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Let $E \to \operatorname{Spec} K$ be an elliptic curve and $S$ be its Neron model. That is, a scheme $S$ over a Dedekind domain $R$ with $E/K$ as the generic fiber of the structure morphism $N \to \operatorname{Spec}R$. Given the addition morphism $n_E:E\to E$, how would one go about showing that this can be extended to a morphism on a dense open subset of $S$?

This fact is used in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, Theorem IV.5.c, but it is stated without proof. I have (so far) been unable to prove it myself or find a reference, although it seems frustratingly simple.

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    $\begingroup$ Let $E$ be an elliptic curve over $K$. Let $N$ (why $S$?) be its Neron model over $R$. Let $f:E\to E$ be a morphism. This is an element of $Hom_K(E,E)$. By the Neron mapping property (i.e., the property which one uses to define the Neron model), as $N$ is smooth finite type over $R$, we have that $Hom_R(N,N) = Hom_K(E,E)$. Thus, $f$ extends (uniquely) to a morphism $F:N\to N$ such that $F_K = f$. No need to restrict to a dense open of $N$. The morphism is defined everywhere. Take $f= n_E$ to answer your question. $\endgroup$ – Ariyan Javanpeykar Jun 19 '18 at 20:40
  • $\begingroup$ In hindsight I agree that this is very easy if you're willing to use the universal property. However the exercise I was trying to do essentially asked for the proof of this fact. $\endgroup$ – DKS Jun 26 '18 at 13:27
  • $\begingroup$ I am confused. What other way is there to define the Neron model besides its universal property? Which exercise in Silverman are you referring to? Maybe, and this is just a guess, you are trying to prove that the smooth locus of the minimal regular model has the Neron mapping property? If so, please have a look at Qing Liu's book "Algebraic Geometry and Arithmetic Curves". The answer you seek can be found in Chapters 9.4 and 10.2.2 $\endgroup$ – Ariyan Javanpeykar Jun 29 '18 at 7:49
  • $\begingroup$ Your guess is more or less correct, the problem came from Hartshorne. I’ll take a look there, thanks. $\endgroup$ – DKS Jun 29 '18 at 10:42
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This is really just a long comment and in no way a complete answer. I hope this is slightly helpful.

Let $X, Y$ be irreducible schemes over $Spec(R)$, where $R$ is a domain. Let $K = Q(R)$. Let $X', Y'$ be the generic fibers. Let $f : X' \rightarrow Y'$ be a morphism of schemes over $Spec(K)$. Then

There exists a multiplicative set $S \subset R$ and a morphism $g : X \times_{R} S^{-1}R \rightarrow Y \times_{R} S^{-1}R$, such that $g \times Spec(K) = f$.

This statement certainly implies the statement required in the question. Hence we are reduced to proving the statement above.

Step 1. Reduce to the case when $Y$ is affine.

Step 2. Reduce to the case of $X$ is affine.

Step 3. Given a map $Spec(B \otimes_R K) \rightarrow Spec(A \otimes_R K)$, where $A, B$ are finitely generated $R-$algebras, then there exists a multiplicative set $S \subset R$, and a homomorphism $g : S^{-1}A \rightarrow S^{-1}B$ such that $g \otimes K = f$.

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  • $\begingroup$ Forgive a possibly dumb question: how does your claim above imply the result? It certainly would imply that $f$ induces a morphism on another fiber, but it's not immediately obvious how this gives a morphism on an open dense set of $N$. $\endgroup$ – DKS Jun 13 '18 at 10:50
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    $\begingroup$ @DKS The sets $X \times_R S^{-1}R$ are definitely open subset of the scheme $X$. That this open set is dense follows since $X$ can be assumed irreducible in the context of the book. So I will make this edit. $\endgroup$ – random123 Jun 13 '18 at 12:25
  • $\begingroup$ Ahh I see, this is a principal open set, instead of a fiber. Must-have gotten the wires crossed somehow. Thanks $\endgroup$ – DKS Jun 13 '18 at 13:26

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