Skip to main content
MathOverflow

Return to Answer

added 159 characters in body
Source Link
nfdc23
nfdc23
  • 968
  • 26
  • 39

And even more is true: the translation action of $\mathcal{E}^{\rm{sm}}$ on itself extends to a left action $$\mathcal{E}^{\rm{sm}} \times \mathcal{E} \rightarrow \mathcal{E}$$ on $\mathcal{E}$ (so at the level of $R$-points, this recovers the action of $\mathcal{E}^{\rm{sm}}(R)$ on $\mathcal{E}$ that could be built via other considerations -- do the comparisong of constructions over $K$). This ultimately comes down to good behavior of the formation of the minimal regular proper model under certain kinds of mild extensions of $R$. See Lemma 2.12(c) in section 10.2 of Qing Liu's book Algebraic Geometry and Arithmetic Curves for a proof (expressed in slightly different terms, but yielding the above by composing with a projection map). This is really the most crucial point, since it allows one to actually harness geometric information: if we pass to the special fibers over the residue field $k$ of $R$ we get an action of the smooth $k$-group $G := \mathcal{E}^{\rm{sm}}_k$ on the typically reducible (and generally not everywhere reduced) $k$-curve $X = \mathcal{E}_k$, and your question is now reduced to showing that the action of the identity component $G^0$ on $X$ preserves each irreducible component of $X$.

We can now forget about the original setup with elliptic curves, and just want to show that if $H$ is a smooth connected group scheme over a field $k$ (such as $G^0$ above) and $X$ is a $k$-scheme of finite type equipped with a left $H$-action then $H$ preserves each irreducible component $C$ of $X$. Since $C_{\overline{k}}$ is a union of irreducible components of $X_{\overline{k}}$, it suffices to prove the result over $\overline{k}$, so we can assume $k$ is algebraically closed. We can also assume $C$ is reduced. The condition on $H$ of preserving $C$ is now represented by the closed subgroup scheme $${\rm{Stab}}_H(C) := \bigcap_{c \in C(k)} \{h \in H\,|\,h.c \subset C\},$$ yet this has finite index at the level of $k$-points since $X$ has only finitely many irreducible components. But a smooth connected group $H$ over aan algebraically closed field $k$ has no proper closed subgroup of finite index apart from itselfat the level of $k$-points (since smooth connected groups are irreducibleas $H$ is the disjoint union of finitely many $H(k)$-translates of that closed subgroup, forcing the closed subgroup to be open and hence to coincide with $H$), so $H$ preserves $C$ and we win.

And even more is true: the translation action of $\mathcal{E}^{\rm{sm}}$ on itself extends to a left action $$\mathcal{E}^{\rm{sm}} \times \mathcal{E} \rightarrow \mathcal{E}$$ on $\mathcal{E}$. This ultimately comes down to good behavior of the formation of the minimal regular proper model under certain kinds of mild extensions of $R$. See Lemma 2.12(c) in section 10.2 of Qing Liu's book Algebraic Geometry and Arithmetic Curves for a proof (expressed in slightly different terms, but yielding the above by composing with a projection map). This is really the most crucial point, since it allows one to actually harness geometric information: if we pass to the special fibers over the residue field $k$ of $R$ we get an action of the smooth $k$-group $G := \mathcal{E}^{\rm{sm}}_k$ on the typically reducible (and generally not everywhere reduced) $k$-curve $X = \mathcal{E}_k$, and your question is now reduced to showing that the action of the identity component $G^0$ on $X$ preserves each irreducible component of $X$.

We can now forget about the original setup with elliptic curves, and just want to show that if $H$ is a smooth connected group scheme over a field $k$ (such as $G^0$ above) and $X$ is a $k$-scheme of finite type equipped with a left $H$-action then $H$ preserves each irreducible component $C$ of $X$. Since $C_{\overline{k}}$ is a union of irreducible components of $X_{\overline{k}}$, it suffices to prove the result over $\overline{k}$, so we can assume $k$ is algebraically closed. We can also assume $C$ is reduced. The condition on $H$ of preserving $C$ is now represented by the closed subgroup scheme $${\rm{Stab}}_H(C) := \bigcap_{c \in C(k)} \{h \in H\,|\,h.c \subset C\},$$ yet this has finite index at the level of $k$-points since $X$ has only finitely many irreducible components. But a smooth connected group over a field has no closed subgroup of finite index apart from itself (since smooth connected groups are irreducible), so $H$ preserves $C$ and we win.

And even more is true: the translation action of $\mathcal{E}^{\rm{sm}}$ on itself extends to a left action $$\mathcal{E}^{\rm{sm}} \times \mathcal{E} \rightarrow \mathcal{E}$$ on $\mathcal{E}$ (so at the level of $R$-points, this recovers the action of $\mathcal{E}^{\rm{sm}}(R)$ on $\mathcal{E}$ that could be built via other considerations -- do the comparisong of constructions over $K$). This ultimately comes down to good behavior of the formation of the minimal regular proper model under certain kinds of mild extensions of $R$. See Lemma 2.12(c) in section 10.2 of Qing Liu's book Algebraic Geometry and Arithmetic Curves for a proof (expressed in slightly different terms, but yielding the above by composing with a projection map). This is really the most crucial point, since it allows one to actually harness geometric information: if we pass to the special fibers over the residue field $k$ of $R$ we get an action of the smooth $k$-group $G := \mathcal{E}^{\rm{sm}}_k$ on the typically reducible (and generally not everywhere reduced) $k$-curve $X = \mathcal{E}_k$, and your question is now reduced to showing that the action of the identity component $G^0$ on $X$ preserves each irreducible component of $X$.

We can now forget about the original setup with elliptic curves, and just want to show that if $H$ is a smooth connected group scheme over a field $k$ (such as $G^0$ above) and $X$ is a $k$-scheme of finite type equipped with a left $H$-action then $H$ preserves each irreducible component $C$ of $X$. Since $C_{\overline{k}}$ is a union of irreducible components of $X_{\overline{k}}$, it suffices to prove the result over $\overline{k}$, so we can assume $k$ is algebraically closed. We can also assume $C$ is reduced. The condition on $H$ of preserving $C$ is now represented by the closed subgroup scheme $${\rm{Stab}}_H(C) := \bigcap_{c \in C(k)} \{h \in H\,|\,h.c \subset C\},$$ yet this has finite index at the level of $k$-points since $X$ has only finitely many irreducible components. But a smooth connected group $H$ over an algebraically closed field $k$ has no proper closed subgroup of finite index at the level of $k$-points (as $H$ is the disjoint union of finitely many $H(k)$-translates of that closed subgroup, forcing the closed subgroup to be open and hence to coincide with $H$), so $H$ preserves $C$ and we win.

Source Link
nfdc23
nfdc23
  • 968
  • 26
  • 39

Where you say "trivial action on $\Gamma_{t0}$" near the end, you mean "carries $\Gamma_{t0}$ into itself" (i.e., doesn't move it), not that the effect on $\Gamma_{t0}$ is the identity.

The global base is a red herring: the core issue here completely concerns the object over the local ring at $t$ (recall that the formation of both the minimal regular proper model and the Neron model of an elliptic curve over the function field of a connected Dedekind scheme commutes with passage to local rings on the base, more-or-less by design). So to clarify matters, let's work in that context.

So let $R$ be any discrete valuation ring, say with fraction field $K$, and let $E$ be an elliptic curve over $K$ with minimal regular proper model $\mathcal{E}$ over $R$. Let $\mathcal{E}^{\rm{sm}}$ denote the maximal $R$-smooth open subscheme of $\mathcal{E}$; since $\mathcal{E}$ is $R$-flat of finite type with smooth generic fiber (and the generic point of ${\rm{Spec}}(R)$ is open), $\mathcal{E}^{\rm{sm}}$ is the complement in $\mathcal{E}$ of the closed non-smooth locus in the closed fiber. In general if $X$ is an $R$-scheme of finite type that is regular, any section $x: {\rm{Spec}}(R) \rightarrow X$ lands inside the maximal smooth open subscheme $X^{\rm{sm}}$ (this is a nice application of the characterization of smoothness in terms of completed local rings; see 3.1/2 in the highly-recommended book Neron Models for a proof). Consequently, $\mathcal{E}(R) = \mathcal{E}^{\rm{sm}}(R)$; this is very important (it is the only thing which ensures that $\mathcal{E}^{\rm{sm}}$ has non-empty special fiber, which is to say that the special fiber of $\mathcal{E}$ isn't everywhere non-smooth, since there is at least the "identity section" = the $R$-point extending the identity section of $E$).

Now finally come to the real substance of the matter: $\mathcal{E}^{\rm{sm}}$ is the Neron model (and in particular has an $R$-group structure!), or more precisely the unique $R$-map $f:\mathcal{E}^{\rm{sm}} \to N(E)$ to the Neron model extending the identification of $K$-fibers is an isomorphism. This is not obvious, and is not perceived when one builds the Neron model through the general methods that are used for abelian varieties since there is no analogue of $\mathcal{E}$ for abelian varieties of higher dimension (i.e., the contact between the Neron model and the minimal regular proper model is a special feature of the 1-dimensional context). To prove that $f$ is an isomorphism, it is harmless to extend scalars to the strict henselization $R^{\rm{sh}}$ of $R$ (since a map which becomes an isomorphism after a faithfully flat base change was an isomorphism originally, and the formation of both the Neron model and minimal regular proper model commute with such scalar extension more-or-less by design); that case in turn is proved as 1.5/1 in Neron Models (the proof uses much of the work involved in the actual construction of $N(E)$ by general principles for abelian varieties).

And even more is true: the translation action of $\mathcal{E}^{\rm{sm}}$ on itself extends to a left action $$\mathcal{E}^{\rm{sm}} \times \mathcal{E} \rightarrow \mathcal{E}$$ on $\mathcal{E}$. This ultimately comes down to good behavior of the formation of the minimal regular proper model under certain kinds of mild extensions of $R$. See Lemma 2.12(c) in section 10.2 of Qing Liu's book Algebraic Geometry and Arithmetic Curves for a proof (expressed in slightly different terms, but yielding the above by composing with a projection map). This is really the most crucial point, since it allows one to actually harness geometric information: if we pass to the special fibers over the residue field $k$ of $R$ we get an action of the smooth $k$-group $G := \mathcal{E}^{\rm{sm}}_k$ on the typically reducible (and generally not everywhere reduced) $k$-curve $X = \mathcal{E}_k$, and your question is now reduced to showing that the action of the identity component $G^0$ on $X$ preserves each irreducible component of $X$.

We can now forget about the original setup with elliptic curves, and just want to show that if $H$ is a smooth connected group scheme over a field $k$ (such as $G^0$ above) and $X$ is a $k$-scheme of finite type equipped with a left $H$-action then $H$ preserves each irreducible component $C$ of $X$. Since $C_{\overline{k}}$ is a union of irreducible components of $X_{\overline{k}}$, it suffices to prove the result over $\overline{k}$, so we can assume $k$ is algebraically closed. We can also assume $C$ is reduced. The condition on $H$ of preserving $C$ is now represented by the closed subgroup scheme $${\rm{Stab}}_H(C) := \bigcap_{c \in C(k)} \{h \in H\,|\,h.c \subset C\},$$ yet this has finite index at the level of $k$-points since $X$ has only finitely many irreducible components. But a smooth connected group over a field has no closed subgroup of finite index apart from itself (since smooth connected groups are irreducible), so $H$ preserves $C$ and we win.

QED

Post Made Community Wiki by nfdc23