By ed van der Geer at al Birkhaeuser

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Now Bdiv has finite Zp-corank. It is clear that Thus, H1(rn,B) has order bounded by [B:Bdiv], which is independent of If we use the fact that B is finite, then ker(h,) has the same order as H"(rn,B), namely IE(Fn)pI. n. 2. Coker(h,) = 0. Proof. The sequence H1(F,, E[pw]) -+ H1(F,, ~ [ p , ] ) ~ --+ H 2 ( r n ,B) is exact, where B = H o(F,, E[pW]) again. But r, % H, is a free pro-p group. Hence H 2 ( r n ,B) = 0. Thus, h, is surjective as claimed. Let v be any prime of F. We will let v, denote any prime of F, lying over v.

Should be in Ax, where T L= (1 + T)-l - 1. The analogue of this statement is true for fE(T). 14. Assume that E is an elliptic curve defined over F with good, ordinary reduction or multiplicative reduction at all primes of F lying over p. Assume that SelE(F,), is A-cotorsion. Then the characteristic ideal of XE(Fm) is fied by the involution L of A induced by ~ ( y = ) y-' for all y E r. A proof of this result can be found in [Gr2] using the Duality Theorems of Poitou and Tate. There it is dealt with in a much more general context-that of Selmer groups attached to "ordinary" padic representations.

Now if one considers the A-module Y = A/( fi (T)ai),where f i (T) is irreducible in A, then Y/TY is infinite if and only if fi(T) is an associate of T. Therefore, if F is an imaginary quadratic field in which p splits and if F, is the cyclotomic Bpextension of F, then TI f (T), where f (T) is a generator of the characteristic ideal of X . One can prove that T2 I( f (T). (This is an interesting exercise. It is easy to show that X/TX has Zp-rank 1. One must then show that X/T2X also has Zp-rank 1.