criterion of Néron-Ogg-Shafarevich

In this entry, we use the following notation. $K$ is a local field, complete with respect to a discrete valuation $\nu$, $R$ is the ring of integers of $K$, $ℳ$ is the maximal ideal of $R$ and $𝔽$ is the residue field of $R$.

Definition.

Let $\mathrm{\Xi }$ be a set on which $\mathrm{Gal}\mathit{}\mathrm{(}\overline{K}\mathrm{/}K\mathrm{)}$ acts. We say that $\mathrm{\Xi }$ is unramified at $\nu$ if the action of the inertia group $I_{\nu }$ on $\mathrm{\Xi }$ is trivial, i.e. ${\zeta }^{\sigma }\mathrm{=}\zeta$ for all $\sigma \mathrm{\in }{I}_{\nu }$ and for all $\zeta \mathrm{\in }\mathrm{\Xi }$.

Theorem (Criterion of N$\stackrel{\mathrm{´}}{𝐞}$ron-Ogg-Shafarevich).

Let $E\mathrm{/}K$ be an elliptic curve defined over $K$. The following are equivalent:

1. $E$ has good reduction over $K$;

2. $E[m]$ is unramified at $\nu$ for all $m\ge 1$, $\gcd (m,\mathrm{char}(𝔽))=1$;

3. The Tate module $T_l(E)$ is unramified at $\nu$ for some (all) l, $l\ne \mathrm{char}(𝔽)$;

4. $E[m]$ is unramified at $\nu$ for infinitely many integers $m\ge 1$, $\gcd (m,\mathrm{char}(𝔽))=1$.

Corollary.

Let $E\mathrm{/}K$ be an elliptic curve. Then $E$ has potential good reduction if and only if the inertia group $I_{\nu }$ acts on $T_l\mathit{}\mathrm{(}E\mathrm{)}$ through a finite quotient for some prime $l\mathrm{\ne }\mathrm{char}\mathit{}\mathrm{(}\mathrm{F}\mathrm{)}$.

Proof of Corollary.

($\Rightarrow$) Assume that $E$ has potential good reduction. By definition, there exists a finite extension of $K$, call it $K^{\prime }$, such that $E/K^{\prime }$ has good reduction. We can extend $K^{\prime }$ (if necessary) so $K^{\prime }/K$ is a Galois finite extension.

Let ${\nu }^{\prime }$ and $I_{{\nu }^{\prime }}$ be the corresponding valuation and inertia group for $K^{\prime }$. Then the theorem above ( (1)$\Rightarrow$(3) ) implies that $T_l(E)$ is unramified at ${\nu }^{\prime }$ for all $l$, $l\ne \mathrm{char}(𝔽)=\mathrm{char}({𝔽}^{\prime })$ (since ${𝔽}^{\prime }$ is a finite extension of $𝔽$). So $I_{{\nu }^{\prime }}$ acts trivially on $T_l(E)$ for all $l\ne \mathrm{char}({𝔽}^{\prime })$. Thus $I_{\nu }\hookrightarrow T_l(E)$ factors through the finite quotient $I_{\nu }/I_{{\nu }^{\prime }}$.

($\Leftarrow$) Let $l\ne \mathrm{char}(𝔽)$, and assume $I_{\nu }\hookrightarrow T_l(E)$ factors through a finite quotient, say $I_{\nu }/J$. Let ${\overline{K}}^J$ be the fixed field of $J$, then ${\overline{K}}^J/{\overline{K}}^{I_{\nu }}$ is a finite extension, so we can find a finite extension $K^{\prime }/K$ so that ${\overline{K}}^J=K^{\prime }{\overline{K}}^{I_{\nu }}$. So the inertia group of $K^{\prime }$ is equal to $J$, and $J$ acts trivially on $T_l(E)$. Hence the criterion ( (3)$\Rightarrow$(1) ) implies that $E$ has good reduction over $K^{\prime }$, and since $K^{\prime }/K$ is finite, $E$ has potential good reduction. ∎

Proposition.

Let $E\mathrm{/}K$ be an elliptic curve. Then $E$ has potential good reduction if and only if its $j$-invariant is integral ( i.e. $j\mathit{}\mathrm{(}E\mathrm{)}\mathrm{\in }R$ ).

Proof.

($\Leftarrow$) Assume $\mathrm{char}(𝔽)\ne 2$, it is easy to prove that we can extend $K$ to a finite extension $K^{\prime }$ so that $E$ has a Weierstrass equation:

$$E:y^2=x(x-1)(x-\lambda )\mathit{\hspace{1em}}\lambda \ne 0,1$$ (1)

Since we are assuming $j(E)\in R$, and:

$${(1-\lambda (1-\lambda ))}^3-j{\lambda }^2{(1-\lambda )}^2=0$$ (2)

then $\lambda \in R$ and $\lambda \ne 0,1mod{ℳ}^{\prime }$ ( $\Rightarrow$ ${\mathrm{\Delta }}^{\prime }\in {(R^{\prime })}^{*}$ ). Hence $E/K^{\prime }$ has good reduction, i.e. $E$ has potential good reduction.

($\Rightarrow$) Assume that $E$ has potential good reduction, so there exists $K^{\prime }$ so that $E/K^{\prime }$ has good reduction. Let ${\mathrm{\Delta }}^{\prime }$, $c_4^{\prime }$ the usual quantities associated to the Weierstrass equation over $K^{\prime }$. Since $E/K^{\prime }$ has good reduction, ${\mathrm{\Delta }}^{\prime }\in {(R^{\prime })}^{*}$, and so $j(E)=\frac{{(c_4{}^{\prime })}^3}{{\mathrm{\Delta }}^{\prime }}\in R^{\prime }$. But since $E$ is defined over $K$, $j(E)\in K$, so $j(E)\in K\bigcap R^{\prime }=R$. ∎

Title	criterion of Néron-Ogg-Shafarevich
Canonical name	CriterionOfNeronOggShafarevich
Date of creation	2013-03-22 17:14:58
Last modified on	2013-03-22 17:14:58
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	4
Author	alozano (2414)
Entry type	Theorem
Classification	msc 14H52
Synonym	criterion of Neron-Ogg-Shafarevich
Related topic	EllipticCurve
Related topic	ArithmeticOfEllipticCurves