criterion of Néron-Ogg-Shafarevich

In this entry, we use the following notation. $K$ is a local field, complete with respect to adiscrete 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 areequivalent:

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 $\nu$ forsome (all) l, $l\ne \mathrm{char}(𝔽)$;

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

Corollary.

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

Proof of Corollary.

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

Let ${\nu }^{\prime }$ and $I_{{\nu }^{\prime }}$ be the corresponding valuationand 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 $I_{{\nu }^{\prime }}$ actstrivially on $T_l(E)$ for all $l\ne \mathrm{char}({𝔽}^{\prime })$. Thus$I_{\nu }\hookrightarrow T_l(E)$ factors through the finitequotient $I_{\nu }/I_{{\nu }^{\prime }}$.

($\Leftarrow$) 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 finiteextension $K^{\prime }/K$ so that ${\overline{K}}^J=K^{\prime }{\overline{K}}^{I_{\nu }}$. Sothe 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, $E$ haspotential good reduction.∎

Proposition.

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

Proof.

($\Leftarrow$) Assume $\mathrm{char}(𝔽)\ne 2$, it is easy to provethat 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 $E/K^{\prime }$ hasgood reduction, i.e. $E$ has potential goodreduction.

($\Rightarrow$) Assume that $E$ has potential good reduction,so there exists $K^{\prime }$ so that $E/K^{\prime }$ has goodreduction. Let ${\mathrm{\Delta }}^{\prime }$, $c_4^{\prime }$ the usual quantities associatedto 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 $j(E)\in K\bigcap R^{\prime }=R$.∎