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Suppose I have two isogenous elliptic curves over $\mathbb{Q}$, $E$ and $E'$. Will the minimal models of $E$ and $E'$ still be isogenous?

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Dear user, what exactly do you mean by the mimimal models being isogenous? Neron models are functorial, and so the isogeny $E \to E'$ will induce a morphism of Neron models $\cal E \to \cal E'$. Are you asking about the kernel of this morphism? Regards, – Matt E Jan 31 '13 at 16:25
@MattE, sorry for not being clear. By minimal model, I mean as defined here. So given two isogenous elliptic curves over $\mathbb{Q}$ by their Weierstrass equation, $E$ and $E'$, they both have minimal models. Denote them by $\widehat{E}$ and $\widehat{E'}$. Are $\widehat{E}$ and $\widehat{E'}$ isogenous? – user60194 Jan 31 '13 at 16:51
Dear user, But isogenous in what sense? In the entry you linked, the minimal model is just another Weierstrass equation for the same curve. The notion of isogeny is certainly independent of Weierstrass equations, and so the answer to your question would then be "yes", for trivial reasons. But do you mean/want something more? Regards, – Matt E Jan 31 '13 at 20:10
The definition of minimal model on planetmath is that of minimal Weierstrass model, so they are schemes over $\mathbb Z$ and it doesn't really make sense to talk about isogeny unless you are asking for finite morphism or use Néron models as suggests MattE. By the way, on planetmath they say the minimal discriminant is smaller than 12, this is incorrect. – user18119 Feb 7 '13 at 22:10
up vote 3 down vote accepted

If I interpret the question correctly, the answer is yes. Each elliptic curve is isomorphic to its minimal model by definition: the isomorphism is given by the change of variables transformation. Remember that isomorphisms are degree $1$ isogenies and that the composition of isogenies is an isogeny. Therefore the minimal models are isogenous (with unchanged degree).

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