By Niels Lauritzen
ISBN-10: 0521826799
ISBN-13: 9780521826792
Concrete summary Algebra develops the idea of summary algebra from numbers to Gr"obner bases, whereas takin in all of the ordinary fabric of a standard introductory direction. moreover, there's a wealthy offer of subject matters resembling cryptography, factoring algorithms for integers, quadratic residues, finite fields, factoring algorithms for polynomials, and platforms of non-linear equations. a distinct characteristic is that Gr"obner bases don't look as an remoted instance. they're totally built-in as a subject matter that may be effectively taught in an undergraduate context. Lauritzen's method of educating summary algebra is predicated on an in depth use of examples, functions, and workouts. the elemental philosophy is that inspiring, non-trivial purposes and examples provide motivation and straightforwardness the training of summary thoughts. This e-book is equipped on a number of years of skilled educating introductory summary algebra at Aarhus, the place the emphasis on concrete and encouraging examples has superior pupil functionality considerably.
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Extra resources for Concrete Abstract Algebra: From Numbers to Gröbner Bases
Example text
Euclid proved that there are infinitely many prime numbers. This proof is more than 2000 years old and still breathtaking. First we need a lemma. 1 Every non-zero natural number n is a product of prime numbers. Proof. The natural number 1 is the empty product of prime numbers by definition. We prove the general statement by induction. Assume that every natural number m < n is a product of prime numbers. Then we have to prove that n is a product of prime numbers. If n is a prime number then it is a product of prime numbers (with one factor).
1 Let X be any integer and k a natural number. Then X k( p−1)(q−1)+1 ≡ X (mod N ). Proof. 11(i) it is enough to prove that X k( p−1)(q−1)+1 ≡ X (mod p), X k( p−1)(q−1)+1 ≡ X (mod q). We will prove the congruence for p (the proof for q is similar). If p | X then X ≡ 0 (mod p). Therefore X k( p−1)(q−1)+1 ≡ 0 (mod p) and X k( p−1)(q−1)+1 ≡ X (mod p). However, if p X , then gcd(X, p) = 1. 2 gives X ϕ( p) = X p−1 ≡ 1 (mod p), and then we compute with congruences: X k( p−1)(q−1) ≡ (X p−1 )k(q−1) ≡ 1 (mod p).
2 Since p (( p − 1)/2)! 4 finishes the proof. 11 Let p be an odd prime. Then 2 is a quadratic residue modulo p if p ≡ 1, 7 (mod 8) and a quadratic non-residue if p ≡ 3, 5 (mod 8). Proof. 10 is the number of elements in the list 1 · 2, 2 · 2, 3 · 2, . . , p−1 ·2 2 that are greater than ( p − 1)/2. To compute µ we consider two cases. If p ≡ 1 (mod 4) then µ= p−1 p−1 p−1 − = . 2 4 4 µ= p−1 p−3 p+1 − = . 12 Exercises since ( p − 1)/4 is even when p ≡ 1 (mod 8) and odd when p ≡ 5 (mod 8) and ( p + 1)/4 is even when p ≡ 7 (mod 8) and odd when p ≡ 3 (mod 8).
Concrete Abstract Algebra: From Numbers to Gröbner Bases by Niels Lauritzen
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