Math 210B: Algebra, Homework 6

Transcription

1 Math 210B: Algebra, Homework 6 Ian Coley February 19, 2014 Problem 1. Let K/F be a field extension, α, β K. Show that if [F α) : F ] and [F β) : F ] are relatively prime, then [F α, β) : F ] = [F α) : F ] [F β) : F ]. We have two towers of extensions Therefore we have F α, β) F α) F and F α, β) F β) F. [F α) : F ] [F α, β) : F α)] = [F α, β) : F ] = [F β) : F ] [F α, β) : F β)]. Since [F β) : F ] is relatively prime to [F α) : F ], we see that [F α) : F ] [F β) : F ] divides [F α, β) : F ]. In particular, we have [F β) : F ] divides [F α, β) : F α)] and in particular [F β) : F ] [F α, β) : F α)]. We claim that we also have [F α, β) : F α)] [F β) : F ]. Suppose that m F β is the minimal polynomial of β over F. Then m β may be viewed as a polynomial over F α) as well, so m F α) β m F α) β, which implies deg mf β m F β. Therefore the degrees of the field extensions satisfies the above inequality. But this implies that [F α, β) : F α)] = [F β) : F ]. Replacing this into our initial equality, we have as required. [F α, β) : F ] = [F β) : F ] [F α) : F ] Problem 2. Find the minimal polynomial of over Q. We first investigate powers of α = We have Therefore α is a root of the polynomial α 4 = and α 2 = fx) = X 4 10X

2 We claim that this polynomial is irreducible. First, under the substitution Y = X 2, we have Y 2 10Y + 1. Applying the quadratic formula, we have This yields Y = 10 ± = 5 ± 2 6. fx) = X ))X )). Another calculation shows that we have a complete factorisation of f as fx) = X 2 + 3))X 2 3))X 2 + 3))X 2 3)). Since the minimal polynomial of divides f, we must choose some combination of these terms. It is easily verified that no smaller combination of these monomials yields a polynomial over Q. Therefore f is minimal and we are done. Problem 3. Find the minimal polynomial over Q of a complex) primitive root of unity of degree p n. We proceed by induction. First, for a primitive pth root of unity, have the minimal polynomial X p 1 + X p X + 1 = Xp 1 X 1 by factorising X p 1 and that the above polynomial is irreducible by the linear substitution X X + 1 and Eisenstein s criterion which we did in-class). Define Φ k by Φ k X) = Φ k 1 X p ), where Φ 1 is as above. Let ζ be a p k th root of unity. Then by induction on k, we see that Φ k ζ) = Φ k 1 ζ p ) = 0, since ζ p is a p k 1 th root of unity. We now prove that Φ k is irreducible, with the base case k = 1 proven. We see that Φ k X) = Φ k 1 X p ) = Φ 1 X pk 1 ). Under the linear substitution X X + 1, we have Φ k X + 1) = Φ 1 X + 1) pk 1) = X + 1) pk 1) p X + 1) p k Now we would like to apply Eisenstein s criterion, so we look at this expression modulo p. In particular, we use X + 1) p = X p + 1. Φ k X + 1) = X + 1) pk 1) p X + 1) p k p 1 = X pk 1 + 1) + + X p k 1 + 1) + 1 p X pk 1 + 1) 1 = X pk 1 + 1) 1 = Xpk = X pk 1 p 1). X pk 1 2

3 Therefore every term but the leading term is divisible by p. To calculate the constant term under the X X + 1 substitution, we take Φ k 0 + 1) = Φ 1 1) = p. Therefore p 2 does not divide the constant term. Applying Eisenstein s criterion, Φ k X + 1) is irreducible, so Φ k X) is as well. Therefore Φ k X) is the minimal polynomial of the p k th root of unity, and we are done. Problem 4. Let X be a variable over a field F, Y = X2. Find [F X) : F Y )]. X 1 We would like to find the degree of the minimal polynomial of X. We see that X 2 X 1)Y = X 2 X 1)X2 X 1) = 0. Therefore X is a root of the polynomial ft ) = T 2 Y T + Y F Y )[T ]. Therefore [F X) : F Y )] 2, depending on whether ft ) is irreducible. First, viewing f as a polynomial in F [Y ][T ], so that the quotient field of F [Y ] is F Y ), f is primitive. Note that Y is an prime element of the PID F [Y ], since F [Y ]/Y ) = F is a domain. Therefore by Eisenstein s criterion with the prime element Y, ft ) F [Y ][T ] is a primitive irreducible polynomial. By Gauss lemma, it is an irreducible polynomial in F Y )[T ] as well. Therefore [F X) : F Y )] = 2. Problem 5. Let K be a splitting field of the polynomial X 3 2 over Q. Find [K : Q]. We see the three roots of this polynomial are α, ζα, ζ 2 α, where α = 3 2 and ζ is a primitive third root of unity. Therefore the splitting field of this polynomial contains Qα, ζα, ζ 2 α). We claim that K = Qα, ζ) is a splitting field for X 3 2. First, X 3 2 clearly splits in K. Further, the splitting field must contain α ζ 2 α) 2 = 2ζ, so it must contain ζ as well. Therefore every splitting field of X 3 2 contains α and ζ, so indeed K is a splitting field of X 3 2. We may apply Problem 1 now. Since [F α) : F ] = 3 is relatively prime to [F ζ) : F ] = 2, the total degree of the extension is 3 2 = 6. Problem 6. Let ζ be a complex) primitive nth root of unity. Show that Qζ) is a splitting field of the polynomial X n 1 over Q. 3

4 It is clear that X n 1 splits in Qζ) since, by definition, a primitive nth root of unity generates all nth roots of unity, and these are exactly the solutions to X n = 1. Further, any splitting field K of X n 1 must contain ζ, since it is a root of X n 1. Therefore Q Qζ) K, so since K Qζ) by the above, we have Qζ) is a splitting field of this polynomial. Problem 7. Find an example of a nontrivial field extension K/F with isomorphic fields K and F. Let X be a variable over F, and consider F X 2 ) F X). This is a proper extension since X / F X 2 ), and in particular it is of degree 2. But under the change of variables X 2 = Y, we see that F X) = F Y ), so the fields are isomorphic. Problem 8. Let K/F be a field extension, α K. Show that if for some non-constant polynomial f F [X] the element fα) is algebraic over F, then α is also algebraic over F. If fα) is algebraic over F, then there exists a polynomial g F [X] so that gfα)) = 0. But this is the same as g f)α) = 0, and since the composition of polynomials is still a polynomial, α is algebraic over F via this polynomial. Problem 9. Let K/F be a field extension, α, β K. Prove that the extension F α, β)/f α + β, αβ) is algebraic. It suffices to show that α and β are both algebraic elements. First, we see that so that α β is a root of the polynomial α + β) 2 4αβ = α 2 2αβ + β 2 = α β) 2, X 2 [ α + β) 2 4αβ ] F α + β, αβ)[x]. Therefore α β is algebraic over F α + β, αβ), so that F α β, α + β, αβ)/f α + β, αβ) is an algebraic extension. Call the larger field E. Then we see that α + β) + α β) = 2α E, so α E. Therefore α + β α = β E as well, so E F α, β). But since α β, α +β, αβ E, we have E F α, β) too. Therefore E = F α, β) and so our extension is algebraic. Problem 10. Find an example of an infinite algebraic extension. 4

5 Let α n = 2 1/n. Then we claim that K = Q {α n } n N ) is an infinite algebraic extension. That it is algebraic is clear, since each α n is a root of the polynomial X n 2 = 0. Further, by Eisenstein s criterion, each of these polynomials are irreducible, so [Qα n ) : Q] = n. Therefore since we have [K : Q] = [K : Q {α n } n p )] [Qα p ) : Q] for each prime p N, every prime number divides the order of [K : Q]. Therefore the least common multiple of all primes divides [K : Q], so we must have [K : Q] =. 5