Some commutative algebra

Before proving Theorem 4.1.1 we recall some well-known lemmas from commutative algebra. Let $M$ be a module over a commutative ring $R$ and let  $\mathfrak{m}$ be a finitely generated prime ideal of $R$ .

Lemma 4.2.1   If $M_\mathfrak{m}$ is Artinian, then $M_\mathfrak{m}\neq 0\iff M[\mathfrak{m}] \neq 0.$

Proof. $(\Longleftarrow)$ We first prove that $M_\mathfrak{m}=0$ implies $M[\mathfrak{m}]=0$ by a slight modification of the proof of [AM69, Prop. I.3.8]. Suppose $M_\mathfrak{m}=0$ , yet there is a nonzero $x\in M[\mathfrak{m}]$ . Let $I={\mathrm{Ann}}_R(x)$ . Then $I\neq (1)$ is an ideal that contains $\mathfrak{m}$ , so $I=\mathfrak{m}$ . Consider $\frac{x}{1} \in M_\mathfrak{m}$ . Since $M_\mathfrak{m}=0$ , we have $x/1=0$ , hence by definition of localization, $x$ is killed by some element of $R-\mathfrak{m}$ (set-theoretic difference). But this is impossible since ${\mathrm{Ann}}_R(x)=\mathfrak{m}$ .

$(\Longrightarrow)$ Next we prove that $M_\mathfrak{m}\neq 0$ implies $M[\mathfrak{m}]\neq 0$ . Since $M_\mathfrak{m}$ is an Artinian module over the (local) ring $R_\mathfrak{m}$ , by [AM69, Prop. 6.8], $M_\mathfrak{m}$ has a composition series:

$$M_\mathfrak{m}= M_0 \supset M_1 \supset \cdots \supset M_{n-1} \supset M_n = 0,$$

where by definition each quotient $M_i/M_{i+1}$ is a simple $R_\mathfrak{m}$ -module. In particular, $M_{n-1}$ is a simple $R_\mathfrak{m}$ -module. Suppose $x\in M_{n-1}$ is nonzero, and let $I={\mathrm{Ann}}_{R_\mathfrak{m}}(x)$ . Then

$$R_\mathfrak{m}/I \cong R_\mathfrak{m}\cdot x \subset M_{n-1},$$

so by simplicity $R_\mathfrak{m}/I \cong M_{n-1}$ is simple. Thus $I=\mathfrak{m}$ , otherwise $R_\mathfrak{m}/I$ would have $\mathfrak{m}/I$ as a proper submodule. Thus $x\in M_{n-1}[\mathfrak{m}]$ is nonzero.

Write $x=[y,a]$ with $y\in M$ and $a\in R-\mathfrak{m}$ , where $[y/a]$ means the class of $y/a$ in the localization (same as $(y,a)$ on page 36 of [AM69]). Since $a\in R-\mathfrak{m}$ , the element $a$ acts as a unit on $M_\mathfrak{m}$ , hence $ax = [y/1] \in M_{n-1}$ is nonzero and also still annihilated by $\mathfrak{m}$ (by commutativity).

To say that $[y/1]$ is annihilated by $\mathfrak{m}$ means that for all $\alpha\in \mathfrak{m}$ there exists $t\in R-\mathfrak{m}$ such that $t\alpha y = 0$ in $M$ . Since $\mathfrak{m}$ is finitely generated, we can write $\mathfrak{m}=(\alpha_1,\ldots, \alpha_n)$ and for each $\alpha_i$ we get corresponding elements $t_1,\ldots, t_n$ and a product $t=t_1\cdots t_n$ . Also $t\not\in\mathfrak{m}$ since $\mathfrak{m}$ is a prime ideal and each $t_i\not\in\mathfrak{m}$ . Let $z=ty$ . Then for all $\alpha\in \mathfrak{m}$ we have $\alpha z = t\alpha y = 0$ . Also $z\neq 0$ since $t$ acts as a unit on $M_{n-1}$ . Thus $z\in M[\mathfrak{m}]$ , and is nonzero, which completes the proof of the lemma. $\qedsymbol$

Lemma 4.2.2   Suppose $0 \to M_1 \to N \to M_2 \to 0$ is an exact sequence of $R$ -modules each of whose localization at $\mathfrak{m}$ is Artinian. Then $N[\mathfrak{m}]\neq 0 \iff (M_1\oplus M_2) [\mathfrak{m}]\neq 0.$

Proof. By Lemma 4.2.1 we have $N[\mathfrak{m}]\neq 0$ if and only if $N_\mathfrak{m}\neq 0$ . By Proposition 3.3 on page 39 of [AM69], the localized sequence

$$0 \to (M_1)_\mathfrak{m}\to N_\mathfrak{m}\to (M_2)_\mathfrak{m}\to 0$$

is exact. Thus $N_\mathfrak{m}\neq 0$ if and only if at least one of $(M_1)_\mathfrak{m}$ or $(M_2)_\mathfrak{m}$ is nonzero. Again by Lemma 4.2.1, at least one of $(M_1)_\mathfrak{m}$ or $(M_2)_\mathfrak{m}$ is nonzero if and only if at least one of $M_1[\mathfrak{m}]$ or $M_2[\mathfrak{m}]$ is nonzero. The latter is the case if and only if $(M_1\oplus M_2)[\mathfrak{m}]\neq 0$ . $\qedsymbol$

Remark 4.2.3   One could also prove the lemmas using the isomorphism $M[\mathfrak{m}]\cong \Hom_R(R/\mathfrak{m}, M)$ and exactness properties of ${\mathrm{Hom}}$ , but even with this approach many of the details in Lemma 4.2.1 still have to be checked.

Remark 4.2.4   In Theorem 4.1.1, we have $R\subset {\mathrm{End}}(C)$ , hence $R$ is finitely generated as a $\mathbb{Z}$ -module, so $R$ is noetherian.

Lemma 4.2.5   Let $G$ be a finite cyclic group, $M$ be a finite $G$ -module that is also a module over a commutative ring $R$ such that the action of $G$ and $R$ commute (i.e., $M$ is an $R[G]$ -module). Suppose $\mathfrak{p}$ is a finitely-generated prime ideal of $R$ , and $H^0(G,M)[\mathfrak{p}]=0$ . Then $H^1(G, M)[\mathfrak{p}]= 0$ .

Proof. Argue as in [Se79, Prop. VIII.4.8], but noting that all modules are modules over $R$ and maps are morphisms of $R$ -modules. $\qedsymbol$

proof This argument was inspired by the proof of [Se79, Prop. VIII.4.8]. Let tex2html_wrap_inline$s$ be a generator for tex2html_wrap_inline$G$, and let tex2html_wrap_inline$D=s-1$. There is an exact sequence equation 0 &rarr#to;M^G &rarr#to;M D M &rarr#to;M_G &rarr#to;0, and each map is a homomorphism of tex2html_wrap_inline$R$-modules (here we use that the action of tex2html_wrap_inline$R$ and tex2html_wrap_inline$G$ commutes, so that the middle map is an tex2html_wrap_inline$R$-module homomorphism). By [AM69, Prop. 3.3], the localization displaymath 0 &rarr#to;(M^G)_p&rarr#to;M_pD M_p&rarr#to;(M_G)_p&rarr#to;0 of () is exact. By hypothesis tex2html_wrap_inline$H^0(G,M)[p]=0$, and by definition tex2html_wrap_inline$M^G = H^0(G,M)$, so by Lemma 4.2.1, tex2html_wrap_inline$(M^G)_p= 0$, and the following sequence is exact: displaymath 0 &rarr#to;M_pD M_p&rarr#to;(M_G)_p&rarr#to;0. But tex2html_wrap_inline$D:M_p&rarr#to;M_p$ is an injective map of finite sets, so it is a bijection, hence tex2html_wrap_inline$(M_G)_p=0$. Again using Lemma 4.2.1 it follows that tex2html_wrap_inline$M_G[p]=0$.

It follows from [Se79, Ch. VIII, §4] that displaymathH^1(G,M) &cong#cong;M[N]/DM         (as functors in $M$), where tex2html_wrap_inline$M[N]$ is the kernel of the map tex2html_wrap_inline$M N M$, and tex2html_wrap_inline$N = &sum#sum;_g &isin#in;Gg$. Since tex2html_wrap_inline$M_G M/DM$, we have an tex2html_wrap_inline$R$-module inclusion tex2html_wrap_inline$H^1(G,M) &rarrhk#hookrightarrow;M_G$. We showed above that tex2html_wrap_inline$M_G[p]=0$, so the lemma follows.

theorem_type[remark][theorem][][remark][][] If tex2html_wrap_inline$p$ were replaced by a prime number tex2html_wrap_inline$p&isin#in;Z$ then the result would be immediate since using Herbrand quotients one shows that tex2html_wrap_inline$#H^0(G,M) = #H^1(G,M)$ (see [Se79, Prop. VIII.4.8]). It is unclear to the authors if the result is true in general, i.e., if tex2html_wrap_inline$G$ is replaced by an arbitrary group.

theorem_type[lemma][theorem][][plain][][] [Grothendieck]Let tex2html_wrap_inline$A$ be an abelian variety over the fraction field tex2html_wrap_inline$K$ of a strictly Henselian discrete valuation ring tex2html_wrap_inline$R$ (e.g. the maximal unramified extension of local field). Let tex2html_wrap_inline$n$ be an integer coprime to the residue characteristic of tex2html_wrap_inline$K$. Let tex2html_wrap_inline$x &isin#in;A(K)$ be a point whose reduction lands in the identity component of the closed fiber of the Néron model of tex2html_wrap_inline$A$. Then tex2html_wrap_inline$x &isin#in;nA(K)$.

theorem_type[proposition][theorem][][plain][][] Let tex2html_wrap_inline$A$ be an abelian variety over the fraction field tex2html_wrap_inline$K$ of a strictly Henselian discrete valuation ring tex2html_wrap_inline$R$. Let tex2html_wrap_inline$A$ be the Néron model of tex2html_wrap_inline$A$ and let tex2html_wrap_inline$A^0$ be the connected component of the identity element of tex2html_wrap_inline$A$. Then the group tex2html_wrap_inline$A^0(R)$ of tex2html_wrap_inline$R$-points on tex2html_wrap_inline$A^0$ is tex2html_wrap_inline$n$-divisible for every tex2html_wrap_inline$n$ that is relatively prime to the characteristic.

theorem_type[remark][theorem][][remark][][] Suppose that tex2html_wrap_inline$P &isin#in;A(K)$ is a point whose reduction lies in the identity component of the special fiber of the Néron model of tex2html_wrap_inline$A$. Since tex2html_wrap_inline$A(K) &sime#simeq;A(R)$ because of the Néron mapping property, the proposition implies that tex2html_wrap_inline$P$ is tex2html_wrap_inline$n$-divisible in tex2html_wrap_inline$A(K)$. Furthermore, tex2html_wrap_inline$P = nP'$ for a point tex2html_wrap_inline$P' &isin#in;A(K)$ whose reduction also lies in the identity component of the special fiber of tex2html_wrap_inline$A$ (although we will not need this last property).