Let B P e 1 1 P e 2 2 P e g g where P 1 P 2 P g are distinct prime ideals of B

Let b p e 1 1 p e 2 2 p e g g where p 1 p 2 p g are

This preview shows page 27 - 29 out of 36 pages.

Let ℘B = P e 1 1 P e 2 2 · · · P e g g , where P 1 ,P 2 ,...,P g are distinct prime ideals of B and e 1 ,e 2 ,...,e g are their ramification indices. As noted in the proof of the Theorem in § 2.1, we have ℘B A = = P e i i A , and we have an isomorphism of A/℘ –vector spaces B/℘B g circleplusdisplay i =1 B/P e i i . Let us set ¯ A = A/℘ and ¯ B = B/℘B . For x B , let ¯ x denote the image of x in ¯ B . Note that we clearly have Tr ¯ B/ ¯ A x ) = Tr L/K ( x ) for all x B. Now if { α 1 ,...,α n } is any K –basis of L contained in B such that { ¯ α 1 ,..., ¯ α n } is an ¯ A –basis of ¯ B , then using the above identity for traces, we see that D ¯ B/ ¯ A α 1 ,..., ¯ α n ) = D L/K ( α 1 ,...,α n ) . (1) Next, we show that if ¯ B ¯ B 1 ... ¯ B g , where the isomorphism is of ¯ A –vector spaces, then we have D ¯ B/ ¯ A = g productdisplay i =1 D ¯ B i / ¯ A . (2) 6 This assumption would always be satisfied in number theoretic applications since A/℘ would usually be a finite field. 26
Image of page 27
To see the above identity, it suffices to consider the case when g = 2 since the general case would follow by induction on g . For convenience of notation, let us denote the element of B corresponding to ( u, 0) ¯ B 1 ¯ B 2 by u itself and, similarly, the element of B corresponding to (0 ,v ) ¯ B 1 ¯ B 2 by v itself. It is clear that we can choose ¯ A –bases { u 1 ,...,u r } and { v 1 ,...,v s } of ¯ B 1 and ¯ B 2 respectively such that { u 1 ,...,u r ,v 1 ,...,v s } is an ¯ A –basis of ¯ B . In view of the above convention, we see that u i v j = 0. Thus Tr ¯ B/ ¯ A ( u i v j ) = 0, and so D ¯ B/ ¯ A ( u 1 ,...,u r ,v 1 ,...,v s ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Tr( u i u i ) | 0 ...... | ...... 0 | Tr( v j v j ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = D ¯ B 1 / ¯ A ( u 1 ,...,u r ) D ¯ B 2 / ¯ A ( v 1 ,...,v s ) . Since ¯ A is a field and the non-vanishing of any of the above discriminants is independent of the choice of the corresponding ¯ A –bases, the desired equality of discriminant ideals follows. Thus we have proved (2). Now suppose is a ramified prime. Then e i > 1 for some i and thus the ring B/P e i i contains a nonzero nilpotent element (which may be taken to be any element of P e i 1 i \ P e i i ), and hence so does ¯ B . Let β B be such that ¯ β ¯ B is a nonzero nilpotent element. Extend { ¯ β } to an ¯ A –basis { ¯ β 1 ,..., ¯ β n } of ¯ B with β 1 = β . Since ¯ β 1 is nilpotent, so is ¯ β 1 ¯ β j for 1 j n . Hence Tr( ¯ β 1 ¯ β j ) = 0 for 1 j n [because if u ¯ B is nilpotent, then 0 is clearly the only eigenvalue of the linear transformation x mapsto→ ux of ¯ B ¯ B and Tr( u ) equals the sum of all eigenvalues of this linear transformation]. Consequently, D ¯ B/ ¯ A ( ¯ β 1 ,..., ¯ β n ) = 0, and so D ¯ B/ ¯ A is the zero ideal.
Image of page 28
Image of page 29

You've reached the end of your free preview.

Want to read all 36 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes