Semistable models of elliptic curves over residue characteristic 2

TL;DR

This paper introduces an algorithm to determine semistable models of elliptic curves over rings of mixed characteristic (0,2), based solely on the valuation of the parameter in Legendre form, with applications to 2-torsion.

Contribution

It provides a novel valuation-dependent algorithm for constructing semistable models of elliptic curves in characteristic 2, with practical examples and implications for torsion points.

Findings

Algorithm depends only on valuation of λ

Examples illustrate the method's effectiveness

Corollary relates to 2-torsion points

Abstract

Given an elliptic curve $E$ in Legendre form $y^2 = x(x - 1)(x - \lambda)$ over the fraction field of a Henselian ring $R$ of mixed characteristic $(0, 2)$, we present an algorithm for determining a semistable model of $E$ over $R$ which depends only on the valuation of $\lambda$. We provide several examples along with an easy corollary concerning $2$-torsion.

Equations

$$y^2=f(x):=x(x-1)(x-\lambda )$$

$$j(E)=2^8\frac{({\lambda }^2-\lambda +1{)}^3}{{\lambda }^2(\lambda -1{)}^2}.$$

$$y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6,$$

$$y^2=F(x)=F_{\alpha ,\beta }(X):=(x+\alpha {\beta }^{-1})(x+\alpha {\beta }^{-1}-{\beta }^{-1})(x+\alpha {\beta }^{-1}-\lambda {\beta }^{-1}).$$

$$G(x)+\frac{1}{4}H(x{)}^2=F(x)=x^3+F^{(2)}(0)x^2+F^{(1)}(0)x+F(0).$$

$$\begin{array}{rl}{a}_3& =2\sqrt{F(0)-a_6}\\ a_1& =\frac{2F^{(1)}(0)-a_4}{a_3}=\frac{F^{(1)}(0)-a_4}{\sqrt{F(0)-a_6}}\\ a_2& =F^{(2)}(0)-\frac{1}{4}{a}_1^2=F^{(2)}(0)-\frac{(F^{(1)}(0)-a_4{)}^2}{4(F(0)-a_6)}\end{array}$$

$$a_1=\frac{F^{(1)}(0)}{\sqrt{F(0)}};a_2=F^{(2)}(0)-\frac{1}{4}{a}_1^2=F^{(2)}(0)-\frac{F^{(1)}(0{)}^2}{4F(0)};a_3=2\sqrt{F(0)}.$$

$$P(X):=3X^4-4(1+\lambda )X^3+6\lambda X^2-{\lambda }^2-\delta ,$$

$$a_2=\frac{\delta }{4\beta \alpha (\alpha -1)(\alpha -\lambda )}.$$

$$v(F(0))=v(\alpha )+v(\alpha -1)+v(\alpha -\lambda )-3v(\beta )=2v(\alpha )-3v(\beta )=m-m-2=-2.$$

$$\begin{array}{rl}& {\beta }^{3}F(0)=\alpha (\alpha -1)(\alpha -\lambda );{\beta }^2{F}^{(1)}(0)=\alpha (\alpha -1)+\alpha (\alpha -\lambda )+(\alpha -1)(\alpha -\lambda );\\ & \beta F^{(2)}(0)=\alpha +(\alpha -1)+(\alpha -\lambda )\end{array}$$

$$\begin{array}{rl}4{\alpha }^2(\alpha -1)(\alpha -\lambda )+& 4\alpha (\alpha -1{)}^2(\alpha -\lambda )+4\alpha (\alpha -1)(\alpha -\lambda {)}^2\\ -& ({\alpha }^2(\alpha -1{)}^2+{\alpha }^2(\alpha -\lambda {)}^2+(\alpha -1{)}^2(\alpha -\lambda {)}^2\\ & +2{\alpha }^2(\alpha -1)(\alpha -\lambda )+2\alpha (\alpha -1{)}^2(\alpha -\lambda )+2\alpha (\alpha -1)(\alpha -\lambda {)}^2)\end{array}$$

$$\begin{array}{rl}& =2{\alpha }^2(\alpha -1)(\alpha -\lambda )+2\alpha (\alpha -1{)}^2(\alpha -\lambda )+2\alpha (\alpha -1)(\alpha -\lambda {)}^2-{\alpha }^2(\alpha -1{)}^2-{\alpha }^2(\alpha -\lambda {)}^2-(\alpha -1{)}^2(\alpha -\lambda {)}^2\\ & =2(2\alpha -\lambda )(\alpha )(\alpha -1)(\alpha -\lambda )-{\alpha }^2(\alpha -\lambda {)}^2+(\alpha -1{)}^2(2\alpha (\alpha -\lambda )-{\alpha }^2-(\alpha -\lambda {)}^2)\\ & =2(2\alpha -\lambda )(\alpha )(\alpha -1)(\alpha -\lambda )-{\alpha }^2(\alpha -\lambda {)}^2-(\alpha -1{)}^2{\lambda }^2\\ & =[4{\alpha }^4-2(2+3\lambda ){\alpha }^3+2(3\lambda +{\lambda }^2){\alpha }^2-2{\lambda }^2\alpha ]-[{\alpha }^4-2\lambda {\alpha }^3+{\lambda }^2{\alpha }^2]-[{\lambda }^2{\alpha }^2-2{\lambda }^2\alpha +{\lambda }^2]\end{array}$$

$$=3{\alpha }^4-4(1+\lambda ){\alpha }^3+6\lambda {\alpha }^2-{\lambda }^2.$$

$$P(X)=3X^4-4(1-{\omega }^2)X^3-6{\omega }^2{X}^2-\omega -\delta .$$

$$y^2-\omega (-3{)}^{-1/4}{2}^{5/3}xy+(-3{)}^{1/4}y=x^3+{\omega }^2(-3{)}^{-1/2}{2}^{1/3}{x}^2$$

$$P(X)=3X^4-12X^3+12X^2-4-\delta .$$

$$y^2+(3^{1/4}+3^{3/4})(1+\frac{2}{\sqrt{3}}{)}^{-1/4}xy+3^{-1/4}(1+\frac{2}{\sqrt{3}}{)}^{1/4}y=x^3$$

$$v(F(0))=v(\alpha )+v(\alpha -1)+v(\alpha -\lambda )-3v(\beta )=2v(\alpha )-6;$$

$$v(F^{(1)}(0))=v(2\alpha (\alpha -1)-\lambda (\alpha -1)+\alpha (\alpha -\lambda ))-2v(\beta )=\min \{v(\alpha )+1,m,2v(\alpha )\}-4=v(\alpha )-3.$$

$$4{\beta }^{4}F(0)a_2=3{\alpha }^4-4(1+\lambda ){\alpha }^3+6\lambda {\alpha }^2-{\lambda }^2.$$

$$\begin{array}{rl}{y}^2+{\bar{a}}_{1}xy+{\bar{a}}_{3}y& =x^3+{\bar{a}}_2{x}^2\\ {\bar{a}}_{1}y& =x^2\\ {\bar{a}}_{1}x+{\bar{a}}_3& =0\end{array}$$

$$\frac{{\bar{a}}_3^4}{{\bar{a}}_1^6}+\frac{{\bar{a}}_3^3}{{\bar{a}}_1^3}-\frac{{\bar{a}}_2{\bar{a}}_3^2}{{\bar{a}}_1^2}=0$$

$$\frac{\bar{a}_{3}}{\bar{a}_{1}}\Big{(}\frac{\bar{a}_{3}}{\bar{a}_{1}^{3}}+1\Big{)}-\bar{a}_{2}=0.$$

$$\begin{split}\frac{a_{3}}{a_{1}}=\frac{4F(0)}{2F^{(1)}(0)}&\equiv\frac{-4\alpha^{2}\beta^{-3}}{-4\alpha\beta^{-2}}=\frac{\alpha}{\beta}\\ \frac{a_{3}}{a_{1}^{3}}=\Big{(}\frac{a_{3}}{a_{1}}\Big{)}\Big{(}\frac{4F(0)}{4F^{(1)}(0)^{2}}\Big{)}&\equiv\Big{(}\frac{\alpha}{\beta}\Big{)}\frac{-4\alpha^{2}\beta^{-3}}{16\alpha^{2}\beta^{-4}}=-\frac{\alpha}{4}\end{split}$$

$$a_2=\frac{{\beta }^{-4}(3{\alpha }^4-4(1+\lambda ){\alpha }^3+6\lambda {\alpha }^2-{\lambda }^2)}{4F(0)}\equiv \frac{{\beta }^{-4}({\alpha }^4-4{\alpha }^3-{\lambda }^2)}{-4{\alpha }^2{\beta }^{-3}}=-\frac{{\alpha }^2}{4\beta }+\frac{\alpha }{\beta }+\frac{{\lambda }^2}{4{\alpha }^2\beta }.$$

$$\frac{a_{3}}{a_{1}}\Big{(}\frac{a_{3}}{a_{1}^{3}}+1\Big{)}-a_{2}\equiv\frac{\alpha}{\beta}\Big{(}-\frac{\alpha}{4}+1\Big{)}+\frac{\alpha^{2}}{4\beta}-\frac{\alpha}{\beta}-\frac{\lambda^{2}}{4\alpha^{2}\beta}=-\frac{\lambda^{2}}{4\alpha^{2}\beta}.$$

$$y^2+3ixy+3iy=x^3+x^2$$

$$y^2+7ixy+14iy=x^3+2x^2$$

Full text

Let $R$ be a Henselian ring of mixed characteristic $(0,2)$ with a discrete valuation $v:K^{\times}\to\mathbb{Q}$ normalized so that $v(2)=1$, and let $K$ be its fraction field. Let $E$ be the elliptic curve over $K$ defined by an equation of the form $y^{2}=f(x)$ for some separable polynomial $f(x)\in K[x]$. After replacing $K$ by a suitable extension and possibly scaling $y$ by an element of $K$ to get an isomorphic elliptic curve, we assume that $\alpha_{1},\alpha_{2},\alpha_{3}\in R$ with $\alpha_{2}-\alpha_{1}\in R^{\times}$. After possibly applying another isomorphism which translates $x$ by $-\alpha_{1}$ and then scales it by $(\alpha_{2}-\alpha_{1})^{-1}$, we further assume that $E$ is in Legendre form; that is, $E$ is a smooth projective model of an affine curve given by an equation of the form

[TABLE]

with $\lambda\in R\smallsetminus\{0,1\}$ (we denote the point at infinity by $\mathcal{O}\in E(K)$). The purpose of this note is to explicitly find a semistable model of $E$ over a finite extension of $K$. More precisely, we will find a finite extension $K^{\prime}/K$ such that $E/K^{\prime}$ has a model $E^{\mathrm{ss}}$ given by explicit formulas with coefficients in the ring of integers of $K^{\prime}$ and which has either good or (split) multiplicative reduction.

It is well known (see for instance [4, §IV.1.2] or [5, Proposition VII.5.5]) that any elliptic curve over a discrete valuation field has good (resp. multiplicative) reduction over some finite algebraic extension of that field if and only if the valuation of its $j$-invariant is nonnegative (resp. negative). The formula for the $j$-invariant of the Legendre curve $E$ is given as in [5, Proposition III.1.7] by

[TABLE]

For simplicity, we assume throughout this paper that $v(\lambda-1)=0$, noting that if $v(\lambda-1)>0$, then we have $v(\lambda)=0$ and the assumption becomes true after replacing $\lambda$ by $1-\lambda$ and applying the isomorphism given by $(x,y)\mapsto(1-x,y)$. It follows from this assumption and the formula in (1) that $v(j(E))=8-2v(\lambda)$ and that therefore any semistable model $E^{\mathrm{ss}}$ has good (resp. multiplicative) reduction if and only if $v(\lambda)\leq 4$ (resp. $v(\lambda)>4$). This explains “why” the formula for the $j$-invariant includes an “extra” factor of $2^{8}$. The equivalence between potential good reduction and integrality of the $j$-invariant over residue characteristic $2$ is derived by Silverman as [5, Corollary A.1.4] by converting $E$ to its Deuring normal form and arguing via manipulations involving the $j$-invariant. In the course of constructing a semistable model of $E$, we will show essentially the same result more directly and without invoking the $j$-invariant.

It appears that the first investigations of how to explicitly determine semistable models of curves which are Galois $p$-covers of projective lines over DVRs of residue characteristic $p$ were done by Coleman in [2, §6]. The general approach proposed by Coleman led to results of Lehr and Matignon in [3], where they investigate superelliptic curves given by equations of the form $y^{p}=f(x)$ under the assumption that the roots of $f$ are equidistant (that is, after scaling $x$ and $y$ by appropriate powers of a uniformizer, the roots of $f$ specialize to distinct elements in the residue field). The “monodromy polynomial” defined in that paper, which is used to define an extenion of the DVR over which the superelliptic curve obtains semistable reduction, is essentially the polynomial $P(x)$ in the statement of Theorem 1. However, the work of both Coleman and of Lehr-Matignon focus on finding stable models of curves of genus greater than $1$, and although the construction in [3] of “stably marked” models using the monodromy polynomial also works for elliptic curves, the equidistance hypothesis assumed there translates to the restrictive assumption that $v(\lambda)=0$ in the context of our paper. Our goal is to describe a completely explicit method of finding semistable models of elliptic curves in Weierstrass form over mixed characteristic $(0,2)$ which shall be presented in a more elementary fashion than the results in [2] and [3]. To the best of the author’s knowledge, such a method for general elliptic curves is not present in the literature, although particular examples are done in [1, §4.1] (indeed, some of the ideas and notation used in this note were inspired by [1]). We believe that the strategy presented here is also applicable to determining semistable models and reduction types for hyperelliptic curves over mixed characteristic $(0,2)$, as suggested in both [2] and [3].

1. Our general set-up

Given a finite extension $R^{\prime}/R$ of Henselian rings with fraction fields $K^{\prime}/K$, by a Weierstrass model of $E$ over $R^{\prime}$ we mean an elliptic curve $E^{\prime}/K^{\prime}$ isomorphic over $K^{\prime}$ to $E/K^{\prime}$, which is determined by an equation of the form

[TABLE]

with all $a_{i}\in R^{\prime}$. We write $\bar{E}^{\prime}$ for the reduction of $E^{\prime}$; it is a projective curve over the residue field of $R^{\prime}$. If this curve is either smooth or has only a single node, we say that the Weierstrass model $E^{\prime}$ is semistable. By [5, Proposition VII.5.4], there is always a finite extension $K^{\prime}/K$ and a semistable Weierstrass model $E^{\mathrm{ss}}$ of $E$ over $R^{\prime}$. We note that at least one of $a_{1}$ and $a_{3}$ must be a unit in $R$ to ensure that $E^{\mathrm{ss}}$ does not have a cusp; that $v(a_{1})>0$ is then sufficient to ensure smoothness of $\bar{E}^{\mathrm{ss}}$; and that $v(a_{3})>0$ on the other hand implies that $E^{\mathrm{ss}}$ has a node at $(0,0)$.

An equation of the form given in (2) can be converted to an equation of the form $y^{2}=F(x)\in K^{\prime}[x]$ for some finite extension $K^{\prime}/K$ by completing the square: we replace $y$ by $y-\frac{1}{2}(a_{1}x+a_{3})$. Then an isomorphism from the curve $E$ given by $y^{2}=f(x)$ to the curve given by $y^{2}=F(x)$ must be of the form $(x,y)\mapsto(\alpha+\beta x,\beta^{3/2}y)$ for some $\alpha,\beta\in K^{\prime}$ (in fact, $K^{\prime}$ will just be the extension given by adjoining the elements $\alpha$ and $\beta^{1/2}$ to $K$). Given such elements $\alpha,\beta$, we first observe that this isomorphism maps $E$ to the curve defined by

[TABLE]

Now we want to find polynomials $G(x)=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}\in R^{\prime}[x]$ and $H(x)=a_{1}x+a_{3}\in R^{\prime}[x]$ such that $F=G+\frac{1}{4}H^{2}$; then the isomorphism $(x,y)\mapsto(x,y+H(x))$ maps the curve given by (3) to the one given by (2).

For each integer $n\geq 1$, we write $F^{(n)}$ for the $n$th derivative of $F$ divided by $n!$, so that $F^{(n)}(0)$ equals the coefficient of the $x^{n}$-term of $F$. We compute formulas for the elements $a_{1},a_{2},a_{3}$, using the fact that

[TABLE]

Our formulas are as follows:

[TABLE]

It will be convenient to fix $a_{4}=a_{6}=0$ so that the elements $a_{1},a_{2},a_{3}$ are completely determined (up to choosing a sign for $a_{3}$) by our choice of $\alpha$ and $\beta$ and are given by the slightly simpler formulas

[TABLE]

2. The $v(\lambda)<4$ case

In this section we assume that $v(j(E))>0$. We then see from the formula in (1) that we have $0\leq m:=v(\lambda)<4$. Since $j(E)$ is integral, the desired model $E^{\mathrm{ss}}$ should have good reduction. For any $\lambda$ with $0\leq v(\lambda)<4$, we now show how to find algebraic elements $\alpha,\beta\in\bar{K}$ such that we get $v(a_{1})>0$, $v(a_{3})=0$ and even allow $a_{2}$ to be any integral element that we choose.

Theorem 1**.**

Assume that $0\leq m<4$. Choose any $\beta\in\bar{K}$ such that $v(\beta)=\frac{1}{3}m+\frac{2}{3}$ (where $v$ is extended uniquely to a discrete valuation on $K(\beta)$). Let $\alpha$ be a root of the polynomial

[TABLE]

where $\delta\in K(\beta^{1/2})$ satisfies $v(\delta)\geq\frac{4}{3}m+\frac{8}{3}$ (e.g. $\delta=0$). Then $E$ is isomorphic over $K^{\prime}:=K(\beta^{1/2},\alpha,F(0)^{1/2})$ to the elliptic curve $E^{\mathrm{ss}}$ given by the equation in (2), where $a_{4}=a_{6}=0$ and the other coefficients $a_{i}$ are given by the formulas in (6). The isomorphism $\varphi:E\stackrel{{\scriptstyle\sim}}{{\to}}E^{\mathrm{ss}}$ is given by composing the map $(x,y)\mapsto(\alpha+\beta x,\beta^{3/2}y)$ with the map $(x,y)\mapsto(x,y+\frac{1}{2}(a_{1}x+a_{3}))$.

We have $v(a_{1})>0$, $v(a_{2})\geq 0$, and $v(a_{3})=0$ (which implies that $E^{\mathrm{ss}}$ has good reduction). Moreover, we have

[TABLE]

Proof.

Assume that we have chosen an algebraic element $\beta$ with $v(\beta)=\frac{1}{3}m+\frac{2}{3}$, an element $\delta\in K(\beta^{1/2})$ satisfying $v(\delta)=\frac{1}{3}m+\frac{2}{3}$, and a root $\alpha$ of the polynomial $P(X)$. The first statement just reaffirms what was shown in the above discussion where the formulas for $a_{1},a_{2},a_{3}\in K^{\prime}$ were derived, so our main task is to demonstrate the desired bounds for the valuations of these elements.

We note first that $v(\lambda^{2})=2m<\frac{4}{3}m+\frac{8}{3}$, so that the constant coefficient of the polynomial $P$ has valuation equal to $2m$ regardless of our choice of $\delta$. Then since the coefficient of $X^{4}$ is a unit and the coefficient of $X^{3}$ (resp. $X^{2}$) has valuation at least $2\geq\frac{1}{2}m$ (resp. equal to $1+m>m$), the Newton polygon of this polynomial consists of a single line segment with slope $\frac{1}{2}m$. It follows that $v(\alpha)=\frac{1}{2}m$. We clearly have $v(\alpha-1)=0$ and $v(\alpha-\lambda)=v(\alpha)=\frac{1}{2}m$ as long as $m>0$. If $m=0$, we claim that these equalities still hold so that $v(\alpha-1)=v(\alpha-\lambda)=0$. To see this, assume that $m=0$ and consider the polynomials $P(X+1)$ and $P(X+\lambda)$; it is straightforward to calculate (using the fact that $v(\lambda)=v(\lambda-1)=0$) that the Newton polygons of these shifted polynomials both coincide with the Newton polygon of $P$, and the claim follows from the fact that $\alpha-1$ and $\alpha-\lambda$ are roots of the respective polynomials. We now have

[TABLE]

The desired equality $v(a_{3})=v(2\sqrt{F(0)})=0$ immediately follows.

Now we treat the requirement that $v(a_{2})\geq 0$, using the formula for $b_{2}$ given in (6). We use the formulas

[TABLE]

to expand $4\beta^{4}F(0)a_{2}=4(\beta F^{(2)}(0))(\beta^{3}F(0))-(\beta^{2}F^{(1)}(0))^{2}$ as

[TABLE]

[TABLE]

[TABLE]

Thus, since $P(\alpha)=0$ can be written as the above expression minus the element $\delta$, we have $\delta=4\beta^{4}F(0)a_{2}$ (implying the claimed formula for $a_{2}$). Now the fact that $v(a_{2})\geq 0$ is equivalent to saying that $v(\delta)\geq 2+4v(\beta)+v(F(0))=\frac{1}{3}m+\frac{2}{3}$, which was indeed our condition for $\delta$.

It remains only to check that $v(a_{1})>0$. Note that $v(\beta F^{(2)}(0))\geq\min\{v(\alpha),v(1)\}=0$. It follows from the formula for $a_{2}$ in (6) that $v(\frac{1}{4}a_{1})\geq\min\{v(F^{(2)}(0)),v(a_{2})\}=v(\beta^{-1})>-2$. Therefore, $v(a_{1})-2>-2$, hence the desired inequality.

∎

Example 2**.**

Suppose we want to find a semistable model $E^{\mathrm{ss}}$ of the elliptic curve $E/\mathbb{Q}_{2}$ given by $y^{2}=x^{3}-1$ at the prime $(2)$. This elliptic curve is well known to be CM, and so any semistable model $E^{\mathrm{ss}}$ should have good reduction; we can also see this by noting that $j(E)=0$. In fact, $E$ is isomorphic (over $K:=\mathbb{Q}_{2}(\omega)$) to the Legendre curve with $\lambda=-\omega^{2}$, where $\omega:=\frac{1}{2}(-1+\sqrt{-3})$ is a primitive cube root of unity; since $m=v(\lambda)=0<4$, we may apply Theorem 1.

We have

[TABLE]

By an easy computation, plugging in $X=\omega$ to the above polynomial yields $8\omega^{2}-\delta$, so we may take $\delta=8\omega^{2}$ (noting that $v(\delta)=3\geq\frac{4}{3}m+\frac{8}{3}$) and $\alpha=\omega$. Then we may choose $\beta$ to be any element with valuation $\frac{1}{3}m+\frac{2}{3}=\frac{2}{3}$, say $\beta=2^{2/3}$. Now evaluating the formulas in (6) yields the following equation for $E^{\mathrm{ss}}$ over the (abelian) extension $K^{\prime}:=K((-3)^{1/4},2^{1/3})$.

[TABLE]

We see that $E$ and $E^{\mathrm{ss}}$ are isomorphic over $K^{\prime}$ and that the reduction $\bar{E}^{\mathrm{ss}}$ is the nonsingular curve given by $y^{2}+y=x^{3}$.

Example 3**.**

Suppose we want to find a semistable model $E^{\mathrm{ss}}$ of the elliptic curve $E/\mathbb{Q}_{2}$ given by $y^{2}=x^{3}-x$ at the prime $(2)$. Just as in the previous example, this elliptic curve is CM, and so any semistable model $E^{\mathrm{ss}}$ should again have good reduction. Moreover, $E$ is isomorphic over $\mathbb{Q}_{2}$ to the Legendre curve with $\lambda=2$, and since $m=v(\lambda)=1<4$, we may apply Theorem 1.

We let $\beta=2$, noting that this choice of $\beta$ satisfies the requirement that $v(\beta)=\frac{1}{3}m+\frac{2}{3}=1$. Then we have

[TABLE]

One can readily check that if we set $\delta=0$, the roots of this polynomial are $1\pm\sqrt{1\pm\frac{2}{\sqrt{3}}}$, where the choices of sign are independent. We take $\alpha=1+\sqrt{1+\frac{2}{\sqrt{3}}}$. Now evaluating the formulas in (6) yields the following equation for $E^{\mathrm{ss}}$, over the extension $K^{\prime}:=\mathbb{Q}_{2}(2^{1/2},3^{1/4},\sqrt{\sqrt{3}+2})$ (which is abelian over $\mathbb{Q}_{2}(i)$ as it is contained in $\mathbb{Q}_{2}(\zeta_{24},3^{1/4})$, where $\zeta_{24}$ is a primitive $24$th root of unity).

[TABLE]

We see that $E$ and $E^{\mathrm{ss}}$ are isomorphic over $K^{\prime}$ and that the reduction $\bar{E}^{\mathrm{ss}}$ is again the nonsingular curve given by $y^{2}+y=x^{3}$.

3. The $v(\lambda)\geq 4$ case

For this section, we adopt exactly the same set-up but treat the complimentary case where $v(j(E))\leq 0$. In this case, we see from the formula in (1) that we have $m:=v(\lambda)\geq 4$. Therefore, under this assumption, any semistable model $E^{\mathrm{ss}}$ should have good reduction if and only if $m=4$; otherwise $E^{\mathrm{ss}}$ has multiplicative reduction. As in §2, we will show how to find algebraic elements $\alpha,\beta\in\bar{K}$ such that evaluating $a_{1},a_{2},a_{3}\in K^{\prime}:=K(\beta^{1/2},\alpha,F(0)^{1/2})$ using the formulas in (6) yields an equation of the form in (2) (with $a_{4}=a_{6}=0$) for an elliptic curve with semistable reduction.

Theorem 4**.**

Assume that $m\geq 4$. Let $\beta\in(K^{\times})^{2}$ be any element such that $v(\beta)=2$ (e.g. $\beta=4$), and choose an element $\alpha\in K$ such that $2\leq v(\alpha)\leq m-2$. Then $E$ is isomorphic over $K^{\prime}:=K(F(0)^{1/2})$ to the elliptic curve $E^{\mathrm{ss}}$ given by the equation in (2), where $a_{4}=a_{6}=0$ and the other coefficients $a_{i}$ are given by the formulas in (6).

We have $v(a_{1})=0$, $v(a_{2})\geq 0$, and $v(a_{3})=v(\alpha)-2$ (when $v(\alpha)>2$, this directly implies that $E^{\mathrm{ss}}$ has multiplicative reduction). The curve $E^{\mathrm{ss}}$ has good reduction if $m=4$ and has multiplicative reduction otherwise.

Proof.

First of all, we note that $v(\alpha-1)=0$. The condition that $v(\alpha)\leq m-2$ ensures that $v(\lambda)>v(\alpha)$, so $v(\alpha-\lambda)=v(\alpha)$. Therefore, we have

[TABLE]

and

[TABLE]

It follows that $v(a_{3})=v(\alpha)-2\geq 0$ and $v(a_{1})=0$; in particular, $v(a_{3})=0$ if and only if $m=4$.

We next check that $v(a_{2})\geq 0$. In order to do so, we recall the formula in (10) which we derived earlier:

[TABLE]

Since $\min\{v(3\alpha^{4}),v(4(1+\lambda)\alpha^{3}),v(6\lambda\alpha^{2}),v(\lambda^{2})\}=\min\{4v(\alpha),3v(\alpha)+2,2v(\alpha)+m+1,2m\}\geq 2v(\alpha)+4$, we have $v(a_{2})=v(4F(0)\beta^{4}a_{2})-v(4F(0))-4v(\beta)\geq 2v(\alpha)+4-(2v(\alpha)-4)-8=0$, as desired.

Finally we assume that $v(\alpha)=2$ and set out to show that the curve $E^{\mathrm{ss}}$ has good reduction if and only if $m=4$. Any singular point $(x,y)$ on $\bar{E}^{\mathrm{ss}}$ satisfies the following set of equations.

[TABLE]

By solving for $x$ and $y$ in the bottom two equations and plugging the results in the top equation, we see that if such a point $(x,y)$ exists, we must have

[TABLE]

which, after dividing by $\frac{\bar{a}_{3}^{2}}{\bar{a}_{1}^{2}}$ and simplifying, yields

[TABLE]

We now show that this is the case if and only if $m>4$. We compute the following equivalences modulo the prime ideal of $R^{\prime}$, using the formulas in (LABEL:eq_formulas3).

[TABLE]

Meanwhile, using what we know from (17), we compute the equivalence

[TABLE]

Putting (21) and (22) together, we get

[TABLE]

Since the valuation of the right-hand term is $2m-2-4-2=2m-8$, the above expression is equivalent to [math] if and only if $m>4$, and we are done.

∎

Remark 5**.**

It was pointed out to the author by Leonardo Fiore that in the situation of Theorem 4, a semistable model can be obtained by choosing $\alpha$ to be any element satisfying $v(\alpha)\geq 2$ (e.g. $\alpha=0$), as long as we allow the possibility that $a_{4}\neq 0$ or $a_{6}\neq 0$. Indeed, there is an isomorphism (defined over $R$) between any two such models induced by translating $x$ by the integral element $\beta^{-1}(\alpha_{1}-\alpha_{2})\in R$, where $\alpha_{1}$ and $\alpha_{2}$ are the choices of $\alpha$ determining the models.

We now recall that the $2$-torsion subgroup $E[2]\subset E(\bar{K})$ is given by $\{\mathcal{O},(0,0),(1,0),(\lambda,0)\}$.

Corollary 6**.**

Assume that $m>4$ and construct the semistable model $E^{\mathrm{ss}}$ of $E$ as in the statement of Theorem 4. The reduction of the $2$-torsion subgroup $E^{\mathrm{ss}}[2]$ coincides with the subset consisting of the infinity point $\bar{\mathcal{O}}$ and the cusp $P$ of $\bar{E}^{\mathrm{ss}}$; the inverse images of $\{\bar{\mathcal{O}}\}$ and $\{P\}$ correspond to the subgroup $\{\mathcal{O},(1,0)\}\subset E[2]$ and its coset $\{(0,0),(\lambda,0)\}\subset E[2]$ respectively.

Proof.

It is clear that the infinity point $\mathcal{O}$ of $E$ gets sent to $\bar{\mathcal{O}}\in\bar{E}^{\mathrm{ss}}$. Now since $\varphi:E\stackrel{{\scriptstyle\sim}}{{\to}}E^{\mathrm{ss}}$ sends the first coordinate of any point $(x,y)\in E(K^{\prime})\smallsetminus\{\mathcal{O}\}$ to $\beta^{-1}(x-\alpha)$, we see that the first coordinate of the image $\varphi((1,0))\in E^{\mathrm{ss}}(K^{\prime})$ (resp. of each image $\varphi((0,0)),\varphi((\lambda,0))\in E^{\mathrm{ss}}(K^{\prime})$) reduces to $\infty$ (resp. $-\frac{\bar{\alpha}}{\bar{\beta}}$). As in the proof of Theorem 4, the cusp $P$ has $x$-coordinate $-\frac{\bar{a}_{3}}{\bar{a}_{1}}=-\frac{\bar{\alpha}}{\bar{\beta}}$. Since $\bar{\mathcal{O}}$ (resp. $P$) is the only point of $\bar{E}^{\mathrm{ss}}$ whose first coordinate is $\infty$ (resp. $-\frac{\bar{\alpha}}{\bar{\beta}}$), we are done.

∎

Remark 7**.**

In a similar fashion to how we proved the above corollary, it is straightforward to show directly from Theorem 4 (resp. Theorem 1) that in the case that $v(\lambda)=4$ (resp. $0\leq v(\lambda)<4$), the elements $\mathcal{O},(1,0)\in E[2]$ are mapped via $\varphi$ composed with reduction to the infinity point $\bar{\mathcal{O}}$ of $\bar{E}^{\mathrm{ss}}$ and the elements $(0,0),(\lambda,0)\in E[2]$ map to another point of $\bar{E}^{\mathrm{ss}}$ (resp. the elements of $E[2]$ are all mapped to the infinity point $\bar{\mathcal{O}}$ of $\bar{E}^{\mathrm{ss}}$). Since the image of $E[2]$ under $\varphi$ composed with reduction must be contained in the $2$-torsion subgroup $\bar{E}^{\mathrm{ss}}[2]$, we see in this way that when $v(\lambda)=4$ (or equivalently, when $j(\bar{E}^{\mathrm{ss}})\neq 0$), the reduced curve $\bar{E}^{\mathrm{ss}}$ is ordinary. This is one direction of the equivalence given in [5, Exercise 5.7], which states that an elliptic curve over a field of characteristic $2$ is supersingular if and only if its $j$-invariant is [math]. Since the other direction of that equivalence implies that $\bar{E}^{\mathrm{ss}}$ is supersingular in the $v(\lambda)<4$ case, we see that $\bar{E}^{\mathrm{ss}}[2]=\{\bar{\mathcal{O}}\}$ coincides with the reduction of $E[2]\cong E^{\mathrm{ss}}[2]$.

Example 8**.**

Consider the elliptic curve $E/\mathbb{Q}_{2}$ given by $y^{2}=x(x-1)(x-16)$. This curve is already in Legendre form with $\lambda=16$. Since $m=v(\lambda)=4$, any semistable model $E^{\mathrm{ss}}$ will have good reduction, and we may apply Theorem 4.

We set $\alpha=\beta=4$, noting that $v(\alpha)=2=m-2$. Now evaluating the formulas in (6) yields the following equation for $E^{\mathrm{ss}}$, over the extension $K^{\prime}:=\mathbb{Q}_{2}(i)$.

[TABLE]

It is easy to check directly that the reduction $\bar{E}^{\mathrm{ss}}$, given by $y^{2}+xy+y=x^{3}+x^{2}$, is nonsingular.

Example 9**.**

Consider the elliptic curve $E/\mathbb{Q}_{2}$ given by $y^{2}=x(x-1)(x-64)$. The curve is again already in Legendre form, this time with $\lambda=64$ so $m=v(\lambda)=6$. Again, we may apply Theorem 4, but in this case, the semistable model $E^{\mathrm{ss}}$ we arrive at will have multiplicative reduction.

As before, we set $\beta=4$, but this time, we let $\alpha=8$, noting that $2<v(\alpha)=3\leq m-4$. Now evaluating the formulas in (6) yields the following equation for $E^{\mathrm{ss}}$, over the extension $K^{\prime}:=\mathbb{Q}_{2}(i)$.

[TABLE]

The reduction $\bar{E}^{\mathrm{ss}}$ is $y^{2}+xy=x^{3}$, which visibly has a node at the point $(0,0)$; hence, $E^{\mathrm{ss}}$ has (split) multiplicative reduction, as expected.

4. Acknowledgements

The author is grateful to the referee, whose suggestions have helped to improve the exposition and to place this work in a broader context.