MAT3702 ASSIGNMENT TWO 2023
Question 1
𝑎𝑝−1 ≡ 24 (𝑚𝑜𝑑 5)
𝑎𝑝−1 ≡ 16 (𝑚𝑜𝑑 5)
15 = 5 × 3
16 − 1 = 5 × 3
5 divides (16 − 1)
Therefore,
𝑎𝑝−1 ≡ 1 (𝑚𝑜𝑑 5)
Question 2
𝑆 = {0, 3} is not a subring of ℤ6 because 𝑆 does not contain 1.
Question 3
𝑅 is a ring and 𝑏 ∈ 𝑅 such that 𝑏𝑏 = 𝑏.
Given that 𝑏 ≠ 0𝑅
Therefore, 𝑏 has a unique inverse 𝑏 −1 ∈ 𝑅 such that 𝑏𝑏 −1 = 1𝑅 = 𝑏 −1 𝑏
𝑏𝑏 = 𝑏
⇒ (𝑏𝑏)𝑏 −1 = 𝑏𝑏 −1
Rings are associative, therefore:
𝑏(𝑏𝑏 −1 ) = 𝑏𝑏 −1
𝑏1𝑅 = 1𝑅
𝑏 = 1𝑅
Question 1
𝑎𝑝−1 ≡ 24 (𝑚𝑜𝑑 5)
𝑎𝑝−1 ≡ 16 (𝑚𝑜𝑑 5)
15 = 5 × 3
16 − 1 = 5 × 3
5 divides (16 − 1)
Therefore,
𝑎𝑝−1 ≡ 1 (𝑚𝑜𝑑 5)
Question 2
𝑆 = {0, 3} is not a subring of ℤ6 because 𝑆 does not contain 1.
Question 3
𝑅 is a ring and 𝑏 ∈ 𝑅 such that 𝑏𝑏 = 𝑏.
Given that 𝑏 ≠ 0𝑅
Therefore, 𝑏 has a unique inverse 𝑏 −1 ∈ 𝑅 such that 𝑏𝑏 −1 = 1𝑅 = 𝑏 −1 𝑏
𝑏𝑏 = 𝑏
⇒ (𝑏𝑏)𝑏 −1 = 𝑏𝑏 −1
Rings are associative, therefore:
𝑏(𝑏𝑏 −1 ) = 𝑏𝑏 −1
𝑏1𝑅 = 1𝑅
𝑏 = 1𝑅
