THE SOLUTION OF FINAL OF ALGEBRA

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THE SOLUTION OF FINAL OF ALGEBRA

1. (1) φ(f + g) =R₁

0 (f + g)(x)dx =R₁

0 [f (x) + g(x)]dx

=R₁

0 f (x)dx +R₁

0 g(x)dx = φ(f )+φ(g).

(2) the index of the subgroup hµiin S₆ is ^6!₄ = 180.

2. (1) φ(10, 3) = φ(10(1, 0) + 3(0, 1)) = φ(1, 0)¹⁰· φ(0, 1)³

= [(1, 7)(6, 10, 8, 9)]²(3, 5)(2, 4) = (6, 8)(10, 9)(3, 5)(2, 4).

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3. (1) Let φ : Z5 → Z be a nontrivial (group) homomorphism with φ(1) = a, a 6= 0, we have 5φ(1) = φ(5 · 1) = φ(0) = 0

⇒ φ(1) = a = 0

hence there is no nontrivial homomorphism.

(2) Let the order of ab is n ab · ab · · · ab = e

⇒ b · (ab · ab · · · ab) · a = bea = ba

⇒ (ba)ⁿ⁺¹ = ba

⇒ (ba)ⁿ = e

the order of ba ≤ n. If the order of ba < n, then the order of ab < n.

That is contradiction. Hence the order of ab is equal to the order of ba.

4. (1) ∗ : G × X → X

∗(e, H) = eHe⁻¹= H and ∀g1, g2 ∈ G

∗(g₁g₂H) = (g₁g₂)H(g₁g₂)⁻¹ = g₁g₂Hg⁻¹₂ g⁻¹₁

= g₁∗ (g₂H)g⁻¹₁ = ∗(g₁∗ (g₂H)) hence X is a G-set

(2) (⇒) LetO(H) = {H}

∀g ∈ G, ∀h ∈ H = O(H)

⇒ ghg⁻¹ ∈ H

⇒ H is normal in G

(⇐) It’s clear that H ⊆ O(H)

∀g ∈ G, ∀h ∈ O(H) ⇒ ghg⁻¹ ∈ H

⇒ h ∈ gHg⁻¹ = H

⇒ O(H) = H

5. (1) The characteristic of M₂(Z_n) is n (2) 1 − 2a ∈ R

(1 − 2a)² = 1 − 2a 1 − 4a + 4a² = 1 − 2a

⇒ a = 0

1 − 2a=1, that is an unit of R

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6. (1) Let H be the set of all idempotent elements of a commutative ring R

∀a, b ∈ H, a² = a and b² = b (ab)² = a²b² = ab

⇒ ab ∈ H

H is closed under multiplication.

(2) {(0, 1), (0, 3), (0, 4), (0, 0), (1, 1), (1, 3), (1, 4), (1, 0)}

7. (1) Let ϕ : D ⇒ Q(D) with ϕ(a) = (a, 1) then we can show that ϕ is an isomorphism (2) There is no zero in Z₇

8. (1) N = {f (x) ∈ R[x] | f (2) = 0}

∀f (x), g(x) ∈ H, f (2) = g(2) = 0⇒ f (2) − g(2) = 0

⇒ f (x) − g(x) ∈ N

and ∀h(x) ∈ R[x], (h · f )(2) = h(2)f (2) = 0, (f · h)(2) = f (2)h(2) = 0

⇒ h(x)f (x) ∈ H, f (x)h(x) ∈ H, hence H is an ideal of R[x]

(2) Let N = hx − 2i

R[x]/hx − 2i = R is a field.

hence N is a maximal ideal of R[x]

9. (1) Let f (x) = x^p + a ∈ Z_p[x]

If a = 0 ⇒ f (x) = x^p ⇒ f (0) = 0, then f (x) is reducible If a 6= 0 ⇒ f (−a) = (−a)^p+ a = 0, then f (x) is reducible (∀a ∈ Z_p, (−a)^p−1= e ⇒ (−a)^p = (−a) )

(2) ∀a, b ∈ R

φ_p(a + b) = (a + b)^p = a^p+ b^p = φ_p(a) + φ_p(b) (the characteristic of R is p)

φp(ab) = (ab)^p = 0 and φp(a)φp(b) = 0 · 0 = 0 hence φ_p(a) = a^p is a (ring) homomorphism.

10. Let φ_m be the endomorphism of the abelian group such that φ_m(1) = m.

Then {φ_m | m ∈ Z} is the entire homomorphism ring.

Define ϕ : End(Z) → Z by ϕ(φ_m) = m Then we can show that ϕ is an isomorphism

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