Smoothness and Surjectivity

In this section, we recall some well-known lemmas that we will use in Section 3.4 to produce unramified cohomology classes. The authors are grateful to B. Conrad for explaining the proofs of these lemmas.

Lemma 3.3   If $G$ is a finite-type smooth commutative group scheme over a strictly henselian local ring $R$ and the fibers of $G$ over $R$ are (geometrically) connected, then the multiplication map

$$n_G:G(R) \rightarrow G(R)$$

is surjective when $n \in R^{\times}$.

Proof. Pick an element $g \in G(R)$ and form the cartesian diagram

$$\xymatrix @=3pc{ Y_g \ar[rr]^{\psi}\ar[d] && {\Spec(R)}\ar[d]^{g}\\ G\ar[rr]^{n_G} && G}$$

We want to prove that $\psi$ has a section. Since $R$ is strictly henselian, by [Gro67, 18.8.1] it suffices to show that $Y_g$ is étale over $R$ with non-empty closed fiber, or more generally that $n_G$ is étale and surjective.

By Lemma 2(b) of [BLR90, §7.3], $n_G$ is étale. The image of the étale $n_G$ must be an open subgroup scheme, and on fibers over $\Spec(R)$ we get surjectivity since an open subgroup scheme of a smooth connected (hence irreducible) group scheme over a field must fill up the whole space [Gro70, VI $_{\rm {A}}$, 0.5]. $\qedsymbol$

Lemma 3.4   Let $A$ be an abelian variety over the fraction field $K$ of a strictly henselian dvr (e.g., $K$ could be the maximal unramified extension a local field). Let $n$ be an integer not divisible by the residue characteristic of $K$. Suppose that $x$ is a point of $A(K)$ whose reduction lands in the identity component of the closed fiber of the Néron model of $A$. Then there exists $z\in A(K)$ such that $nz=x$.

Proof. Let $\mathcal{A}$ denote the Néron model of $A$ over the valuation ring $R$ of $K$, and let $\mathcal{A}^0$ denote the ``identity component'' (i.e., the open subgroup scheme obtained by removing the non-identity components of the closed fiber of $\mathcal{A}$). The hypothesis on the reduction of $x \in A(K) = \mathcal{A}(R)$ says exactly that $x \in \mathcal{A}^0(R)$. Since connected schemes over a field are geometrically connected when there is a rational point [Gro65, Prop. 4.5.13], the fibers of $\mathcal{A}^0$ over $\Spec(R)$ are geometrically connected. The lemma now follows from Lemma 3.3 with $G=\mathcal{A}^0$. $\qedsymbol$

Remark 3.5   M. Baker noted that this argument can also be formulated in terms of formal groups when $R$ is the strict henselization of a complete dvr.

Lemma 3.6   Let $\mathcal{J}\stackrel{\phi}{\rightarrow } \mathcal{C}$ be a smooth surjective morphism of schemes over a strictly Henselian local ring $R$. Then the induced map $\mathcal{J}(R) \rightarrow \mathcal{C}(R)$ is surjective.

Proof. The argument is similar to that of the proof of Lemma 3.3. Pick an element $g \in \mathcal{C}(R)$ and form the cartesian diagram

$$\xymatrix @=3pc{ Y_g \ar[rr]^{\psi}\ar[d] && {\Spec(R)}\ar[d]^{g}\\ \mathcal{J}\ar[rr]^{\phi} && \mathcal{C}}$$

We want to prove that $\psi$ has a section. Since $\phi$ is smooth, $\psi$ is also smooth. By [Gro67, 18.5.17], to show that $\psi$ has a section, we just need to show that the closed fiber of $\psi$ has a section (i.e., a rational point). But this closed fiber is smooth and non-empty (since $\phi$ is surjective); also its base field is separably closed since $R$ is strictly Henselian. Hence by [BLR90, Cor. 2.2.13], the closed fiber has an $R$-rational point. $\qedsymbol$