As we saw in example 515 if m is a monic polynomial

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As we saw in Example 5.15, if m is a monic polynomial over K of degree ‘ > 0, then the elements of quotient ring S = D/mD are in one-to-one correspondence with the polynomials of degree less than . More precisely, every element of S can be expressed uniquely as [ a mod m ], where a is a polynomial of degree less than . As we saw in Example 5.20, the ring S contains an isomorphic copy of K . Now, if m happens to be irreducible, then S is a field , since every [ a mod m ] with a 6≡ 0 (mod m ) has a multiplicative inverse. If K = Z p for a prime number p , then we see that S is a field of cardinality p . 47
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Chapter 7 The Structure of Z * n We study the structure of the group of units Z * n of the ring Z n . As we know, Z * n consists of those elements [ a mod n ] Z n such that a is an integer relatively prime to n . Suppose n = p e 1 1 · · · p e r r is the factorization of n into primes. By the Chinese Remainder Theo- rem, we have the ring isomorphism Z n = Z p e 1 1 × · · · × Z p er r which induces a group isomorphism Z * n = Z * p e 1 1 × · · · × Z * p er r . Thus the problem of studying the group of units of modulo an arbitrary integer reduces to the studying the group of units modulo a prime power. Define φ ( n ) to be the cardinality of Z * n . This is equal to the number of integers in the interval { 0 , . . . , n - 1 } that are relatively prime to n . It is clear that φ ( p ) = p - 1 for prime p . Theorem 7.1 If n = p e 1 1 · · · p e r r is the prime factorization of n , then φ ( n ) = p e 1 - 1 1 ( p 1 - 1) · · · p e r - 1 r ( p r - 1) . Proof. By the Chinese Remainder Theorem, we have φ ( n ) = φ ( p e 1 1 ) · · · φ ( p e r r ), so it suffices to show that for a prime power p e , φ ( p e ) = p e - 1 ( p - 1). Now, φ ( p e ) is equal to p e minus the number of integers in the interval { 0 , . . . , p e - 1 } that are a multiple of p . The integers in the interval { 0 , . . . , p e - 1 } that are multiplies of p are precisely 0 , p, 2 p, 3 p, . . . , ( p e - 1 - 1) p , of which there are p e - 1 . Thus, φ ( p e ) = p e - p e - 1 = p e - 1 ( p - 1). 2 Next, we study the structure of the group Z * n . Again, by the Chinese Remainder Theorem, it suffices to consider Z * p e for prime p . We consider first consider the simpler case Z * p . Theorem 7.2 Z * p is a cyclic group. This theorem follows from the more general theorem: Theorem 7.3 Let K be a field and G a subgroup of K * of finite order. Then G is cyclic. 48
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Proof. Let n be the order of G , and suppose G is not cyclic. Then by Theorem 4.32, we have that the exponent m of G is strictly less than n . It follows that for all α G , α m = 1 K . That is, all the elements of G are roots of the polynomial T m - 1 K K [ T ]. But since a polynomial of degree m over a field has at most m roots, this contradicts the fact that m < n . 2 Now we consider more generally the structure of Z * p e . The situation for odd p is described by the following theorem. Theorem 7.4 Let p be an odd prime and e 1 . Then Z * p e is cyclic.
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