Let $K$ be a complete local field of characteristic 0 with residue field $k$ of characteristic $p>0$ and ring of integers $R=\mathcal{O}_{K}$. Suppose we have an isogeny of elliptic curves: $$f:F\rightarrow E$$ over $K$ and let $\mathcal{C}$ be the minimal proper regular model of $E$ over $R$. Suppose that $f$ has degree $d$ and so corresponds to a finite extension of function fields of the same degree: $$f^{*}:K(E)\rightarrow K(F)$$ Let $\mathcal{D}$ be the normalisation of $\mathcal{C}$ in $K(F)$ so that the normalisation map: $$\phi:\mathcal{D}\rightarrow \mathcal{C}$$ extends $f$ to $\text{Spec}(R)$.

I would like to know what can be said about the special fibre $\mathcal{D}_{k}$ of $\mathcal{D}$.

In particular, I am interested in the case where $E$ has bad multiplicative reduction so that $\mathcal{C}_{k}$ is a nodal curve with reduced components and I would like to know when $\mathcal{D}_{k}$ is also nodal with reduced components.

My feeling is that the answer should depend on how ramified $f$ is and potentially depend on whether $f$ is Galois (i.e. when $\text{ker}(f)$ is defined over $K$).

Apologies if this question is slightly naïve, I do not have a lot of familiarity with taking normalisations in finite extensions and so any resources which give examples of such normalisations of curves where more can be said would also be greatly appreciated.


1 Answer 1


Edit: The argument in this answer is only complete in the case that the degree $d$ is prime to $p$. I think the statement is still true in the other case and will think more about it.

All isogenies have $\operatorname{ker}(f)$ defined over $K$, since it is the inverse image of a $K$-point (the identity) under a map of schemes defined over $K$ ($f$). However, not all isogenies are Galois as finite étale coverings. The covering is Galois if and only if the action of $\operatorname{Gal}(K)$ on $\operatorname{ker}(f)$ is trivial. (This is because the group of automorphisms of $F_{\overline{K}}$ over $E_{\overline{K}}$ is exactly $\operatorname{ker}(f)$, which acts simply transitively on the fibers, and the group of automorphisms of $F$ over $E$ is the subset of these that are defined over $K$, which is simply transitive if and only if all elements of $\operatorname{Aut}(F_{\overline{K}}/E_{\overline{K}})$ are defined over $K$.)

There exists a unique isogeny of degree $d$ such that $\mathcal D$ is a nodal curve with reduced components. A necessary condition is that $\ker(f)$ have trivial action of the inertia group of $K$, and if the multiplicative reduction is split then trivial action of the Galois group is also necessary, but not quite sufficient. The $n$-torsion of an elliptic curve with multiplicative reduction caries a canonical $\mu_n$-subspace, the reduction mod $n$ of the unique nontrivial proper Galois-invariant subspace of the Tate module, and a necessary and sufficient condition is that $\operatorname{ker}(f)$ must have trivial intersection with this subspace.

Why is this? The map $\mathcal D \to \operatorname{Spec} R$ factors through $\mathcal D \to \mathcal C$, so the special fiber of $\mathcal D$ is the inverse image of the special fiber of $\mathcal C$. Hence each component of the special fiber of $\mathcal D$ is reduced if and only if the covering $\mathcal D\to \mathcal C$ is unramified at its generic point.

Since the covering $\mathcal D \to \mathcal C$ is unramified on the generic fiber, if it is also unramified at the generic point of each component of the special fiber then it is unramified outside a set of codimension $2$. Because $\mathcal C$ is regular, purity of the branch locus guarantees the covering is unramified everywhere, i.e. finite étale. Conversely, finite étaleness certainly guarantees the special fiber is reduced and nodal.

Finite étale covers of proper schemes over Henselian local rings are equivalent to finite étale covers of the special fiber. The special fiber of $\mathcal C$ is a loop of $\mathbb P^1$'s. An unramified covering is constant over each $\mathbb P^1$ and thus a geometrically connected unramified covering of $\mathcal C$ must geometrically consist of a loop of $\mathbb P^1$s winding around that loop multiple times. There are multiple ways to descend this covering to $k$, but only one where the identity lifts to a $k$-rational point, as it must in an isogeny. This condition is also sufficient to obtain an isogeny since a finite étale cover of a genus $1$ curve over $K$ where the identity lifts to a $K$-rational point is an isogeny.

We have shown there exists a unique isogeny of degree $d$ with reduced special fiber. Let's now characterize it. The kernel of the isogeny consists of lifts of $d$ points on $d$ different components of the special fiber of $\mathcal D$ (i.e. the $d$ components arising from the identity component), which is the minimal regular model of $F$ (since if $\mathcal C$ has an $I_n$ special fiber then $\mathcal D$ has $I_{nd}$, which is also minimal). All isogenies with kernels of this type arise this way, since you can view them as quotienting the minimal regular model of the covering elliptic curve by a group of automorphisms with no fixed points and thus they give finite étale covers.

So it suffices to characterize which subgroups of the torsion points of an elliptic curve with multiplicative reduction map injectively to the components. It is convenient here to use the Tate curve model where a curve with split multiplicative reduction is $\mathbb G_m / \langle q \rangle$ for $q$ an element of $K$ with positive $p$-adic valuation and its $n$-torsion points are the $n$th roots of unity times powers of the $n$th roots of $q$. The components are given by the different possible valuations modulo the valuation of $q$. A torsion point lies in any component at all if and only if it has an integer valuation, and the elements of the subgroup lie on distinct components if and only if they have distinct integer valuations.

Since elements of the Galois group send an element to another element of the same valuation, this forces all elements of $\ker(f)$ to be fixed by the Galois group, and since none of them have valuation $0$, none of them lie in the special subgroup of valuation $0$ elements, which is the canonical $\mu_n$-subspace mentioned above. Conversely, a Galois-invariant subgroup which doesn't intersect the $\mu_n$-subspace must not have two elements of the same valuation and thus must have only elements of integral valuation since an element of rational valuation would be sent to a different element of the same valuation by the Galois group.

For curves with non-split multiplicative reduction, we base-change to a unramified extension of $K$ where the covering spits and apply the criterion there.

  • $\begingroup$ Hi Will, thanks for the very complete answer! Apologies, my intended meaning was that $\text{ker}(f)$ has its geometric points defined over $K$ i.e. $\text{Gal}(\bar{K}/K)$ acts on the set of geometric points trivially, but of course what I have written above seems to only ask that $\text{ker}(f)$ is a $K$-scheme. $\endgroup$ Commented May 22 at 10:44
  • $\begingroup$ a question regarding the map $\mathcal D \to \mathcal C$: Why a component of the special fiber of $\mathcal D$ is reduced iff $\mathcal D \to \mathcal C$ is unramified at generic point, once we know that the special fiber $\mathcal D$ is inverse image of special fiber of $\mathcal C$? $\endgroup$
    – user267839
    Commented May 26 at 22:00
  • $\begingroup$ @user267839 Thanks for pointing this out, there is a subtlety here. Restricted to the generic point, the map $\mathcal D \to \mathcal C$ is a map of regular local rings. This is unramified if and only if the inverse image of a uniformizer is a uniformizer and not a higher power of a uniformizer (i.e. the inverse image of the special fiber is reduced) and the extension on residue fields is separable. The inseparable case can only happen if the degree is divisible by $p$. I think in fact the inseparable case doesn't happen but the argument needs to proceed a different way. $\endgroup$
    – Will Sawin
    Commented May 27 at 1:34
  • $\begingroup$ but let's stick on separable case, I'm not sure if I understood the geometric picture correctly. So assume that the restriction of $\mathcal D \to \mathcal C$ to generic point - that's just the original map $f:F \to E$, right? - is unramified, so as you wrote uniformizer maps to uniformizer in local rings - but everyting over base $K$. The claim is that any component of special fiber $D_k$ is reduced, so over base $k =O_K/m$. For sake of reaching contradiction, let $D^1_k$ nonreduced. Any closed point of $F$ ($=D_K$) gives rise (by taking closure) to a "horizontal" divisor intersecting $\endgroup$
    – user267839
    Commented May 27 at 7:29
  • 1
    $\begingroup$ @user267839 It probably depends a bit on how you define "multiplicative reduction". One knows various definitions are equivalent because of the structure theory of elliptic fibrations, so one doesn't need to specify a definition. If one knows the special fiber is reduced and topologically a loop of $\mathbb P^1$, then one computes the intersection of the special fiber with one of the $\mathbb P^1$s: On the one hand this is zero because the fiber is movable, and on the other hand this is two minus the intersection multiplicites of the two singularities, so the multiplicities are both $1$. $\endgroup$
    – Will Sawin
    Commented May 27 at 15:15

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