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Let $\pi:X\to E$ be a finite étale morphism, where $E$ is an elliptic curve over a number field $K$. Assume $X$ to be connected, and to be of genus 1.

Edit: Assume $X$ and $E$ have semi-stable reduction over $O_K$.

Is the minimal discriminant of $X$ equal to the minimal discriminant of $E$ multiplied by $\deg \pi$?

What if the genus of $X$ is bigger than 1. (This can't happen by QiL's comment below.)

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What is the stable discriminant ? For last part of your question, $X$ has always genus $1$ by Riemann-Hurwitz formula. –  user18119 Nov 23 '11 at 17:07
    
See the edited question. –  Konoyaro Nov 25 '11 at 8:57
    
In general the answer must be no. For instance you can consider the map $E \to E$ given by multiplication by $n$; this is an étale cover of degree $n^2$, but the minimal discriminant does not change. –  Dan Petersen Nov 25 '11 at 10:11
    
Hmmm...I just have a feeling one one should be able to say "something" about the minimal discriminants...I guess it was too much to hope for. –  Konoyaro Nov 25 '11 at 13:32
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1 Answer

up vote 1 down vote accepted

If $E_1 \to E_2$ is an isogeny of elliptic curves over a number field $K$, then $E_1$ and $E_2$ have the same places of bad reduction, and so their minimal discriminants are divisible by the same primes.

As an example of how minimal discriminants can change under isogeny, consider the map $X_1(11) \to X_0(11)$ of elliptic curves over $\mathbb Q$, which is an isogeny of degree $5$. Up to possible $\pm$ signs (which I don't remember off the top of my head), the minimal discriminant of $X_1(11)$ is $11$, and of $X_0(11)$ is $11^5$.

In both cases the conductor is the same (this is a general feature of isogenous elliptic curves) --- namely $11$. The power of $11$ dividing the minimal discriminant relates to the size of the connected component group of the fibre over $11$ of the Neron model. (This fibre is connected for $X_1(11)$ --- in fact it is general result of Conrad, Edixhoven, and Stein that the Neron model of the Jacobian of $X_1(p)$ over $\mathbb Z$ has connected fibre at $p$ for all primes $p$ --- and has component group of order $5$ for $X_0(11)$.)

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Is the modular curve $X_1(n)$ defined over $\mathbf{Q}$? I thought any field of definition should contain $\mathbf{Q}(\zeta_n)$. –  Harry Mar 18 '12 at 18:09
    
@Harry: Dear Harry, Yes, $X_1(n)$ is defined over $\mathbb Q$ (although there are two natural models, depending on whether you think of it as classifying embeddings $\mathbb Z/n \hookrightarrow E$ or embeddings $\mu_n \hookrightarrow E$; the two models are easily related by an appropriate twist). Perhaps you are thinking of the curve $X(n)$ of full level $n$, which is naturally defined over $\mathbb Q(\zeta_n)$? Regards, –  Matt E Mar 18 '12 at 19:06
    
I thought "the" field of definition of a modular curve given by a congruence subgroup $\Gamma$ can be "computed" as follows. Let $H$ be a subgroup of $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$ such that $\Gamma$ is the inverse image of $H$ under the natural map $\mathrm{SL}_2(\mathbf{Z})\to \mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$. Then $H$ acts on $\mathbf{Q}(\zeta_n)$ via the determinant map $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})\to (\mathbf{Z}/n\mathbf{Z})^\ast$. Then the modular curve $X_\Gamma$ can be defined over the field of invariants $\mathbf{Q}(\zeta_n)^H$. But maybe this isn't minimal? –  Harry Mar 18 '12 at 21:00
    
Just to be precise, the action of $(\mathbf{Z}/n\mathbf{Z})^\ast$ on $\mathbf{Q}(\zeta_n)$ is given by $a \cdot \zeta_n = \zeta_n^a$. So the field of invariants of $H$ is $\mathbf{Q}(\zeta_n)$ when $\Gamma = \Gamma_1(n)$ and $H$ is the group of matrices with first column $(1 \ 0)^t$ in $\mathrm{GL}_2(\mathbf{Z}/n\mathbf{Z})$. –  Harry Mar 18 '12 at 21:02
    
@Harry: Dear Harry, You have the right general principle, but the wrong $H$. For $\Gamma_1(N)$, the $H$ should be the subgroup of $GL_2(\mathbb Z/n)$ consisting of matrices of the form $\begin{pmatrix} * & * \\ 0 & 1 \end{pmatrix}$, and $\det$ induces a surjection from this $H$ to $(\mathbb Z/n)^{\times}$. Regards, –  Matt E Mar 19 '12 at 2:56
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