Unformatted text preview: MATH 3360 – Applicable Algebra. HW 5 Solutions
(1) Problem 2 on page 130.
Solution: By deﬁnition −(−a) is the inverse of (−a). Since a + (−a) = 0 and inverses are unique,
we must have −(−a) = a.
(2) Problem 7 on page 130.
Solution: We know that 1 + 0 = 1 ⇒ b · (1 + 0) = b · 1 ⇒ b · 1 + b · 0 = b · 1 ⇒ b · 1 + b · 0 − b · 1 = b · 1 − b · 1
⇒ b · 0 = 0.
(3) Problem 11 on page 130.
Solution: Since a is a nonzero element of a ﬁeld, it has multiplicative inverse, say a−1 . So we
have ar = as ⇒ a−1 · ar = a−1 · as ⇒ 1 · r = 1 · s ⇒ r = s.
(4) Problem 23 on page 138. (graded)
Solution: Since a is a zero divisor, ∃b 0 such that a · b = 0. Also, a · 0 = 0 by problem 7 of this
assignment. Thus, there is more than one solution to ax = 0 when a is a zero divisor.
(5) Problem 31 on page 139. (graded)
Solution: The inverse of [9] is [3], the inverse of [11] is [19], the inverse of [17] is [23]. Note that
22 does not have an inverse, since (22, 26) = 2.
(6) Problem 35 on page 139.
Solution:
(i) 0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i.
(ii) Pair the elements with their inverses.
(iii) The order of 1 + i is 8.
(7) Problem 38 on page 146.
Solution: Since a is a unit of R, ∃a−1 ∈ R such that a · a−1 = 1. Since f is a homomorphism, we
have 1 = f (1) = f (a · a−1 ) = f (a) · f (a−1 ). Hence f (a−1 ) is the inverse of f (a) in S . Therefore,
f (a) is a unit with f (a)−1 = f (a−1 ).
(8) Problem 40 on page 146. (graded)
Solution: We just need to check the three properties from the deﬁnition of ring homomorphism:
g( f (r + s)) = g( f (r) + f ( s)) = g( f (r)) + g( f ( s))
g( f (r · s)) = g( f (r) · f ( s)) = g( f (r)) · g( f ( s)) and g( f (1)) = g(1) = 1.
(9) Problem 2 on page 174. (graded)
Solution: The units are: 1 has order 1; 2 has order 6; 4 has order 3; 5 has order 6; 7 has order 3
and 8 has order 2. 1 ...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell.
 Spring '08
 BILLERA
 Algebra

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