Abelian varieties defined over their Fields of Moduli, I†
Abelian varieties defined over their Fields of Moduli, I†
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在其 Moduli 域上定义的阿贝尔簇，I†
DOI：
10.1112/blms/4.3.370
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发表时间：
1972
影响因子：
0.9
通讯作者：
J. Milne
中科院分区：
文献类型：
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作者：
J. Milne
公牛伦敦数学。SOC，4(1972)，370-372.定理的证明包含一个错误。设K/k是m次循环Galois扩张，σ生成Galal(K/k)，(A，I，θ)定义在K上.设K上存在同构λ：(A，I，θ)→(Aσ，Iσ，θσ))使得vλσm−1…λσλ=1，其中v是正则同构(Aσm，Iσm，θσm)→(A，I，θ)。则(A，I，θ)在k上有一个模型，它同构于K上的(A，I，θ)。设G是交换准有限群，Φ：G→q/Z是G的连续特征标，G的像是p阶的，则：(A)存在G的子群G‘和H，使得H对某个m是PM阶循环的，且Φ(G’)=0，且G=G‘×H，或者(B)对任意m&gT；0，存在G的连续特征标Φm，使得PmΦm=Φ.Proof.如果(B)对给定的元为假，则存在一个σɛG，阶为Pr，对≤m，使得Φ(σ)？0.(考虑序列DUAL到0→Ker(PM)→G→PMG)。存在一个开子群GoofG，使得Φ(G0)=0且σ有序Print G/G0。选择H为σ生成的G的子群，然后将有限交换群理论的一个简单应用于G/G0，证明了G‘的存在性(请注意，Φ(σ)α0意味着σ不是AP-TH。我们现在证明这个定理。直至陈述(Iv)为止，证明是正确的(但(I)应为：F‘⊂K1⊂F’ab)。为了消除(Iv)证明中的一个小歧义，选择σ作为Gal1(F‘ab/K2)的一个元素，它的像在Gal1(K1/K2)中生成最后一个群。错误出现在规范的mapv：aσp→通过发送aσp↦a来作用于点的语句中；当然，Sendsa↦A。证明是正确的，然而，在可以选择σ使得σp=1(在Gal(F‘/K2)中)的情况下。通过应用引理2tog=Gal(F’ab/K2)和映射G→Gal(K1/K2)，我们看到只需要考虑以下两种情况：(A)对于某些M和GT，可以选择σ使得σPm=1，并且g=G‘×HG’其中对k1起平凡作用，并且H是由σ生成的；0存在域K，F‘ab⊃K⊃K1⊃k2是度的循环伽罗瓦扩张.在第一种情况下，我们设K⊂F’为G‘的不动域.那么(A，I，θ)，被认为是被定义的Overk，具有一个模型Overk2。在第二种情况下，设λ：(A，i，θ)→(A，i，θ))是定义在λσ…上的同构λσp−1λ=αɛμ(R)。如果对于某个λautk1((A，I，λγ))用γɛ替换λ，那么α就被αγP替换。因此，由于μ(R)是有限的，我们可以假设对于某个α，−Pm−1=1。如(B)所述，选择K为学位2。设σ为Gal(K/K2)的生成元，其限制令牌为1。则λ：(A，i，θ)→(A，i，θ=(Am，Im，θm)是定义在KandvλσMPM−1，…上的同构λσmλ=αpm−1=1，因此，通过引理1，(A，I，θ)有一个模型2同构于(A，I，θOver K)。证明现在可以像以前一样完成。补充件：Shimura教授向我指出，第371页第25行和第26行的声明，即μ(R)是∏R*t的纯子群，并不是对所有环R都成立。因此，这一似乎对定理的有效性至关重要的条件应该包括在假设中。例如，如果μ(R)是μ(F)的直和，则它成立。
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) → (Aσ, Iσ, θσ) over K such that vλσm−1…λσλ = 1, where v is the canonical isomorphism (Aσm, Iσm, θσm) → (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 pmfor 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 pmΦm= Φ.Proof. If (b) is false for a givenm, then there exists an element σ ɛG, of orderprfor somer≤m, such that Φ(σ) ¦ 0. (Consider the sequence dual to 0 → Ker (pm) →G→pmG). There exists an open subgroupGoofGsuch that Φ(G0) = 0 and σ has orderprinG/G0. ChooseHto be the subgroup ofGgenerated by σ, and then an easy application toG/G0of the theory of finite abelian groups shows the existence ofG′ (note that Φ(σ) ¦ 0 implies that σ is not ap-th. power inG).We now prove the theorem. The proof is correct up to the statement (iv) (except that (i) should read:F′ ⊂k1⊂F′ab). To remove a minor ambiguity in the proof of (iv), choose σ to be an element of Gal (F′ab/k2) whose imagein Gal (k1/k2) generates this last group. The error occurs in the statement that the canonical mapv : AσP→Aacts on points by sendingaσp↦a; it, of course, sendsa ↦ a.The proof is correct, however, in the case that it is possible to choose σ so that σp= 1 (in Gal (F′/k2)).By applying Lemma 2 toG= Gal (F'ab/k2) and the mapG→ Gal (k1/k2) one sees that only the following two cases have to be considered.(a) It is possible to choose σ so that σpm= 1, for somem, andG=G′ × HwhereG′ acts trivially onk1andHis generated by σ.(b) For anym> 0 there exists a fieldK, F′ab⊃K⊃k1⊃k2is a cyclic Galois extension of degreepm.In the first case, we letK ⊂ F′abbe the fixed field ofG′. Then (A, I, θ), regarded as being defined overK, has a model overk2. Indeed, ifm= 1, then this was observed above, but whenm> 1 the same argument applies.In the second case, let λ : (A, I, θ) → (A, I, θ) be an isomorphism defined overk1and letvλσ… λσp−1λ = α ɛ μ(R).If λ is replaced by λγ for some γ ɛ Autk1((A, I, θ)) then α is replaced by αγP. Thus, as μ(R) is finite, we may assume that αpm−1= 1 for somem. ChooseK, as in (b), to be of degreepmoverk2. Let σmbe a generator of Gal (K/k2) whose restriction tok1is. Thenλ : (A, I, θ) → (A, I, θ= (Am, Im, θmis an isomorphism defined overKandvλσmpm−1, … λσmλ = αpm−1= 1, and so, by) Lemma 1, (A, I, θ) has a model overk2which becomes isomorphic to (A, I, θ overK.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 ringsR. 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).