Exercises on Field Extensions

  1. If F is a field, then the intersection of any nonempty collection ofsubfields of F is also a subfield of F.

  2. Determine whether the polynomial x2 − 5 is irreducible over ℚ and ℚ(√3)

  3. Let K an extension of F and let a ∈ K, an algebraic element over F. Let p(x) its minimal polynomial over F (that is the monic polynomial ∈ F[x such that p(a) = 0). Prove that p(x) is irreducible over F. Conversely, if p(x) is a monic and irreducible polynomial of F[x], prove this is the minimal polynomial for each one of its roots.

  4. Let K an extension of F and let a ∈ K. We have seen that if a is algebraic over F, then

    F(a) = F[a]

    Caluclate the inverse of an element g(a) = ∈ F(a), g ≠ 0, expressing it in the form α0 + α1 a + α2 a2 + ... + αn − 1 an − 1, with n the degree of the minimal polynomial p(x) of a over F.

  5. Let K = ℚ(∛2), a subfield of ℂ and let a = 1/(1 + 2∛2). Prove that a ∈ K and a ∉ ℚ. Study ℚ(a).

  6. Check whether the followin are algebraic numbers, and in case calculate the minimal polynomial:

    • √2 + √3 over ℚ

    • √2 + √3 over ℚ(√2)

    • (π + 2) /(π + √2)

    • ∛2 + √5

    • (4 + √5)(1/3)

  7. Determine a basis for the following extensions

    • ℚ(√2√3), ℚ(√2,√3) and ℚ(√2,∛2) over ℚ

    • ℚ(√3 + √5) over ℚ(√5)

  8. Let p a prime number and ξ a p-th root of the unit. Study the extension ℚ(ξ).

  9. Prove that ℚ(√i) = ℚ(√2, i)

  10. Verify whether the element √3√7 is algebraic over ℚ, and in case determine the minimal polynomial.

  11. Prove that √2 + i is algebraic of degree 4 over ℚ, and of degree 2 over ℝ, then determine the minimal polynomial.

  12. Prove that the order of a finite field is a prime power.

  13. Prove that {1, √5} is a basis of ℚ(√5, √7).

Solutions

  1. Let S = ⋂i ∈ I Fi where the Fi are subfields of F. Then by the subfield criterion, 0,1 ∈ Fi. Furthermore, for any a,b,c ∈ S with c ≠ 0, we have a,b,c ∈ Fi for all i. Thus, a − b ∈ Fi and a·c−1 ∈ Fi for all i by the subfield criterion, and therefore a− b ∈ S and a ⋅ c−1 ∈ S, so S is a subfield. ■

  2. If x2 − 5 were reducible it should have a 1st degree factor in its factorization; But by Ruffini's theorem since there are no roots it cannot be divisible by x − a, so it's irreducible in ℚ.

    Suppose that x2 − 5 has the root √5 ∈ ℚ(√3); A basis of ℚ(√3) is {1, √3}. Then

    ∃ a,b ∈ ℚ: √5 = a + b√5  ⟹  5 = a2 + 2ab√5 + 5b2   ⟹   5 − a2 − 5b2 = 2ab√5

    but the L.H.S is rational and the R.H.S is irrational, a contradiction. Thus √5 ∉ ℚ(√3). ■

  3. Let p(x) = f(x)g(x), with f(x), g(x) ∈ F[x], a factorization of p(x), with ∂g(xc) and ∂f(x) less than ∂p>(x). Then 0 = p(0) = f(a)g(a), this being a relation in a field without zero divisors implies either f(a) = 0 or g(a) = 0, contradicting the fact that p(x) is the minimal polynomial with root a.

    Conversely, let f(x) a monic and irreducible polynomial in F[x]. Let a one of the roots of f(x), we must prove that f(x) is the minimal polynomial of a. To this end it suffices to prove, that given the ideal I of all polynomials in F[x] zeroed by a, it results I = (f(x)). Let I = (g(x)). Then f(x) must be a multiple of g(x). Since f(x) is irreducible we have f(x) = αg(x), with α ∈ F. But then (f(x)) = (g(x)).  □

  4. Let g(a) = α0 + α1 a + α2 a2 + ... + αn − 1 an − 1n − 1. Consider the polynomial g(x) = α0 + α1 x + α2 x2 + ... + αn − 1 xn − 1n − 1. Since g(a) ≠ 0, g(x) is not a multiple of p(x). Then

    (g(x),p(x)) = 1

    and thus there exist s(x), t(x) in F[x] such that

    1 = s(x)g(x) + t(x)p(x)

    By letting x = a, we obtain

    1 = s(a)g(a) + t(a)p(x) = s(a)g(a)

    Since s(x) = p(x)q(x) + r(x) with ∂r(x) < n. Evaluating it in a we have s(a) = r(a), and we can write the inverse as g(a) ⋅ r(a) = 1.  ■

  5. 1 + 2 ⋅ ∛2 ∈ ℚ(21/3) ⟶ 1 / (1 + 2 ⋅ ∛2) ∈ ℚ(∛2) since we are dealing with a field. (1 + 2⋅∛2) ∉ ℚ, otherwise ∛2 would be inside ℚ (but x3 − 2 is irreducible over ℚ), without calculating (1 + 2 ⋅ ∛2 we can conclude that ℚ(a) = ℚ(∛2), because [ℚ(∛2): ℚ] = 3, (prime see Corollary 8.1.5), thus there are no interdmediate fields. (if the dimension weren't prime for example, for a = the extensions ℚ < ℚ(21/2) < ℚ(21/4) and the biggest field there has degree 4 over ℚ).  ■

    • Let x = √2 + √3. Note that x2 = 2 + √6 + 3 and therefore x2 − 5 = 2√6. We can square both sides of this equation and obtain (x2 − 5)2 = 24. By expanding this equation, we obtain a monic polynomial p of degree 4 such that p(x) = 0, x4 + 25 − 10x2 + 1 = 0.

    • We can write the equivalent expression: π = (2 − a√2) / (a − 1); This is an expression made up of algebraic numbers only, which make up a field i.e. algebraic + algebraic = algebraic, algebraic * algebraic = algebraic, algebraic / algebraic = algebraic.

      let x = ∛2 + √5, then (x − √5)3 = 2 so x3 + 15x − (3x2 + 5)√5 = 2, so (x3 + 15x − 2)2 = 5⋅(3x2 + 5)2. By developing the remaining powers we eventually get x6 −15x4 −4x3 + 75x2 −60 x − 121, which is indeed the minimal polynomial, since ∛2 + √5 ∉ ℚ(∛2). ℚ(∛2) + √5) = ℚ(∛2), √5) and ℚ(∛2), √5) is as an extension of degree 6.

    • x = (4 + √5)1/3; ⟶ x3 = 4 + √5 ⟶ x3 − 4 = √5 ⟶ x6 −8x3 + 11 (minimal polynomial).

      x = (2 + √2)1/2 ⟶ x2 = (2 + √2) ⟶ x2 − 2 = √2 ⟶ x4 − 4x2 + 2, which is irreducible by Eisenstein’s irreducibility criterion.

    • Note that ℚ(√2√3) = ℚ(√6) this is because √2√3 = √6, so basis is {1,√6}. We have that x2 − √6 = 0 so [Q(√2√3) : Q] = 2; This matches with |{1, √6}| = 2.

    • The the min. poly of ℚ(√2) over ℚ is x2 − √2 = 0 and similarly for Q(√3): x2 − √3 = 0. Since Q(√2, √3) = (Q√2))(√3), we have

      [Q(√2, √3) : ℚ] = [ℚ(√2, √3) : ℚ(√2)][Q(√3) : ℚ] = 2 ⋅ 2 = 4

      Since a basis for ℚ(√2) is {1, √2} is a basis and {1, √3} is a basis for ℚ(√3), a basis for ℚ(√2, √3) is given by Theorem 8.1.4 doing doing all the possible multiplications {1, √2, √3, √6}.

    • [ℚ(√2, 21/3) : ℚ] = 6. We have that a basis for ℚ(√2) is {1, √2} and one for ℚ(21/3) is {1, ∛2, 22/3 by Theorem 8.1.4 doing doing all the possible multiplications {1, √2, ∛2, 22/3, √221/3, √222/3}.

    • We look for the minimal polynomial first by letting (x − 3)2 = 5; So [ℚ(√3 + √5) : ℚ(√5)] = 2. We know from Theorem 8.1.19 that ℚ(√3 ± √5), ℚ(√3√5), ℚ(√3/√5) are all inside ℚ(√3,√5). Q(sqrt(3), sqrt(5)) has basis {1, √3, √5, √3√5} over ℚ, so ℚ(√3 + √5) has the basis {1,√3} over ℚ(√5).  ■

  8. ξ satisfies the polynomial xp − 1, which is not the minimal polynomial for ξ since it is not irreducible: xp − 1 = (x − 1) (xp−1 + xp−2 + ... + x + 1). Now ξ does not satisfy x − 1, but it satisfies xp−1 + xp−2 + ... + x + 1 and this latter is irreducible over ℚ (see the properties of cyclotomic polynomials), which is thus the minimal polynomial of ξ. We thus have [ℚ(ξ) : ℚ] = p − 1, and a basis of ℚ(ξ) is given by {1, ξ, ξ2, ..., ξp − 2}. ■

  9. We have that

    √i = (cos π/2 + isin π/2)1/2 = cos π/4 + isin π/4 = 1/√2 + i/√2.  (*)

    In addition, i√i = −1/√2 + i/√2 which by multiplying by √2√i yields to

    √i − i√i = √2

    and since (√i)2 = i, it follows that ℚ(i, √2) ⊆ ℚ(√i). From (*) we've also that ℚ(√i) ⊆ ℚ(i, √2), thus the equality of the two extensions.  ■

  10. Let α = √3 + √7; We have α − √3 = √7 ⟶ (α − √3)2 = 7 ⟶ α2 + 3 − 2α√3 = 7, thus (α2 − 4)2 = 12α2 ⟶ a4 − 20α2 + 16 = 0. The minimal polynomial is thus x4 −20x2 + 16. Indeed α canno satisfy a polynomial of lower degree; it should have degree 2, but in this case α should be contained in ℚ(√3) and thus √7 as well should belong to ℚ(√3), which is impossible. ■

  11. Let α = √2 + i, α ∉ ℝ and α − √2 = i. By squaring both sides of the equation, α2 + 2 −2√2α = −1. Thus x2 −2√2x + 3 is the minimal polynomial of α on ℝ.

    To find the minimal polynomial over ℚ, we isolate the √2 monomial in the polynomial just obtained and square it to obtain x4 − 2x2 + 9 = 0. We now have to show this is an irreducible polynomial. We know K = ℚ(i,√2) is a field extension of L = ℚ(√2) which is a field extension of F = ℚ. In a previous result, we showed that [K : L] = 2 and since [L : ℚ] = 2 it follows [K : ℚ] = 4. Therefore, the minimum polynomial must have degree 4.  ■

  12. A finite field has characteristic a prime number p, thus is an extension of ℤp (see Theorem 4.15.5). The field F, considered as a vector space over its prime field ℤp contains a finite basis of n elements. Each element of F can be expressed as a unique linear combination of the n basis elements with coefficients in ℤp: (a1)b1 + ...+ (ak)bk with bi ∈ B = {b1,..., bn} (basis of F) and ai ∈ ℤp. Therefore there are pn elements in F. ■

  13. It suffices to show that both √5 and √7 can be written as linear combination of 1 and √5. Indeed √5 = α ⋅1 + β ⋅ √7 with α = √5 and β = 0; √7 = γ ⋅ 1 + δ √7, with γ = 0 and δ = 1.  ■

«Field Extensions Index Splitting Field»