By Frances Kirwan, Jonathan Woolf

A grad/research-level advent to the ability and sweetness of intersection homology concept. obtainable to any mathematician with an curiosity within the topology of singular areas. The emphasis is on introducing and explaining the most principles. tough proofs of significant theorems are passed over or simply sketched. Covers algebraic topology, algebraic geometry, illustration concept and differential equations.

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Let now p be an odd prime and suppose that (a, p) = 1. Further, let a j be the numerically least residue of a j (mod p) for j = 1, 2, . . Then Gauss’ lemma states that a = (−1)l , p where l is the number of j ≤ 12 ( p − 1) for which a j < 0. 30 Quadratic residues For the proof we observe that the numbers |a j | with 1 ≤ j ≤ r , where r = are simply the numbers 1, 2, . . , r in some order. For certainly we have 1 ≤ |a j | ≤ r , and the |a j | are distinct since a j = − ak , with k ≤ r , would give a( j + k) ≡ 0 (mod p) with 0 < j + k < p, which is impossible, and a j = ak gives a j ≡ ak (mod p), whence j = k.

Finally we deduce that, for any natural number n, there exists a primitive root (mod n) if and only if n has the form 2, 4, p j or 2 p j , where p is an odd prime. Clearly 1 and 3 are primitive roots (mod 2) and (mod 4). Further, if g is a primitive root (mod p j ) then the odd element of the pair g, g + p j is a primitive root (mod 2 p j ), since φ(2 p j ) = φ( p j ). Hence it remains only to prove the necessity of the assertion. Now if n = n 1 n 2 , where (n 1 , n 2 ) = 1 and n 1 > 2, n 2 > 2, then there is no primitive root (mod n).

Xiii) Prove that, for any prime p, the sum of all the distinct primitive roots (mod p) is congruent to μ( p − 1) (mod p). (xiv) Prove that, for a prime p > 3, the product of all the distinct primitive roots (mod p) is congruent to 1 (mod p). p (xv) Prove that if p is a prime and k is a positive integer then n=1 n k is congruent (mod p) to −1 if p − 1 divides k and to 0 otherwise. (xvi) Determine all the solutions of the congruence y 2 ≡ 5x 3 (mod 7) in integers x, y. (xvii) Prove that, for any prime p > 3, the numerator of 1 + 12 + · · · + 1/( p − 1) is divisible by p 2 (Wolstenholme’s theorem).