3
$\begingroup$

I am trying to understand Example 7.5/9 from the book "Neron models". There one has a discrete valuation ring $R'$ that is the localization of $\mathbb{Z}[\zeta_p]$ at $p$, so that the absolute ramification index of $R'$ is $p - 1$. One knows that there is a homomorphism $u'\colon E'\rightarrow F'$ of abelian $R'$-schemes such that its generic fiber $u'_K$ is a closed immersion but that $u'$ itself is not a monomorphism (so $u'$ is quite pathological, e.g., it is not flat).

The claim is that in the situation above the induced map $\mathrm{Lie}(u')\colon \mathrm{Lie}(E') \rightarrow \mathrm{Lie}(F')$ on Lie algebras (which are finite free $R'$-modules) is not an inclusion of a direct summand. My question is: how does one see this?

I've tried passing to the special fiber and then checking that the induced map on Lie algebras is not injective but I don't know how to rule out the possibility that the kernel of the special fiber of $u'$ is etale.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

If I am not mistaken, the claim that $\mathrm{Lie}(u')$ is not an inclusion of a direct summand is erroneous. Here is why.

Let $G'$ be the abelian $R'$-scheme with generic fiber $F'_K/E'_K$. Then according to Raynaud's Theorem A.1 of the 1996 Compositio paper of Abbes and Ullmo the sequence $$ E' \rightarrow F' \rightarrow G' $$ of abelian $R'$-schemes induces an exact sequence $$ 0 \rightarrow \mathrm{Lie}(E') \rightarrow \mathrm{Lie}(F') \rightarrow \mathrm{Lie}(G') \rightarrow (\mathbb{Z}/p\mathbb{Z})^r \rightarrow 0 $$ for some $r \ge 0$. In particular, $\mathrm{Lie}(F')/\mathrm{Lie}(E')$ is a free $R'$-module, so $\mathrm{Lie}(E')$ is a direct summand of $\mathrm{Lie}(F')$, contradicting the claim in question.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Raynaud's theorem also requires that $F'$ be semi-abelian. As Bosch-Lütkebohmert-Raynaud say at the end of example 7/5.9, their example shows that this hypothesis cannot be relaxed. $\endgroup$
    – ACL
    Commented Dec 21, 2014 at 17:49
  • $\begingroup$ But $F'$ is semiabelian in the present situation; it is even an abelian scheme. I am not applying Raynaud's theorem to the restrictions of scalars (to which it would not apply). $\endgroup$
    – Question Mark
    Commented Dec 21, 2014 at 17:54
  • $\begingroup$ you can not use Raynaud's result since this requires an exact sequence between the Néron model: in particular, the Néron model of the quotient $F_K'/E_K'$ should be semi-abelian. $\endgroup$
    – Tong
    Commented Jan 3, 2015 at 21:11
  • $\begingroup$ @Tong: Raynaud's result does not require an exact sequence between the Neron models. Its whole point is to quantify failure of such exactness (under suitable assumptions). Besides, the Neron model of the quotient $F_K'/E_K'$ is semi-abelian---it is even an abelian scheme---see Lemma 7.4/2 in "Neron models." $\endgroup$
    – Question Mark
    Commented Jan 4, 2015 at 2:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .