Note that OEB W k\u0141 OEB T p i W k\u0141 D P OEB p i W k\u0141 r Therefore B has only

# Note that oeb w kł oeb t p i w kł d p oeb p i w kł

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Note that OEB W OEB= T p i W D P OEB= p i W r: Therefore B has only finitely many prime ideals, say p 1 ;:::; p g where g OEB W , and T p i D 0 . When we take r D g in (*) we 107
find that B D Y g i D 1 B= p i : For each i , B= p i is a field, and it is a finite extension of k . Because k is perfect, it is even a separable extension of k . Now we can apply (2.26) to deduce that disc ..B= p i /=k/ ¤ 0 , and we can apply the preceding lemma to deduce that disc .B=k/ ¤ 0 . 2 We now prove the theorem. From the first lemma, we see that disc .B=A/ mod p D disc ..B= p B/=.A= p //; and from the last lemma that disc ..B= p B/=.A= p // D 0 if and only B= p B is not reduced. Let p B D Q P e i i . Then B= p B ' Q B= P e i , and Q B= P e i is reduced each B= P e i is reduced each e i D 1: R EMARK 3.39 (a) In fact there is a precise, but complicated, relation between the power of p dividing disc .B=A/ and the extent to which p ramifies in B . It implies for example that ord p . disc .B=A// P f i .e i 1/ , and that equality holds if no e i is divisible by the characteristic of A= p . See Serre 1962, III 6. (b) Let A be the ring of integers in a number field K , and let B be the integral closure of A in a finite extension L of K . It is possible to define disc .B=A/ as an ideal without assuming B to be a free A -module. Let p be an ideal in A , and let S D 108
A p . Then S 1 A D A p is principal, and so we can define disc .S 1 B=S 1 A/ . It is a power . p A p / m. p / of p A p . Define disc .B=A/ D Y p m. p / : The index m. p / is nonzero for only finitely many p , and so this formula does define an ideal in A . Clearly this definition agrees with the usual one when B is a free A -module, and the above proof shows that a prime ideal p ramifies in B if and only if it divides disc .B=A/: E XAMPLE 3.40 (For experts on Riemann surfaces.) Let X and Y be compact connected Riemann surfaces, and let ˛ W Y ! X be a nonconstant holomorphic mapping. Write M .X/ and M .Y / for the fields of meromorphic functions on X and Y . The map f 7! f ı ˛ is an inclusion M .X/ , ! M .Y / which makes M .Y / into a field of finite degree over M .X/ ; let m be this degree. Geometrically, the map is m W 1 except at a finite number of branch points. Let P 2 X and let O P be the set of meromorphic functions on X that are holomorphic at P — it is the discrete valua- tion ring attached to the discrete valuation ord P , and its max- imal ideal is the set of meromorphic functions on X that are zero at P . Let B be the integral closure of O P in M .Y / . Let ˛ 1 .P / D f Q 1 ;:::;Q g g and let e i be the number of sheets of Y over X that coincide at Q i . Then p B D Q q e i i where q i is the prime ideal f f 2 B j f .Q i / D 0 g : 109
Finding factorizations The following result often makes it very easy to factor an ideal in an extension field. Again A is a Dedekind domain with field of fractions K , and B is the integral closure of A in a finite separable extension L of K .

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