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Abstract Algebra

Exercise 3.1.13 (Homomorphism $r \mapsto e^{4\pi i r}$)

Repeat the preceding exercise with the map φ replaced by the map φ : r ↦ e 4 πir .

Answers

Proof. Since

e ix = 1 ⟺ ∃ ⁡ k ∈ ℤ , x = 2 kπ ,

Equivalently, for all real r ∈ ℝ ,

e 4 πir = 1 ⟺ ∃ ⁡ k ∈ ℤ , r = k 2 ⟺ r ∈ 1 2 ℤ . (1)

If φ : G → H is the homomorphism φ = r ↦ e 4 πir , then by (1)

ker ⁡ ( φ ) = 1 2 ℤ

is the set of half integers.

If z 0 = e 4 πi r 0 ( r 0 ∈ ℝ ), we obtain

r ∈ φ − 1 ( z 0 ) ⟺ e 4 πir = e 4 πi r 0 ⟺ e 4 πi ( r − r 0 ) = 1 ⟺ r − r 0 ∈ 1 2 ℤ ⟺ r ∈ r 0 + 1 2 ℤ ,

so

φ − 1 ( { e 4 iπ r 0 } ) = r 0 + 1 2 ℤ .

Since − 1 = e iπ = e 4 iπ ⋅ ( 1 ∕ 4 ) , i = e iπ ∕ 2 = e 4 iπ ⋅ ( 1 ∕ 8 ) , e 4 iπ ∕ 3 = e 4 iπ ⋅ ( 1 ∕ 3 ) , this gives the fibers

φ − 1 ( { − 1 } ) = 1 4 + 1 2 ℤ , φ − 1 ( { i } ) = 1 8 + 1 2 ℤ , φ − 1 ( { e 4 iπ ∕ 3 } ) = 1 3 + 1 2 ℤ . □
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2025-11-15 09:32
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