THE CANONICAL SUBGROUP OF E IS SpecR[x]/(x p +
نویسنده
چکیده
Let p be a prime. In this note we make explicit some results on the canonical subgroup of an elliptic curve E over the ring of integers Rp of Cp implicit in [K-pPMF]. In particular, if ω generates Ω1E/Rp and E has a canonical subgroup CE , knowlege of the Hasse invariant of the reduction of (E,ω) modulo p is equivalent to knowledge of the pair (CE , ω|CE). 1. Group Schemes of order p. Let μ denote the group of (p− 1)-st roots of unity in Zp and A the subring of Qp {r ∈ Zp: ∃n ∈ N, pnr ∈ Z[μ, 1/(p− 1)]}. Suppose R is an A-algebra, e.g., a p-adically complete ring with identity. For a ∈ R, let Ra = R[x]/(xp + ax) and Ba = SpecRa and for ǫ ∈ μp−1(R), [ǫ]a the automorphism of Ba corresponding to x 7→ ǫx. If a 6= 0, the automorphisms α of Ba such that α ◦ [ǫ]a = [ǫ]a ◦ α for ǫ ∈ μ are the [γ]a for γ ∈ μp−1(R). Suppose ∃b ∈ R such that ab = p. Because then, d(xp+ax) = a(1+bxp−1)dx and (1+bxp−1)(1−bxp−1/(1−p)) = 1, Ω1Ba/R ∼= Ba/aBa. Proposition 1.1. Suppose G is a group scheme of order p over R. Then the R-module of invariant differentials ΩG/R on G over R is cyclic and if ω is a generator, there are a, b ∈ R such that ab = p and a unique isomorphism of schemes h:Ba → G such that h ◦ [ǫ]a = [ǫ]G ◦ h, for ǫ ∈ μ, and h∗ω = (1 + bxp−1)−1dx. Moreover, ΩG/R ∼= R/aR, a is determined modulo a2R and b is determined modulo pR. In particular, if R is integrally closed and G is self-dual both a and b are determined modulo pR. Proof. We know from [O-T, pp.13-14] that there are universal constants wi ∈ A, i ≥ 1, such that w1 = 1, wj ∈ Z∗p, j < p, wp = pwp−1 and there are u, v ∈ R such that uv = wp, an isomorphism g:B−u → G over R for which the pullback of the group law on G to B−u is
منابع مشابه
Finite groups with $X$-quasipermutable subgroups of prime power order
Let $H$, $L$ and $X$ be subgroups of a finite group$G$. Then $H$ is said to be $X$-permutable with $L$ if for some$xin X$ we have $AL^{x}=L^{x}A$. We say that $H$ is emph{$X$-quasipermutable } (emph{$X_{S}$-quasipermutable}, respectively) in $G$ provided $G$ has a subgroup$B$ such that $G=N_{G}(H)B$ and $H$ $X$-permutes with $B$ and with all subgroups (with all Sylowsubgroups, respectively) $...
متن کاملSupport and Injective Resolutions of Complexes over Commutative Rings
Examples are given to show that the support of a complex of modules over a commutative noetherian ring may not be read off the minimal semi-injective resolution of the complex. These also give examples of semiinjective complexes whose localization need not be homotopically injective. Let R be a commutative noetherian ring. Recall that the support of a finitely generated R-module M is the set of...
متن کاملGroups in which every subgroup has finite index in its Frattini closure
In 1970, Menegazzo [Gruppi nei quali ogni sottogruppo e intersezione di sottogruppi massimali, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. 48 (1970), 559--562.] gave a complete description of the structure of soluble $IM$-groups, i.e., groups in which every subgroup can be obtained as intersection of maximal subgroups. A group $G$ is said to have the $FM$...
متن کاملFUZZY GRADE OF I.P.S. HYPERGROUPS OF ORDER 7
i.p.s. hypergroups are canonical hypergroups such that$[forall(a,x),a+xni x]Longrightarrow[a+x=x].$i.p.s. hypergroups were investigated in [1], [2], [3], [4] and it was proved thatif the order is less than 9, they are strongly canonical (see [13]). In this paperwe obtain the sequences of fuzzy sets and of join spaces determined (see [8])by all i.p.s. hypergroups of order seven. For the meaning ...
متن کاملOn Marginal Automorphisms of a Group Fixing the Certain Subgroup
Let W be a variety of groups defined by a set W of laws and G be a finite p-group in W. The automorphism α of a group G is said to bea marginal automorphism (with respect to W), if for all x ∈ G, x−1α(x) ∈ W∗(G), where W∗(G) is the marginal subgroup of G. Let M,N be two normalsubgroups of G. By AutM(G), we mean the subgroup of Aut(G) consistingof all automorphisms which centralize G/M. AutN(G) ...
متن کامل