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

This preview shows pages 52–55. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern