Correction : Abelian Varieties defined over their Fields of Moduli, I

Bull London Math. Soc, 4 (1972), 370–372. The proof of the theorem contains an error. Before giving a correct proof, we state two lemmas. LEMMA 1. Let K/k be a cyclic Galois extension of degree m, let σ generate Gal (K/k), and let (A, I, θ) be defined over K. Suppose that there exists an isomorphism λ:(A,I,θ) → (A σ , I σ , θ σ ) over K such that vλ σm−1 …λ σ λ = 1, where v is the canonical isomorphism (A σm , I σm , θ σm ) → (A, I, θ). Then (A, I, θ) has a model over k, which becomes isomorphic to (A, I, θ) over K. Proof. This follows easily from [7], as is essentially explained on p. 371. LEMMA 2. Let G be an abelian pro‐finite group and let Φ : G → Q/Z be a continuous character of G whose image has order p. Then either: (a) there exist subgroups G′ and H of G such that H is cyclic of order p m for some m, Φ(G′) = 0, and G = G′ × H, or (b) for any m > 0 there exists a continuous character Φm of G such that p m Φ m = Φ. Proof. If (b) is false for a given m , then there exists an element σ ɛ G , of order p r for some r ⩽ m , such that Φ(σ) ¦ 0. (Consider the sequence dual to 0 → Ker ( p m ) → G → pm G ). There exists an open subgroup G o of G such that Φ( G 0 ) = 0 and σ has order p r in G/G 0 . Choose H to be the subgroup of G generated by σ, and then an easy application to G/G 0 of the theory of finite abelian groups shows the existence of G ′ (note that Φ(σ) ¦ 0 implies that σ is not a p ‐th. power in G ). We now prove the theorem. The proof is correct up to the statement (iv) (except that (i) should read: F ′ ⊂ k 1 ⊂ F′ab ). To remove a minor ambiguity in the proof of (iv), choose σ to be an element of Gal ( F′ ab /k 2 ) whose image σ ¯ in Gal ( k 1 / k 2 ) generates this last group. The error occurs in the statement that the canonical map v : A σP → A acts on points by sending a σp ↦ a ; it, of course, sends a ↦ a . The proof is correct, however, in the case that it is possible to choose σ so that σ p = 1 (in Gal ( F′/k 2 )). By applying Lemma 2 to G = Gal ( F' ab /k 2 ) and the map G → Gal ( k 1 /k 2 ) one sees that only the following two cases have to be considered. (a) It is possible to choose σ so that σ pm = 1, for some m , and G = G′ × H where G ′ acts trivially on k 1 and H is generated by σ. (b) For any m > 0 there exists a field K, F′ab ⊃ K ⊃ k 1 ⊃ k 2 is a cyclic Galois extension of degree p m . In the first case, we let K ⊂ F′ ab be the fixed field of G ′. Then ( A, I , θ), regarded as being defined over K , has a model over k 2 . Indeed, if m = 1, then this was observed above, but when m > 1 the same argument applies. In the second case, let λ : ( A, I , θ) → ( A σ ¯ , I σ ¯ , θ σ ¯ ) be an isomorphism defined over k 1 and let v λ σ … λ σp−1 λ = α ɛ μ( R ). If λ is replaced by λγ for some γ ɛ Aut k 1 (( A , I, θ)) then α is replaced by αγ P . Thus, as μ( R ) is finite, we may assume that α pm−1 = 1 for some m . Choose K , as in (b), to be of degree p m over k 2 . Let σ m be a generator of Gal ( K / k 2 ) whose restriction to k 1 is σ ¯ . Then λ : (A, I, θ) → ( A σ ¯ , I σ ¯ , θ σ ¯= ( A σ ¯ m, I σ ¯ m, θ σ ¯ m is an isomorphism defined over K and v λ σm pm−1 , … λ σm λ = α pm−1 = 1, and so, by) Lemma 1, ( A, I , θ) has a model over k 2 which becomes isomorphic to ( A, I , θ over K . The proof may now be completed as before. Addendum: Professor Shimura has pointed out to me that the claim on lines 25 and 26 of p. 371, viz that μ( R ) is a pure subgroup of ∏ R * t , does not hold for all rings R . Thus this condition, which appears to be essential for the validity of the theorem, should be included in the hypotheses. It holds, for example, if μ( R ) is a direct summand of μ( F ).