This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PMATH 441/641 ALGEBRAIC NUMBER THEORY 1. Lecture: Wednesday, January 5, 2000 NO TEXT References: Number Fields (Marcus) Algebraic Number Theory (Lang) (Stewart & Hill) Marks: Final Exam 65% Midterm 25% Assignments 10% Definition. An algebraic integer is the root of a monic polynomial in Z [ x ]. An algebraic number is the root of any nonzero polynomial in Z [ x ]. We are interested in studying the structure of the ring of algebraic integers in an algebraic number field. A number field is a finite extension of Q . Well assume that the number fields we consider are all subfields of C . Definition. Suppose that K and L are fields with K L . Then K is a subfield of L and L is an extension field of K . We denote the dimension of L as a vector space over K by [ L : K ] . If [ L : K ] < , we say L is a finite extension of K . Definition. Suppose that H is a field with K H L . Then we say H is an intermediate field of K and L . Recall that [ L : K ] = [ L : H ][ H : K ]. Definition. A polynomial f K [ x ] is said to be irreducible over K iff whenever f = gh with g,h K [ x ], we have g or h constant. Recall that K [ x ] is a Principal Ideal Domain. Definition. Let K be a subfield of C and let C be an algebraic number. We denote by K ( ) the smallest subfield of C containing K and , Definition. Let K be a subfield of C and let C to be algebraic over K . A polynomial in K [ x ] is said to be a minimal polynomial of over K if it is monic, has as a root, and has degree as small as possible with these properties. Theorem 1. Let K C . Let C be algebraic over K . Then there is a unique minimal polynomial of over K . Proof: Plainly there is at least one. Suppose that p 1 ( x ) and p 2 ( x ) are minimal polynomials for over K . Consider p 1 ( x ) p 2 ( x ). Since p 1 ,p 2 are monic and of minimal degree, the degree of p 1 ( x ) 1 2 PMATH 441/641 ALGEBRAIC NUMBER THEORY p 2 ( x ) is strictly smaller than the degree of p 1 ( x ), or p 1 ( x ) = p 2 ( x ). In the former case, we contradict the minimality of the degree since p 1 ( ) p 2 ( ) = 0. Thus p 1 = p 2 and the result follows. Thus we can speak of the minimal polynomial of over K . Definition. Let K C . Let be algebraic over K . The degree of over K is the degree of the minimal polynomial of over K . Remark. Let K C and C with algebraic over K . Let p ( x ) be the minimal polynomial of over K . Suppose f K [ x ] with f ( ) = 0. Then p  f in K [ x ]. To see this note that by the Division Algorithm in K [ x ], f ( x ) = q ( x ) p ( x ) + r ( x ) with r = 0 or deg r < deg p and q,r K [ x ]. But f ( ) = q ( ) p ( ) + r ( ) Thus r ( ) = 0. If r is not identically zero, then p would not have minimal degree and so r = 0. Then p = f in K [ x ]....
View
Full
Document
This note was uploaded on 07/09/2011 for the course PMATH 441 taught by Professor Nonelisted during the Winter '00 term at Waterloo.
 Winter '00
 NoneListed

Click to edit the document details