By Niels Lauritzen

ISBN-10: 0521826799

ISBN-13: 9780521826792

Concrete summary Algebra develops the speculation of summary algebra from numbers to Gr"obner bases, whereas takin in the entire ordinary fabric of a conventional introductory direction. moreover, there's a wealthy provide of subject matters reminiscent of cryptography, factoring algorithms for integers, quadratic residues, finite fields, factoring algorithms for polynomials, and structures of non-linear equations. a distinct function is that Gr"obner bases don't seem as an remoted instance. they're absolutely built-in as an issue that may be effectively taught in an undergraduate context. Lauritzen's method of educating summary algebra relies on an intensive use of examples, functions, and workouts. the fundamental philosophy is that inspiring, non-trivial functions and examples supply motivation and straightforwardness the educational of summary techniques. This ebook is equipped on numerous years of skilled instructing 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**

The key point is the following lemma. 4 Let p be a prime number and x ∈ Z. If x 2 ≡ 1 (mod p) then x ≡ ±1 (mod p). Proof. By assumption, p | x 2 − 1 = (x + 1)(x − 1). 3, p | x + 1 or p | x − 1. This completes the proof. 28 1 Numbers Consider, say, N = 341. Using repeated squaring we compute 2340 ≡ 1 (mod 341). From this we cannot deduce that 341 is composite. 4 we can drag 2 through some more questioning that tell us whether 341 really is composite. 4 gives that 2170 ≡ ±1 (mod 341), since (2170 )2 = 2340 .

28 1 Numbers Consider, say, N = 341. Using repeated squaring we compute 2340 ≡ 1 (mod 341). From this we cannot deduce that 341 is composite. 4 we can drag 2 through some more questioning that tell us whether 341 really is composite. 4 gives that 2170 ≡ ±1 (mod 341), since (2170 )2 = 2340 . Again one computes that 2170 ≡ 1 (mod 341). Now we reach the crucial question. 4 implies that 285 ≡ ±1 (mod 341). In this step, 2 breaks down and tells us that 285 ≡ 32 (mod 341) and therefore that 341 cannot be a prime number.

4. 4. 5 Let p be an odd prime. Then the Legendre symbol satisfies ab p = a p b . 2 is the following. 6 Let p be an odd prime. Then −1 is a quadratic residue modulo p if p ≡ 1 (mod 4) and a quadratic non-residue if p ≡ 3 (mod 4). Proof. 2 we get −1 p = (−1)( p−1)/2 . Now the result follows, since ( p − 1)/2 is even if p ≡ 1 (mod 4) and odd if p ≡ 3 (mod 4). 11 Quadratic residues 39 We move on to a celebrated lemma due to Gauss. 2. First we need some notation. Let p be an odd prime number. Then every integer a ∈ Z such that p a is congruent to precisely one number (its remainder) from M = {1, 2, 3, .

### Concrete Abstract Algebra: From Numbers to Gröbner Bases by Niels Lauritzen

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