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ASPECTS OF IWASAWA THEORY OVER FUNCTION FIELDS 1 ...

ASPECTS OF IWASAWA THEORY OVER FUNCTION FIELDS 1 ...

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<strong>IWASAWA</strong> <strong>THEORY</strong> <strong>OVER</strong> <strong>FUNCTION</strong> <strong>FIELDS</strong> 18Lemma 5.6. For any finite Galois extension L/F , the mapis surjective.Proof. The sequenceDiv(L) Gal(L/F ) ⊗ Z p −→ M(L) Gal(L/F )(5.12) L ∗ ⊗ Z p ↩→ Div(L) ⊗ Z p ↠ M(L)is exact because Z p is flat and |F ∗ L| is prime with p. The claim follows by taking the Gal(L/F )-cohomology of (5.12) and applying Lemma 5.5 and Hilbert 90.□For any finite subextension L of F/F let p L be the unique prime lying above p. In thefollowing lemma, we identify p L with its image in Div(L) ⊗ Z p . Moreover for any elementx ∈ M(F) we let x L denote its image in M(L) via the canonical norm map.Lemma 5.7. Let x ∈ M(F) Γ : then, for any L as above, x L is represented by a Γ-invariantdivisor supported in p L .Proof. For any L, let y L be the image of p L in M(L). Since Z p y L is a closed subset of M(L),to prove the lemma it is enough to show that ( x L + p n M(L) ) ∩ Z p y L ≠ ∅ for any n.For any finite Galois extension K/L we have the mapsι K L : Div(L) ⊗ Z p −→ Div(K) ⊗ Z pandNL K : Div(K) ⊗ Z p −→ Div(L) ⊗ Z prespectively induced by the inclusion and the norm. For any divisor whose support is unramifedin K/L we haveNL K (ι K L (D)) = [K : L]D .Also, Lemma 5.5 yields(Div(K) ⊗ Z p ) Gal(K/L) = Div(K) Gal(K/L) ⊗ Z p = ι K L (Div(L) ⊗ Z p )(since in a Gal(K/L)-invariant divisor all places of K above a same place of L occur with thesame multiplicity).Choose n and let K ⊂ F be such that [K : L] ≥ p n . By Lemma 5.6, there exists aGal(K/L)-invariant E K ∈ Div(K) ⊗ Z p having image x K . Write E K = D K + a K p K , wherea K ∈ Z p and D K has support disjoint from p K . Then D K is Galois invariant, so D K = ι K L (D L)and (using Assumption 5.1)N K L (E K ) = [K : L]D L + a K p L .Projecting into M(L) we get x L ∈ a K y L + p n M(L) .Corollary 5.8. A(F d ) Γd d−1 = 0.Proof. Taking Γ d d−1 -invariants in (5.2) (with L = F d), one finds a similar sequence□(5.13) A(F d ) Γd d−1 M(F d ) Γd d−1deg Fd Z p .Lemma 5.7 holds, with exactly the same proof, also replacing F and Γ with F d and Γ d . Thereforeany x = (x L ) L ∈ M(F d ) Γd d−1 can be represented by a sequence (aL p L ) L . FurthermoreNL K(a Kp K ) = a L p L implies that the value a L is independent of L: call it a. Thendeg Fd(x) = lim ( a L deg L (p L ) ) = a deg F (p) .Hence x ∈ Ker(deg Fd) = A(F d ) Γd d−1 only if a = 0. □

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