2
$\begingroup$

Of course, the primes dividing the conductor are precisely those dividing the minimal discriminant. But I cannot find any source that addresses the possibility of a prime appearing to the first power in the minimal discriminant but appearing to the second power in the conductor. Put another way, imagine a square-free discriminant. Is it possible for some $p$ dividing the discriminant that the curve has additive reduction at $p$?

And to clarify, an elliptic curve over $\mathbb{Q}$.

$\endgroup$
2
$\begingroup$

The answer to the question in the title is yes. You are looking for Ogg's formula. See Silverman's "Advanced topics in the arithmetic of elliptic curves", Sections 9, 10, and 11, and in particular Ogg's fomula in 11.1:

Let $K/\mathbb{Q}_p$ be a local field and let $E/K$ be an elliptic curve, and let

  • $\nu_K(\mathcal{D}_{E/K})$ = the valuation of the minimal discriminant of $E/K$,

  • $f(E/K)=$ the exponent of the conductor of $E/K$,

  • $m(E/K)=$ the number of components on the special fiber of $E/K$ (thus $m(E/K)\geq 1$).

Then: $$\nu_K(\mathcal{D}_{E/K})=f(E/K)+m(E/K)-1.$$ In particular, $\nu_K(\mathcal{D}_{E/K})\geq f(E/K)$.

$\endgroup$
0
$\begingroup$

Having tried to work it out some more, I think I have figured it out - the answering being yes (at least in characteristic not $2$ or $3$ and ignoring the primes $2$ and $3$).

Write $$E: y^{2} = x^3 + Ax + b$$ and let $p$ be a prime $> 3$. Then the discriminant having bad reduction means $4A^{3} + 27B^{2} \equiv 0$ (mod $p$). Furthermore having additive reduction means $c_{4} \equiv A \equiv 0$ (mod $p)$. Yet this means that $B \equiv 0$ (mod $p$) so that indeed the discriminant is divisible by $p^{2}$ and the conductor divides it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.