Math 343, Introduction to Algebraic Structures, Spring 2013
Homework 17, Solutions
20.2 We observe that in Z11 , 21 = 2, 22 = 4, 23 = 8, 24 = 5, 25 = 10, 26 = 9, 27 =
7, 28 = 3, 29 = 6, 210 = 1. Therefore 2 is a generator of the multiplicative
∗
cyclic group Z11
.
20.4 We note that 47 = 22 · 2 + 3. Using Fermat’s theorem, 347 = 322·2+3 =
(322 )2 · 33 ≡ 27 ≡ 4 (mod 23). So the remainder is 4 when 347 is divided
by 23.
20.27 In the field Zp , 1 and p − 1 are the only elements that are their own multiplicative inverses.
Proof. We note that 1 is obviously its own multiplicative inverse. Since
p − 1 ≡ −1 (mod p), p − 1 is also its own multiplicative inverse.
Now suppose that x ∈ Zp is its own multiplicative inverse. Then x2 = 1,
so x2 − 1 = (x + 1)(x − 1) = 0. Since Zp has no 0 divisors, either x + 1 = 0
or x − 1 = 0. So either x = −1 = p − 1 or x = 1.
20.28 If p is a prime number, then (p − 1)! ≡ −1 (mod p).
Proof. Let p be a prime number. Since 1 ≡ −1 (mod 2), the result holds
for p = 2. So we assume p ≥ 3. Consider the distinct nonzero elements
of Zp : 1, 2, · · · , p − 1. By the previous exercise, for each element k with
1 < k < p − 1, k −1 6= k. So in the product (p − 1)!, k and k −1 appear
as distinct elements. Therefore in Zp , (p − 1)! = 1(p − 1). Hence in Z,
(p − 1)! ≡ 1(p − 1) ≡ −1 (mod p).
If n > 1 is an integer such that (n−1)! ≡ −1 (mod n), then n is a prime.
Proof. Let n > 1 be an integer such that (n − 1)! ≡ −1 (mod n). Suppose
that a is an integer such that 1 ≤ a ≤ n − 1 and a|n. By assumption n|(n −
1)!+1, so a|(n−1)!+1. Since a|(n−1)!, we have a|(n−1)!+1−(n−1)! = 1.
Hence a = 1. We conclude that n is a prime.
1
2
20.29 For any positive integer n, n37 − n is divisible by 383838.
Proof. Let n be a positive integer. We start by noting the prime factorization of 383838 is 37 · 19 · 13 · 7 · 3 · 2. We will repeatedly use the fact that
np ≡ n (mod p) for any prime p. We note the following:
n37 ≡ n (mod 37),
n37
n37
n37 = n19 · n18 ≡ n · n18 ≡ n19 ≡ n (mod 19),
n37 = (n2 )13 · n11 ≡ n2 · n11 ≡ n13 ≡ n (mod 13),
n37 = (n5 )7 · n2 ≡ n5 · n2 ≡ n7 ≡ n (mod 7),
= (n12 )3 ·n ≡ n12 ·n ≡ n13 ≡ (n4 )3 ·n ≡ n4 ·n ≡ n5 ≡ n3 ·n2 ≡ n·n2 ≡ n3 ≡ n (mod 3),
= (n18 )2 ·n ≡ n19 ≡ (n9 )2 ·n ≡ n10 ≡ (n5 )2 ≡ n5 ≡ (n2 )2 ·n ≡ n3 ≡ n2 ·n ≡ n2 ≡ n (mod 2).
We conclude that n37 − n is divisible by all of the prime factors of 383838.
Hence 383838 divides n37 − n.
21.1 Let F denote the field of quotients of the integral subdomain D = {n +
mi | n, m ∈ Z} of C. Then F = {a + bi | a, b ∈ Q}.
Proof. Let m + ni, p + qi ∈ D with p + qi 6= 0. Then
m + ni p − qi
mp + nq np − mq
m + ni
=
·
= 2
+ 2
i ∈ {a + bi | a, b ∈ Q}.
p + qi
p + qi p − qi
p + q2
p + q2
As m, n, p, q vary through Z we see that any element a + bi with a, b ∈ Q
can be expressed as a quotient of D. Therefore F = {a + bi | a, b ∈ Q}.
21.2 Let F denote the field of quotients of the integral subdomain D = {n +
√
√
m 2 | n, m ∈ Z} of R. Then F = {a + b 2 | a, b ∈ Q}.
√
√
√
Proof. Let m + n 2, p + q 2 ∈ D with p + q 2 6= 0. Then
√
√
√
√
m+n 2
m+n 2 p−q 2
mp − 2nq np − mq √
√ =
√ ·
√ = 2
+ 2
2 ∈ {a+b 2 | a, b ∈ Q}.
2
2
p − 2q
p − 2q
p+q 2
p+q 2 p−q 2
√
As m, n, p, q vary through Z we see that any element a + b √2 with a, b ∈ Q
can be expressed as a quotient of D. Therefore F = {a + b 2 | a, b ∈ Q}.