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Fix a conductor. Then

1) Do the elliptic curves in the same isogeny class have the same reduction type at a prime of bad reduction of the curve ?

2) Do the elliptic curves belonging to two different isogeny classes corresponding to the fixed conductor, have the same reduction type at a prime of bad reduction of the curve ?

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1) yes if you just seperate them into "additive", "multiplicative" and "good". no if you mean the Kodaira type. –  Chris Wuthrich Nov 11 '13 at 9:20
2) as for 1) because the conductor knows about the reduction type, but wouldn't know if split or non-split multiplicative reduction for instance. –  Chris Wuthrich Nov 11 '13 at 9:21
@ChrisWuthrich for (1), what about split & non-split multiplicative reduction ? also suggest a reference where i can have the necessary ideas behind your comments. –  Suman Nov 11 '13 at 10:01
Have you tried to do some examples? The formula for a 2-isogeny is very simple, and if you play with a few numeric examples (even just looking at prime $p\ge5$), I expect that you'll be able to answer your question (1) on your own. –  Joe Silverman Nov 11 '13 at 14:19
For multiplicative reduction there's a nice pattern in the Kodaira types of curves related by an isogeny of prime degree $p$: if they have type I$_m$ and I$_n$ then either $m=pn$ or $n=pm$. (Can be proved via Tate curves.) –  Noam D. Elkies Nov 12 '13 at 2:30

1 Answer 1

up vote 5 down vote accepted

Let $E$ be an elliptic curve over a $p$-adic field $k$. Let $\varphi: E \to E'$ be an isogeny defined over $k$. Write minimal Weierstrass equations with integer coefficients for both curves. Write $\varphi$ with fractions of polynomials with integer coefficients and consider the reduction of that map on the reduced equations. It should be clear that $\varphi$ sends a singular point to singular point. Also it sends tangents defined over $k$ to tangents defined over $k$. From looking at the possible cases, using the dual isogeny if needed, one should be able to conclude that if $E$ has good, split multiplicative, non-split multiplicative or additive reduction then $E'$ has the same of these four types. (Of course, one better looks at the isogeny between Néron models, but I tried to keep it elementary here.)

It is much harder to know what the exact Kodaira type do, especially in the potentially good, additive cases. In the multiplicative case, the types will be $I_n$ and $I_{pn}$; in this order or in the other order.

The conductor $N$ is defined such that ord$(N) = 0$, $1$ or $\geq 2$ depending on whether $E$ has good, multiplicative or additive reduction. So two curves with the same conductor will share the same reduction among these three types. However the conductor does not know if the multiplicative reduction is split or non-split. For instance the curves 37a1 and 37b1 will illustrate that.

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Nice answer, you certainly shouldn't delete it. I expect that it would be quite challenging to figure out how the Kodaira-Neron reduction type changes under a $p$-isogeny when there is additive reduction at $p$, especially for $p=2$ and $p=3$. –  Joe Silverman Nov 11 '13 at 18:49
In principle, the Kodaira types are worked out here, Table on page 2, except when degree(isogeny)=prime<5 and the curve has wild reduction. In all the tame additive cases, the Kodaira type either stays the same or becomes opposite (III->III* etc.), depending on the prime mod 12. –  Tim Dokchitser Nov 11 '13 at 19:38
I also like this argument: whether the reduction at $p$ is good/split mult./non-split mult./additive is encoded in the local factor of the $L$-function at $p$; it is then quadratic/1-T/1+T/1. Since the $L$-function is invariant under isogeny, these reduction types must be as well. –  Tim Dokchitser Nov 11 '13 at 19:41
@Tim: Yep, it's the wild additive cases that are hard. I messed up once on the similar problem of calculating what happens to the Kodaira type of twists, when I tried to compute it for p=2 and a quadratic twist by an extension ramified at 2. –  Joe Silverman Nov 12 '13 at 1:04
Yes, the wild ones are awful. For quadratic twists, I know that in the potentially multiplicative case Dino Lorenzini worked out the types exactly. The potentially good case remains open, I think(?) –  Tim Dokchitser Nov 12 '13 at 8:33

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