By Serge Lang (auth.)

ISBN-10: 0387908757

ISBN-13: 9780387908755

ISBN-10: 1441985344

ISBN-13: 9781441985347

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**Example text**

Then L c K. Every automorphism of the universal domain leaving L fixed leaves L Pi invariant, and hence K fixed. Hence K is purely inseparable over L. Since each Pi has coefficient 1, the extension L(P1 , ••. , P n) of L is separable, and hence K is separable over L, whence K = L. • , P n) leaving K fixed permute the Pi transitively. L LEMMA 5. Let a be a divisor of degree 0 on C, and let P v ... , P g be g independent generic points of C over a field K over which a is rational. Then there exists one and only one positive divisor on C linearly equivalent to a + L Pi' and this divisor is of type L Qi where the Qi are independent generic points of Cover K.

We have dim U = dim W - 1, and the rational map f induces a rational map f' on V X U, such that f'(M, N) = f(M, N). Naturally, f' is defined over k(t). By the induction hypothesis, there exists a rational map fo : V X T --+ A, where T is the locus of t over k, such that fo(M, t) = f(M, N). We have thus reduced the proof to the case of curves, which has already been dealt with. k(M, N) k(M, t)/ "" k(M)/ ""k(N) k(t) / ""/ k There are many ways of proving the existence of the desired field k(t) used above.

K(P1> ' , " Pg) k(PI> ' , "/ Prj '" '" K(x) = K(x k(x) '" / K k / = + y) k(Pr+1' ' , " Pg) Furthermore, K(x) and k(P I , ' , " P r ) are linearly disjoint over k(x) because K and k(P I , ' ,"P r ) are linearly disjoint over k, I t will therefore suffice to show that k (PI>' , " P g) is an extension of degree r! over K (x), We see immediately that K (x + y) is contained in the subfield of k(p1> ' , " P g) left invariant by the automorphisms which permute PI' ' , "P r and leave P r+1' ' , " P g fixed, It is equal to this field, because on the one handk(PI> ' , "P g) is separable algebraic over k(x + y) and hence over K(x + y), and on the other hand every isomorphism of k(P I " , "P g ) which leaves K(x + y) invariant leaves P r+1' ' , " P g invariant, We can now use Lemma 1.

### Abelian Varieties by Serge Lang (auth.)

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