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

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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 . We’ll 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 θ C be an algebraic number. We denote by K ( θ ) the smallest subfield of C containing K and θ , Definition. Let K be a subfield of C and let θ C to be algebraic over K . A polynomial in K [ x ] is said to be a minimal polynomial of θ over K if it is monic, has θ as a root, and has degree as small as possible with these properties. Theorem 1. Let K C . Let θ C be algebraic over K . Then there is a unique minimal polynomial of θ over K . Proof: Plainly there is at least one. Suppose that p 1 ( x ) and p 2 ( x ) are minimal polynomials for θ 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
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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 ( θ ) - p 2 ( θ ) = 0. Thus p 1 = p 2 and the result follows. Thus we can speak of “the” minimal polynomial of θ over K . Definition. Let K C . Let θ be algebraic over K . The degree of θ over K is the degree of the minimal polynomial of θ over K . Remark. Let K C and θ C with θ algebraic over K . Let p ( x ) be the minimal polynomial of θ over K . Suppose f K [ x ] with f ( θ ) = 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 ( θ ) = q ( θ ) p ( θ ) + r ( θ ) Thus r ( θ ) = 0. If r is not identically zero, then p would not have minimal degree and so r = 0. Then p = f in K [ x ]. Theorem 2. Let K C . If f K [ x ] is irreducible in K [ x ] of degree n , then f has n distinct roots in C . Proof: Suppose that in C [ x ], f ( x ) = a n ( x - α ) 2 f 1 ( x ) with a n C , α C , f 1 C [ x ]. Then f 0 ( x ) = 2 a n ( x - α ) f 1 ( x ) + a n ( x - α ) 2 f 0 1 ( x ) In particular, f 0 ( α ) = 0.
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