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Let $R$ be a D.V.R. with the fraction field $K$, $A$ a $K$- abelian variety, $\mathfrak{A} \to \operatorname{Spec}R$ the Neron model of $A$. We say $A$ has good reduction if there exists a smooth proper scheme $\mathfrak{X} \to \operatorname{Spec}R$ with the generic fibre $\mathfrak{X}_{\eta} = A$.

I want to show that $A$ has good reduction $\iff$ $\mathfrak{A}$ is proper.

The "if" part is easy. Conversely, if $A$ has good reduction and $\mathfrak{X} \to \operatorname{Spec}R$ is the one, then we have an $R$-morphism $f : \mathfrak{X} \to \mathfrak{A}$ induced by the identity of $A$, from the universal property of the Neron model. (This is proper since a Neron model is separated.) Thus if $f(\mathfrak{X})$ is dense, we have that $\mathfrak{A}$ is proper, and done. But I can't.

Now $A$ is open in $\mathfrak{A}$, and contained in $f(\mathfrak{X})$. Thus if a Neron model is irreducible, I can show it. So please show the irreducibility of a Neron model, or that $f(\mathfrak{X})$ is dense.

P.S. I study Neron models by a non-scheme-theoritical book and translate it, so my definitions are maybe incorrect. If there are some incorrect points, please correct.

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    $\begingroup$ The proof of of this direction requires some work: you must show that the abelian $R$-scheme $\mathfrak{X}$ satisfies the Neron mapping property. This is carefully explained in Proposition 2.2.4 of these notes. $\endgroup$
    – msteve
    Commented Oct 6, 2018 at 19:05
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    $\begingroup$ $\mathfrak A$ is smooth over a dvr => its generic fibre is dense in $\mathfrak A$ (also $\mathfrak A$ is irreducible since you can check irreducibility on an open dense subset). I think that this is sufficient for your purposes. But actually you can show more, any smooth proper model of an abelian variety is isomorphic to its Neron model. This is explained in Proposition 1.4/2 in the book "Neron models". $\endgroup$
    – gdb
    Commented Oct 6, 2018 at 19:13
  • $\begingroup$ Thank you very much! So, does every smooth proper model of an abelian variety have a group scheme structure? $\endgroup$
    – k.j.
    Commented Oct 6, 2018 at 19:23
  • $\begingroup$ @k.j. Yes, it does. Moreover, there is exactly one such model (up to a canonical isomorphism). $\endgroup$
    – gdb
    Commented Oct 6, 2018 at 19:29

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