PMATH 441 - PMATH 441/641 ALGEBRAIC NUMBER THEORY 1....

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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 non-zero 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 ]....
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This note was uploaded on 07/09/2011 for the course PMATH 441 taught by Professor Nonelisted during the Winter '00 term at Waterloo.

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PMATH 441 - PMATH 441/641 ALGEBRAIC NUMBER THEORY 1....

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