Abstract algebra #1 - Questions and Answers

Question 1 Let H be a normal subgroup of a group G, and let m = (G: H). Expose that a^m ∈ H for every a ∈ G Answer Since H is a normal group of G and m = (G: H) then we have that the order of G / H is m. Hence for every a in G we gave that (aH)^m = H, since the order of every element divides the order of the group, which suggests a^m ∈ H Answer logically follows from the assumption that |G / H| = m Question 2 Factor x⁴ - 2 into irreducible polynomials. Answer 1 Step 1 You had a wrong sign in the factorization in R[x] (typo): (x² + √2)(x - √2)(x + √2) is the factorization: the quadratic factor has no roots in ℝ, so it's irreducible. The polynomial is indeed irreducible in ℚ[x]: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in ℝ[x] and the only way you can combine the previous factorization in the required way would be (x² + √2)(x² - √2) and the factors are not in ℚ[x]. Or you can use Eisenstein's criterion, which applies to x⁴ - 2 for the '2. Sorry, but the factorization in ℂ[x] is wrong: it should be a product of linear factors and you get them from the factorization in ℝ[x] I'll do the case ℤ₃[x]. The polynomial has no roots in ℤ₃, but it could be a product of two (monic) quadratic factors: x⁴ - 2 = (x² + ax + b)(x² + cx + d) Since bd = -2 = 1, you can have b = 1 and d = 1 or b = 2 and d = 2. In the first case we have. (x² + ax + 1)(x² + cx + 1) = x⁴ + (a + c)x³ + (2 + ac)x² + (a + c)x + 1 and so you need a + c = 0 and ac + 2 = 0. Hence c = -a and -a² = 2, which is impossible. In the second case we have (x² + ax + 2)(x² + cx + 2) = x⁴ + (a+c)x³ + (1 + ac)x² + (2a + 2c)x + 1 Again a + c = 0 and 1 + ac = 0. Since c = -a, we obtain -a² = 1. Well, this is indeed possible: we have a = 1 or a = 2. Thus the factorization is (x² + x + 2)(x² + 2x + 2) Answer 2 Step 1 The ℤ₁₇ case remains. The units of ℤ₁₇ form a cyclic group of order 16. The order of 2 is 8, as 2⁴ = -1. Thus √2 ∈ ℤ₁₇, and your polynomial factorises as: x⁴ - 2 = (x² + √2)(x² - √2) It cannot factorise further as if there were a linear factor, then ℤ₁₇ would contain an element a with a⁴ = 2, so a¹⁶ = -1, which is impossible as for any element a ∈ ℤ₁₇ we have a¹⁶ = 1. To complete the question you should work out √2 in ℤ₁₇. The easiest way to do this is check values of 2 + 17k, for k = 0, 1, 2, 3, … till you get a perfect square. Question 3 Convoluted definition of a set: H(S) = {x ∈ G|∃n ∈ ℕ, ∃{x₁, x₂,…, xₙ} ⊆ S ∪ S⁻¹, x = x₁…xₙ}. Answer No. H(S) is the set of all elements g of G such that g can be expressed as a finite product of elements of S ∪ S⁻¹. Maybe lets have a look at example. Consider G = {0, 1, 2, 3} with addition modulo 4 as group operation, i.e. G = ℤ₄. Now let S = {2}. Since we are dealing with addition, then S⁻¹ = {2} because 2 + 2 = 0 in our group. Therefore S ∪ S⁻¹ = {2}. Next H(S) is the set of all elements g of G such that g = x₁ + … + xₙ for some x₁,…, xₙ ∈ S ∪ S⁻¹. Since our last set has exactly one element then only possibilities for g are: 2, 2+2, 2+2+2, and which we can write in a compact way as n⋅2 for some n∈ ℕ. By removing duplicates (remember that 2 + 2 = 0) this gives us H(S) = {0,2} Question 4 Generalization of Rational Root Theorem [closed] Can the rational root theorem in univariate variable polynomial equations be extended to more than one variable? For instance could we apply it to f(x,y) = 1 + x + xy - 2y² + 4x² to find all possible rational roots? I am sure the obvious answer is no but is there any hint on how to find all possible rational roots say of this example? Answer 1 As said in the comments: the answer is NO in general. For your particular case, you can do the following thing. We have f(x, y) = -2y² + xy + 4x² + x + 1 = -2(y² - (x/2)y) + 4x² + x + 1 = -2(y² - (x/2)y + x²/16) + 2x²/16 + 4x² + x + 1 = -2(y - x/4)² + 2x²/16 + 4x² + x + 1 = -2(y - x/4)² + 33x²/8 + x + 1 = -2(y - x/4)² + (1/18)(33x²/8 * 18 + 18x + 18) = -2(y - x/4)² + (1/18)(594x² + 18x + 18) = -2(y - x/4)² + (1/18)(1 + x(1 + 18 + (3318x)/8 + 18x) + 18) = -2(y - x/4)² + (1/18)((1 + x(1 + 18)) + 18 + (594x²)/8) = -2(y - x/4)² + 33x²/8 + x + 1 = -2(y - x/4)² + (33/8)(x² + (8/33)x + 8/33) = -2(y - x/4)² + (33/8)(x + 4/33)² - (33/8)(16/1089) + 1 = -2(y - x/4)² + 33/8(x + 4/33)² + 33/544 - 1 = -2(y - x/4)² + (33/8)(x+4/33)² -511/544 Solving f(x,y) = 0 where x,y ∈ ℚ is the same as solving -2u² + (33/8)v² - 511/544 = 0, setting u = y - x/4 and v = x + 4/33 (it is easy to derive x,y from u, v. This is equivalent to -36u² + 81v² = -17, that is (3u - 9v)(3u + 9v) = 17, which is easy to solve. For t ∈ ℚ*, this 17 is equivalent to solve 3u - 9v = t, 3u + 9v = 1/t. You then get u,v as a rational fraction of t, then you get x,y. [The idea is that you have an affine conic, so you can reduce it to a "diagonal form", and then reduce to the easier case to a rational conic ax² + by² = c. This situation is well understood from quadratic form specialist, and you have systematic methods to solve this even if it can be a bit tricky. Here we are in a particularly nice situation.] Answer 2 Commenters have pointed to examples of why this won't generalize. I'll give a more theoretical reason. The trick of the rational root theorem is that it takes a polynomial of one variable and converts it to a homogenous polynomial of two variables, and then deduce things about the integer solutions of homogenous polynomial. For example, rational roots of 2x² + 5x + 3 = 0 correspond to integer solutions to 2x² + 5xy + 3y² = 0, with gcd (x, y) = 1. From there, we see y must be a factor of 2x². But the GCD condition means y must divide 2. Similarly, x must divide 3. With multiple variables we can still homogenize. We get, in your equation: z² + xz + (xy - 2y² + 4x²) = 0 But we can only restrict to integers with gcd (x, y, z) = 1, not pairwise relatively '. For example, if in the original equation, the roots are (x, y) = (1/2, 1/3), the roots in the homogenous polynomial would be (x, y, z) = (3, 2, 6). And, in any event, we don't know a factor of xz + (xy - 2y² + 4x²), so we don't have a factor of x². Indeed, gcd(x, xy - 2y² + 4x²) = gcd(x, 2y²), which are relatively if gcd(x, 2y) = 1. So, the heart of it is that homogenous integer polynomials of two variables, x,y, have one term of the form axⁿ and all other terms are divisible by y. There is no other common divisor for the other terms with more than one variable. But even if there is a common divisor, say: 3z² + 2xy + xy = 0. you do get y|3z², but you don't know all the solutions can reduce to an example with gcd(y,z) = 1. So you can't deduce y|3. One final note is that homogeneous polynomials of more than 2 variables need not even have "lead coefficients." For example: zx₂² + z₂x₃² + z₃x₁² has no terms of the form axⁿ. We can get similar examples with two variables, but they factor: xy² + yx² = xy(y + x). Question 5 An nth root of unity epsilon is an element such that e^n = 1. We say that epsilon is primitive if every nth root of unity is e^k for some k. Show that there are primitive nth roots of unity εₙ ∈ ℂ for all n, and find the degree of ℚ(εₙ) for 1 ≤ n ≤ 6 Answer An nth root of unity ε is an element such that ε^n = 1. It is said that ε is primitive if every nth root of unity is ε^k for some k. To show: There are primitive nth roots of unity: εₙ ∈ ℂ for all n As it is given ε is primitive and ε^k is the nth root of unity, by definition of nth root of unity we can say: (ε^k)^n = 1 Denote the nth root of unity ε = εₙ by the complex number: e^(2πi/n), for some k Obtain the value of ε^k ε^k = (e^(2πi/n))^k = (e^(2πik/n)) = e^(2πi) = 1(∵ e^(2πi) = cos 2π + i sin 2i = 1) As for some k becomes the primitive nth roots of unity. Hence it is proved that there are primitive nth roots of unity εₙ ∈ ℂ for all n Find the degree of ℚ(εₙ) for 1 ≤ n ≤ 6 Here, ℚ(εₙ) is the field extension of the rational numbers generated over by primitive nth root of unity εₙ