$ (if it exists) of $K$ such that $v_{\mathfrak{p}^{\prime}}(a)<0$ and $3\mid v_{\mathfrak{p}^{\prime}}(a)$, and perform the same process with initial generator $y$ and equation $y^{3}+ay+a^{2}=0$, obtaining some $y^{\prime}$ and $a^{\prime}$ such that

[TABLE]

In addition to satisfying the requisite valuation conditions at $\mathfrak{p}$, the element $a^{\prime}$ also satisfies these conditions at $\mathfrak{p}^{\prime}$, i.e., $v_{\mathfrak{p}^{\prime}}(a^{\prime})\geq 0$ or $v_{\mathfrak{p}^{\prime}}(a^{\prime})<0$ and $(v_{\mathfrak{p}^{\prime}}(a^{\prime}),3)=1$. By construction, all other valuations at finite places remain nonnegative or stable in this algorithm. We therefore repeat this process until we find a $b\in\mathbb{F}_{q}(x)$ and generator $z$ of $L/\mathbb{F}_{q}(x)$ as desired.∎

We may now determine $(\partial_{L/\mathbb{F}_{q}(x)})_{\mathbb{F}_{q}[x]}$ in terms of the element $b\in\mathbb{F}_{q}(x)$ of Lemma 1.2.

Lemma 1.3**.**

Let $b\in\mathbb{F}_{q}(x)$ be as in Lemma 1.2. We denote

[TABLE]

where $\xi_{1},\xi_{2},\beta\in\mathbb{F}_{q}[x]$, $\xi_{1}$ is square-free, and $(\xi_{1}\xi_{2},\beta)=1$. Then

[TABLE]

Proof.

By [7, Theorem 3.19], whenever $v_{\mathfrak{q}}(b)<0$ and $(v_{\mathfrak{q}}(b),3)=1$, $\mathfrak{q}$ is fully ramified, and by [7, Theorem 3.20], $e(\mathfrak{Q}|\mathfrak{q})=2$ for a single place $\mathfrak{Q}|\mathfrak{p}$ whenever $v_{\mathfrak{q}}(b)>0$ and $(v_{\mathfrak{q}}(b),2)=1$. For the fully ramified places, by [7, Lemma 3.26], we have

[TABLE]

We also have by [7, Lemma 3.18] that $d(\mathfrak{P}|\mathfrak{p})=1$ whenever $v_{\mathfrak{p}}(b)>0$ and $(v_{\mathfrak{p}}(b),2)=1$. These are the two cases where a finite place occurs in the discriminant of $L/\mathbb{F}_{q}(x)$. In either case, there exists only one place $\mathfrak{P}$ of $L$ above $\mathfrak{p}$ such that $d(\mathfrak{P}|\mathfrak{p})\neq 0$, and for such $\mathfrak{P}$, we again have $f(\mathfrak{P}|\mathfrak{p})=1$. Thus we find from (1) that

[TABLE]

∎

2 Explicit triangular integral bases

In this section, we demonstrate the existence of an explicit triangular integral basis for a separable cubic extension $L/\mathbb{F}_{q}(x)$. We divide the argument between $p\neq 3$ and $p=3$, as the construction in each case is different.

2.1 $p\neq 3$

Let $y$ be a generator of $L/\mathbb{F}_{q}(x)$ which satisfies the equation $y^{3}-3y-a=0$, where $a\in\mathbb{F}_{q}(x)$, which exists by [7, Corollary 1.3]. We let $a=\alpha/(\gamma^{3}\beta)$, where $(\alpha,\beta\gamma)=1$, $\beta$ is cube-free, and $\beta=\beta_{1}\beta_{2}^{2}$, where $\beta_{1}$ and $\beta_{2}$ are square-free. As in §1.1, we let $\omega=\gamma\beta_{1}\beta_{2}y$ and $(4\gamma^{6}\beta^{2}-\alpha^{2})=\eta_{1}\eta_{2}^{2}$, where $\eta_{1}$ is square-free. The element $\omega$ is integral over $\mathbb{F}_{q}[x]$, and the minimal polynomial of $\omega$ over $\mathbb{F}_{q}(x)$ is equal to

[TABLE]

We let $r$ denote a root of the quadratic resolvent of this cubic polynomial. We also let:

• $\cdot$

If $p\neq 2$, $I=\beta_{1}^{2}\eta_{2}^{2}$,

• $\cdot$

If $p=2$, $I=\beta_{1}\alpha_{I}$, where

• —

$\alpha_{I}=\prod_{i=1}^{s}\alpha_{i}^{v_{\mathfrak{p}_{i}}(\alpha)-\frac{1}{2}(\ell_{\mathfrak{p}_{i}}+1)}$, when $L/\mathbb{F}_{q}(x)$ is not Galois,

• —

$\alpha_{I}=\alpha$, when $L/\mathbb{F}_{q}(x)$ is Galois,

where we write $\alpha$ as a product of irreducible polynomials $\alpha_{i}$ ($i=1,\ldots,s$) as

[TABLE]

where $\alpha_{i}$ corresponds to the (finite) place $\mathfrak{p}_{i}$ of $\mathbb{F}_{q}(x)$ and $\ell_{\mathfrak{p}_{i}}$ is defined as in Lemma 1.1.

This machinery allows us to obtain the following description of the explicit triangular integral basis of $\mathcal{O}_{L,x}/\mathbb{F}_{q}[x]$.

Theorem 2.1**.**

Let $p\neq 3$. The set

[TABLE]

forms an integral basis of $L/\mathbb{F}_{q}(x)$, if $S,T,V\in\mathbb{F}_{q}[x]$, $V\equiv T^{2}-3(\gamma\beta_{1}\beta_{2})^{2}\mod I$, and $T$ is chosen in the following manner, which may be done by the Chinese remainder theorem:

When $p\neq 2$,

[TABLE] 2. 2.

When $p=2$,

1. (a)

If $L/\mathbb{F}_{q}(x)$ is Galois,

[TABLE] 2. (b)

If $L/\mathbb{F}_{q}(x)$ is not Galois,

[TABLE]

Proof.

Since

When $p\neq 2$, the result is a direct consequence of [4, Lemma 6.3] and §1.1. 2. 2.

Suppose that $p=2$. By [8, Lemma 3.1, Corollary 3.2], which remain valid in characteristic 2, and Lemma 1.1, a basis of the form of $\mathfrak{B}$ as in the statement of the theorem exists if, and only if, $T,V,S\in\mathbb{F}_{q}[x]$ are such that

[TABLE]

and

[TABLE]

We prove that a choice of $T$ which satisfies the conditions in the statement of the theorem also satisfies (3). We note that as $(\beta_{1},\alpha_{I})=1$, the first condition in (3),

[TABLE]

is equivalent to the two conditions

[TABLE]

and the second condition in (3)

[TABLE]

is equivalent to the two conditions

[TABLE]

First, we note that if $T\equiv 0\mod\beta_{1}$, then $T^{3}+(\gamma\beta_{1}\beta_{2})^{2}T+\alpha\beta_{1}^{2}\beta_{2}\equiv 0\mod\beta_{1}^{2}.$ As $T$ is invertible mod $\alpha_{I}$ from $(\alpha_{I},\gamma\beta_{1}\beta_{2})=1$, the condition

[TABLE]

is equivalent to

[TABLE]

We have

[TABLE]

Thus, the condition

[TABLE]

is equivalent to

[TABLE]

Since $(\gamma^{2}\beta_{2},\alpha_{I})=1$, this condition is in turn equivalent to

[TABLE]

Setting $T^{\prime}=\gamma^{2}\beta_{2}T$, we obtain

[TABLE]

We now note that the quadratic resolvent of the cubic polynomial $X^{3}-3\gamma^{2}\beta_{1}^{2}\beta_{2}^{2}X-\alpha\beta_{1}^{2}\beta_{2}$ is equal to

[TABLE]

We denote by $r$ a root of the quadratic resolvent. Taking $X^{\prime}=\frac{X}{\beta_{1}^{2}\beta_{2}}$, we have that

[TABLE]

is an integral polynomial over $\mathbb{F}_{q}[x]$. We now divide the argument into the cases where $L/\mathbb{F}_{q}(x)$ is Galois or not.

1. (a)

If $L/\mathbb{F}_{q}(x)$ is Galois, then $r\in\mathbb{F}_{q}(x)$, setting $r^{\prime}=\frac{r}{\beta_{1}^{2}\beta_{2}}$, $S(r^{\prime})=0$ and

[TABLE]

Choosing $T\equiv\frac{r}{\gamma^{2}\beta_{1}^{2}\beta_{2}^{2}}\mod\alpha_{I}$ and $T\equiv 0\mod\beta_{1}$ via the Chinese remainder theorem, $T$ satisfies (3), proving the theorem. 2. (b)

If $L/\mathbb{F}_{q}(x)$ is not Galois, then $r\notin\mathbb{F}_{q}(x)$, and the base change $X^{\prime\prime}=\frac{X^{\prime}}{\alpha}$ transforms $S(X^{\prime})$ into the Artin-Schreier polynomial

[TABLE]

Via [9, Example 5.8.9], we let $c$ be chosen such that

[TABLE]

if $\mathfrak{p}_{i}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ is ramified in $\mathbb{F}_{q}(x)(r)$ and $v_{\mathfrak{p}}(c^{2}+c+\frac{\gamma^{6}\beta_{1}^{2}\beta_{2}^{4}}{\alpha^{2}}+1)\geq 0$ for all other finite places $\mathfrak{p}$ of $\mathbb{F}_{q}(x)$. It follows that $c^{\prime}=\alpha c$ is such that

[TABLE]

if $\mathfrak{p}_{i}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ is ramified in $\mathbb{F}_{q}(x)(r)$, and

[TABLE]

for all other finite places $\mathfrak{p}$ of $\mathbb{F}_{q}(x)$. As $\ell_{\mathfrak{p}_{i}}>0$ if $\mathfrak{p}_{i}\mid(\alpha)_{\mathbb{F}_{q}[x]}$ is ramified in $\mathbb{F}_{q}(x)(r)$, it follows for such $\mathfrak{p}_{i}$ by the non-Archimedean triangle inequality and (7) that

[TABLE]

For all other places $\mathfrak{p}$ of $\mathbb{F}_{q}(x)$ dividing $(\alpha)_{\mathbb{F}_{q}[x]}$, we have $\ell_{\mathfrak{p}}=-1$. By the non-Archimedean triangle inequality and (8), we obtain

[TABLE]

Thus, for this this choice of $c^{\prime}$, we have that

[TABLE]

We now choose $d=\frac{c^{\prime}}{\gamma^{2}\beta_{2}}$, which yields

[TABLE]

So that $d$ is a solution of 6. Finally, via the Chinese remainder theorem, we choose

[TABLE]

so that (3) is again satisfied, proving the theorem.

∎

3 $p=3$

We now consider the construction of an explicit triangular integral basis when $p=3$. We let $z$ be a generator of $L/\mathbb{F}_{q}(x)$ with minimal polynomial

[TABLE]

where $b$ is as in Lemma 1.2. We write

[TABLE]

where $p_{i}\in\mathbb{F}_{q}[x]$ distinct irreducible polynomials ($i=1,\ldots,s$), with $p_{i}$ corresponding to the (finite) place $\mathfrak{p}_{i}$ of $\mathbb{F}_{q}(x)$, where also $\xi_{1},\xi_{2}\in\mathbb{F}_{q}[x]$, $\xi_{1}$ is square-free, and $(\xi_{1}\xi_{2},\beta)=1$, and $\ell_{\mathfrak{p}_{i}}>0$, $(\ell_{\mathfrak{p}_{i}},3)=1$ for each $i=1,\ldots,s$.

Theorem 3.1**.**

Let $p=3$. For $j=1,2$, let

[TABLE]

where $\left\lfloor\frac{j2\ell_{\mathfrak{p}_{i}}}{3}\right\rfloor$ is the integral part of $\frac{j2\ell_{\mathfrak{p}_{i}}}{3}$. Then the set

[TABLE]

is an integral basis for $L/\mathbb{F}_{q}(x)$.

Proof.

Let $\mathfrak{y}=\frac{z}{\xi_{2}}$ and $\gamma=\frac{\xi_{1}}{\beta}$. We have

[TABLE]

The discriminant of $\mathfrak{y}$ is equal to $\Delta(\mathfrak{y})=2\gamma^{3}$, and by Lemma 1.3, we have

[TABLE]

We may thus write

[TABLE]

We let $\beta=\beta_{1}\beta_{2}^{2}$, where $\beta_{1}$ is square-free. Let $g$ be chosen in the algebraic closure of $\mathbb{F}_{q}(x)$ to satisfy $g^{2}=-\gamma$. By Kummer theory (see [9, Example 5.8.9]), the finite places $\mathfrak{p}$ of $\mathbb{F}_{q}(x)$ ramified in $\mathbb{F}_{q}(x)(g)$ are those such that $\mathfrak{p}|(\xi_{1}\beta_{1})_{\mathbb{F}_{q}[x]}$. In what follows, $\mathfrak{p}$ denotes a finite place of $\mathbb{F}_{q}(x)$, $\mathfrak{p}_{g}$ a place of $\mathbb{F}_{q}(x)(g)$ above $\mathfrak{p}$, $\mathfrak{P}$ a place of $L$ above $\mathfrak{p}$ and $\mathfrak{P}_{g}$ a place of $L(g)$ above $\mathfrak{p}_{g}$ and $\mathfrak{P}$.

By definition, the element $\frac{\mathfrak{y}}{g}$ satisfies the following Artin-Schreier equation above $\mathbb{F}_{q}(x)(g)$:

[TABLE]

Let $\mathfrak{p}|(\beta_{1})_{\mathbb{F}_{q}[x]}$. Thus, $v_{\mathfrak{p}_{g}}(g)=v_{\mathfrak{p}}(\gamma)<0$ and $(v_{\mathfrak{p}_{g}}(g),3)=1$. Hence $e(\mathfrak{P}_{g}|\mathfrak{p}_{g})=3$ by Artin-Schreier theory, and by the non-Archimedean triangular inequality, $v_{\mathfrak{P}_{g}}(\frac{\mathfrak{y}}{g})=v_{\mathfrak{p}}(\gamma)$, so that

[TABLE]

and

[TABLE]

Then, for $j=1,2$,

[TABLE]

Suppose that $\mathfrak{p}|(\beta)_{\mathbb{F}_{q}[x]}$ and $\mathfrak{p}\nmid(\beta_{1})_{\mathbb{F}_{q}[x]}$. Then $v_{\mathfrak{p}_{g}}(g)=\frac{v_{\mathfrak{p}}(\gamma)}{2}<0$ and $(v_{\mathfrak{p}_{g}}(g),3)=1$. Hence again $e(\mathfrak{P}_{g}|\mathfrak{p}_{g})=3$ by Artin-Schreier theory, and by the non-Archimedean triangle inequality, $v_{\mathfrak{P}_{g}}(\frac{\mathfrak{y}}{g})=\frac{v_{\mathfrak{p}}(\gamma)}{2}$, so that

[TABLE]

and

[TABLE]

Then, for $j=1,2$,

[TABLE]

If $\mathfrak{p}|(\xi_{1})_{\mathbb{F}_{q}[x]}$, then $v_{\mathfrak{p}_{g}}(g)=v_{\mathfrak{p}}(\gamma)=v_{\mathfrak{p}}(\xi_{1})=1$, as $\xi_{1}$ is square free. By the non-Archimedean triangle inequality, we find that either $v_{\mathfrak{P}_{g}}(\frac{\mathfrak{y}}{g})=0$ or $v_{\mathfrak{P}_{g}}(\frac{\mathfrak{y}}{g})=v_{\mathfrak{p}}(\gamma\xi_{2})$. In both cases, $v_{\mathfrak{P}_{g}}(\mathfrak{y})\geq v_{\mathfrak{P}_{g}}(g)>0$, whence $v_{\mathfrak{P}}(\mathfrak{y})>0$ and $v_{\mathfrak{P}}(\mathfrak{y})\geq 1=v_{\mathfrak{p}}(\xi_{1})$. Thus, we obtain

[TABLE]

where $e(\mathfrak{P}|\mathfrak{p})=2$ by [7, Theorem 3.20].

Finally, we obtain

[TABLE]

We have

[TABLE]

whence

[TABLE]

We therefore find via the previous argument and Lemma 1.3 that

[TABLE]

which proves that $\mathfrak{B}$ is an integral basis.

∎