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 acorrect proof, we state two lemmas. LEMMA 1. Let K/k be a cyclic Galois extension of degree m, let generate Gal (K/k), and let (A, I, ) be defined over K. Supposethat there exists an isomorphism :(A,I,) (A, I, ) over K suchthat v^m–1 ... = 1, where v is the canonical isomorphism(A^m, I^m, ^m) (A, I, ). Then (A, I, ) has a model over k, whichbecomes isomorphic to (A, I, ) over K. Proof. This follows easily from 7], as is essentially explainedon p. 371. LEMMA 2. Let G be an abelian pro-finite group and let : 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 cyclicof order p^m for some m, (G') = 0, and G = G' x H, or (b) for any m > 0 there exists a continuous character m ofG such that p^m _m = . Proof. If (b) is false for a given m, then there exists an element G, of order p^r for some r m, such that () ¦ 0. (Considerthe sequence dual to 0 Ker (p^m) G ^pm G). There exists an opensubgroup G_o of G such that (G₀) = 0 and has order p^r in G/G₀.Choose H to be the subgroup of G generated by , and then aneasy application to G/G₀ of the theory of finite abelian groupsshows the existence of G' (note that () ¦ 0 implies that 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' k₁ F'ab). To removea minor ambiguity in the proof of (iv), choose to be an elementof Gal (F'_ab/k₂) whose image $$\stackrel{\¯}{\sigma}$$ in Gal (k₁/k₂) generates this last group. The error occursin the statement that the canonical map v : A^P A acts on pointsby sending a^p a; it, of course, sends a a. The proof is correct, however, in the case that it is possibleto choose so that ^p = 1 (in Gal (F'/k₂)). By applying Lemma 2 to G = Gal (F'_ab/k₂) and the map G Gal(k₁/k₂) one sees that only the following two cases have to beconsidered. (a) It is possible to choose so that ^pm = 1, for some m, andG = G' x H where G' acts trivially on k₁ and H is generatedby . (b) For any m > 0 there exists a field K, F'ab K k₁ k₂is 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, ), regarded as being defined over K, has a modelover k₂. Indeed, if m = 1, then this was observed above, butwhen m > 1 the same argument applies. In the second case, let : (A, I, ) (A$$\stackrel{\¯}{\sigma}$$, I$$\stackrel{\¯}{\sigma }$$, $$\stackrel{\¯}{\sigma}$$) be an isomorphism defined over k₁ and let v ... ^p–1 = µ(R). If is replaced by for some Aut_k1((A, I, )) then is replacedby ^P. Thus, as µ(R) is finite, we may assume that ^pm–1= 1 for some m. Choose K, as in (b), to be of degree p^m overk₂. Let _m be a generator of Gal (K/k₂) whose restriction tok₁ is $$\stackrel{\¯}{\sigma }$$. Then : (A, I, ) (A^{$$\stackrel{\¯}{\sigma }$$}, I^{$$\stackrel{\¯}{\sigma}$$}, ^{$$\stackrel{\¯}{\sigma }$$} = (A^{$$\stackrel{\¯}{\sigma}$$m}, I^{$$\stackrel{\¯}{\sigma }$$m}, ^{$$\stackrel{\¯}{\sigma}$$m} is an isomorphism defined over K and v ^mpm–1, ... ^m =^pm–1 = 1, and so, by) Lemma 1, (A, I, ) has a model overk₂ which becomes isomorphic to (A, I, over K. The proof may now be completed as before. Addendum: Professor Shimura has pointed out to me that the claimon lines 25 and 26 of p. 371, viz that µ(R) is a puresubgroup 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).