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Let $K$ be a number field with Galois group $G$ and $N$ be a finitely generated abelian group which is also a discrete $G$-module. Let $D(N)$ be the algebraic group defined as

$D(N)(R)=Hom_{\mathbb{Z}}(N,R^{*})$ for any $k$-algebra $R$.

Then $D(N)$ is an algebraic group of multiplicative type over $K$. Does the Neron model for $D(N)$ exist ? I understand that if $N$ is torsion free as an abelian group, then $D(N)$ is an algebraic torus and such a Neron model exists. But I am not sure about the general case. Thank you very much.

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I'm not sure about the state of the general theory, but one approach to this kind of question is to write the group as the kernel of a morphism of tori, to bootstrap this to a morphism of the corresponding Neron models, and then to take the kernel of this morphism.

E.g. in the case of $\mu_n$, which is the kernel of $\mathbb G_m \to \mathbb G_m$ given by raising to the $n$th power, which extends to raising to the $n$th power on Neron models. The kernel is then just $\mu_n$ again. If $n$ is coprime to the residue characteristic, this seems pretty natural. Otherwise (e.g. if $n = p =$ the residue characteristic) then it still seems reasonable, but note that we can no longer really test it by the Neron mapping property (because it is not smooth in this case).

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  • $\begingroup$ Thank you for your answer ! However I am a little confused. I understand that the Neron model for $\mathbb{G}_m$ is the unique smooth group scheme $G$ over $O_K$ locally of finite type which represents the sheaf $j_{*}\mathbb{G}_m$ over the smooth topology. So it is not $\mathbb{G}_m$ but has $\mathbb{G}_m$ as the identity connected component. Then the Neron model for $\mu_n$ is not $\mu_n$. $\endgroup$ – raynor14 Feb 13 '15 at 2:47
  • $\begingroup$ Yes, but it is $\mathbb G_m \times \mathbb Z$ on the special fibre, and multiplication by $n$ is injective on $\mathbb Z$. So the kernel is just $\mu_n$, if I'm not mistaken. This is related to the fact that the point of the $\mathbb Z$ factor is to give a place for units in char. zero that are divisible by non-zero powers of $p$ to specialize to mod $p$; but an element of $\mu_n$ is always an integral unit, so it just specializes to an element of $\mathbb G_m$; we don't need to have an extra "$\mathbb Z$-like" factor. $\endgroup$ – tracing Feb 13 '15 at 2:49
  • $\begingroup$ Thank you ! I got it now. So the fact is that $\mu_n=j_{*}\mu_n$ on $O_K$. On the other note, you are saying for any $G$-module $N$, we take a free resolution $ \hat{T}_2 \to \hat{T}_1 \to N \to 0 $ which induces $0 \to D(N) \to T_1 \to T_2$. Let $\mathcal{T}_i$ be the Neron model for $T_i$. We can take the kernel $G$ of $\mathcal{T}_1 \to \mathcal{T}_2$ to be the "Neron model" of $D(N)$. $\endgroup$ – raynor14 Feb 13 '15 at 3:10
  • $\begingroup$ Yes, that's what I'm suggesting. It's a standard method for studying finite groups (another is to resolve them by abelian varieties rather than tori); whether it works in the context you have in mind, I don't know. (I also don't know whether it's been studied in the Neron model context, although I'd be surprised if it hadn't. If you ask on MO you might get more input about this approach --- there are certainly experts on Neron models who participate there but not here.) $\endgroup$ – tracing Feb 13 '15 at 5:31

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