Solutions to the Algebra problems on the Comprehensive Examination of January 30, 2009

Solutions to the Algebra problems on the Comprehensive Examination of January 30, 2009 1. (25 Points). Let G be a group, and let H ⊆ G be a subgroup. Suppose that for every x, y ∈ G satisfying xy ∈ H, we have yx ∈ H. Prove that H is a normal subgroup of G. Solution: Given g ∈ G, h ∈ H, g −1 (gh) = (g −1 g)h = h ∈ H. Thus by the given property of H, ghg −1 = (gh)g −1 ∈ H so H is normal as desired. QED 2. (25 points). Recall that S7 denotes the group of permutations of the set {1, 2, . . . , 7}. Define the permutations σ, τ ∈ S7 by σ = (1 2)(4 6 7 5) and τ = (1 3 5)(2 7 4 6). (a) (10 points). Write στ as a product of disjoint cycles in S7 . Solution: στ = (1 3 4 7 6)(2 5) (b) (10 points). Find the order of each of σ, τ , and στ . Solution: The order of σ is the lcm of the orders of each individual (disjoint) cycle (and the order of an n-cycle is n): o(σ) = lcm(2, 4) = 4. Similarly, o(τ ) = lcm(3, 4) = 12 and o(στ ) = lcm(2, 5) = 10. (c) (5 points). Determine whether each of σ, τ , and στ is even or odd. Solution: σ = (1 2)(4 6)(6 7)(7 5) which is the product of 4 transpositions so σ is even. τ = (1 3)(3 5)(2 7)(7 4)(4 6) which is the product of 5 transpositions so τ is even. στ is therefore the product of 4 + 5 = 9 transpositions so it is odd. 3. (25 points). Let R be a ring. (a) (10 points). Define what it means for a subset I ⊆ R to be an ideal of R. If you use other terms like “closed” or “coset” or “subgroup” or “subring” or “maximal” in your definition, you must define those terms as well. Solution: I ⊆ R is an ideal of R if (I, +) is a subgroup of (R, +) and ∀x ∈ I, r ∈ R, xr ∈ I and rx ∈ I. If G is a group, a subgroup is a set H ⊆ G such that H forms a group under G’s group operation. (b) (15 points). Let I ⊆ R be an ideal of R, and suppose that xy − yx ∈ I for every x, y ∈ R. Prove that the quotient ring R/I is commutative. Solution: Given I + a, I + b ∈ R/I, by the given property of I ab − ba ∈ H. Thus (I + a)(I + b) = I + ab = I + ba = (I + b)(I + a), so R/I is commutative as desired. QED 4. (25 points). Let F2 = {0, 1} denote the field of two elements, and let F5 = {0, 1, 2, 3, 4} denote the field of five elements. (a) (15 points). Prove that f (X) = X 4 + X 2 + 1 is reducible in the polynomial ring F2 [X]. Solution: To be reducible, f must either have a degree 1 factor (and hence have a root in F2 ), or f must be a product of two irreducible degree-2 polynomials. But f (0) = f (1) = 1 so f has no roots, which means that f is the product of two irreducible degree-2 factors. The possible degree-2 polynomials are X 2 , X 2 + 1, X 2 + X and X 2 + X + 1. Over F2 we have X 2 + 1 = (X + 1)2 , so X 2 + X + 1 is the only degree-2 irreducible polynomial. Then an explicit calculation (made easier by using (a + b)2 = a2 + b2 ) shows that f (X) = (X 2 + X + 1)(X 2 + X + 1). (b) (10 points). Prove that g(X) = X 3 + 2X 2 + 2X + 3 is irreducible in the polynomial ring F5 [X]. Solution: Since deg g = 3, if g is reducible, then it must have a degree 1 factor and hence a root in F5 . There are only 5 elements of F5 , so let’s check them (remembering that we are working modulo 5): g(0) = 03 + 2(02 ) + 2(0) + 3 = 0 + 0 + 0 + 3 = 3 6= 0 g(1) = 13 + 2(12 ) + 2(1) + 3 = 1 + 2 + 2 + 3 = 3 6= 0 g(2) = 23 + 2(22 ) + 2(2) + 3 = 3 + 3 + 4 + 3 = 3 6= 0 g(3) = 33 + 2(32 ) + 2(3) + 3 = 2 + 3 + 1 + 3 = 4 6= 0 g(4) = 43 + 2(42 ) + 2(4) + 3 = 4 + 2 + 3 + 3 = 2 6= 0 Thus g has no roots in F5 , it is irreducible. QED 2