criterion of NéronOggShafarevich
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 $K$, $\mathcal{M}$ is the maximal ideal of $R$ and $\mathbb{F}$ 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$\stackrel{\mathrm{\xb4}}{\mathbf{e}}$ronOggShafarevich).
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 $m\ge 1$, $\mathrm{gcd}(m,\mathrm{char}(\mathbb{F}))=1$;

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

4.
$E[m]$ is unramified at $\nu $ for infinitely many integers $m\ge 1$, $\mathrm{gcd}(m,\mathrm{char}(\mathbb{F}))=1$.
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 $l$, $l\ne \mathrm{char}(\mathbb{F})=\mathrm{char}({\mathbb{F}}^{\prime})$ (since ${\mathbb{F}}^{\prime}$ is a finite extension of $\mathbb{F}$). So ${I}_{{\nu}^{\prime}}$ acts trivially on ${T}_{l}(E)$ for all $l\ne \mathrm{char}({\mathbb{F}}^{\prime})$. Thus ${I}_{\nu}\hookrightarrow {T}_{l}(E)$ factors through the finite quotient ${I}_{\nu}/{I}_{{\nu}^{\prime}}$.
($\Leftarrow $) Let $l\ne \mathrm{char}(\mathbb{F})$, 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.
Proof.
($\Leftarrow $) Assume $\mathrm{char}(\mathbb{F})\ne 2$, it is easy to prove that we can extend $K$ to a finite extension ${K}^{\prime}$ so that $E$ has a Weierstrass equation:
$$E:{y}^{2}=x(x1)(x\lambda )\mathit{\hspace{1em}}\lambda \ne 0,1$$  (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 $\lambda \ne 0,1mod{\mathcal{M}}^{\prime}$ ( $\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 NéronOggShafarevich 

Canonical name  CriterionOfNeronOggShafarevich 
Date of creation  20130322 17:14:58 
Last modified on  20130322 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 NeronOggShafarevich 
Related topic  EllipticCurve 
Related topic  ArithmeticOfEllipticCurves 