## Download Abelian l-adic representations and elliptic curves by Jean-Pierre Serre PDF

By Jean-Pierre Serre

This vintage publication comprises an creation to platforms of l-adic representations, an issue of serious significance in quantity conception and algebraic geometry, as mirrored through the awesome fresh advancements at the Taniyama-Weil conjecture and Fermat's final Theorem. The preliminary chapters are dedicated to the Abelian case (complex multiplication), the place one reveals a pleasant correspondence among the l-adic representations and the linear representations of a few algebraic teams (now known as Taniyama groups). The final bankruptcy handles the case of elliptic curves without advanced multiplication, the most results of that is that clone of the Galois team (in the corresponding l-adic illustration) is "large."

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**Sample text**

This isomorphism [ can be norp ~ malized so that it takes p E Qp c A to p(poo) = (p,p, ... ,p, l,p, ... ,p) E Zx. Proof. We see that ~ E QX n (ZX x ~~) means that ~ = ~oo > 0 and the numerator and the denominator of ~ = ~p is prime to p for all p; so, ~ = l. Thus we conclude QX n (ZX x ~~) = {I}. By the isomorphism theorem: ¢ : AX /Qx~~ = ZXQx~~/Qx~~ ~ ZX /Qx n (ZX x ~~) = Zx. = p(poo)ppPoo = 1 in AX /Qx~~ for Pp = (1, ... , 1,P, 1, ... ,1) E AX 00 and Poo = (p, 1, ... , 1) E A x, the above isomorphism brings Pp in Q; C A x to (p(POO))-I.

26 Let VI K and V; K be smooth projective curves. Then we have a canonical isomorphism HomK(V, V') ~ HomK(K(V'), K(V)) given by K(V') :3 ¢ f--t ¢ 0 J Jor a morphism J : V --+ V' oj projective curves, where HomK(V, V') is the collection oj morphisms oj projective curves defined over K and HomK(K(V'), K(V)) is the set oj all field homomorphisms over K. 4 Algebraic Curves over a Field For each f E HomK(V, VI), 4:3 taking the corresponding field homomorphism C5 : K(V/) '-+ K(V), we define the degree of the morphism f by the field extension degree [K(V) : C5(K(V/))].

Since L = UMM for finite Galois extensions of M/LH, we find that h = 0- and H = Gal(L/LH). We now prove LGal(L/M) = M. 18, H = Gal(L/M) E Sj. By definition, M C LH. Supposing that LH -=J M, we try to get a contradiction. Pick ~ E LH - M. Then M[~l/M is a finite extension. Thus M[~lgal C L because L/K is normal. Since ~ tf. M, Gal(M[~lgal/M) -=J {I}. Pick 0- E Gal(M[~lgal /M) with o-(~) -=J ~. 14, we find 7 E Gal(L/M) = H such that 7(0 = o-(~) -=J ~. This is wrong since o-(~) has to be ~ because ~ E LH.