Minimal Polynomial of 2 - Cube Root of 5 over Q

If α is algebraic over F with minimal polynomial of degree D, then the set {1, α, α², ..., α^(D-1)} forms a basis for F(α) as a vector space over F. This provides an explicit basis for all simple algebraic extensions. For Q(∛2), the basis is {1, ∛2, (∛2)²}. For Q(√5), the basis is {1, √5}. For Q(ζ₃), the basis is {1, ζ₃}. The number of basis elements equals the degree of the minimal polynomial.

For a simple algebraic extension L = K(α) over K, the degree [L:K] equals the degree of the minimal polynomial of α over K. If the minimal polynomial has degree n, then a basis for L as a K-vector space is {1, α, α², ..., α^(n-1)}. This provides complete information about the extension's structure. For Q(√2) over Q, the minimal polynomial is t² - 2, degree 2, and basis is {1, √2}. Complex extensions can be studied as towers of simple extensions: for K ⊆ M ⊆ L, if L/M and M/K are finite, then [L:K] = [L:M] × [M:K]. For Q(√2, √3) over Q, we have Q(√2) over Q (degree 2) and Q(√2, √3) over Q(√2) (degree 2), giving total degree 4 with basis {1, √2, √3, √6}.

For α = 2 + ∛5 with degree 3, the field extension Q(α) has dimension 3 over Q. A basis is constructed by taking powers of α from 0 to n-1, where n is the degree. The basis elements are 1, α, and α². This basis spans the entire field extension and allows any element in Q(α) to be expressed as a linear combination of these basis elements with rational coefficients.

Field extensions naturally form vector spaces over their base fields. The set {1, α, α²} forms a basis for Q(α) as a vector space over Q when the minimal polynomial of α has degree 3. This basis spans the entire field (every element is a linear combination) and is linearly independent (only trivial combination gives zero). The degree [E:F] equals the dimension of E as a vector space over F. A finite degree means the dimension is finite, not that there are finitely many elements added. For Q(∛5), the degree is 3.

The basis of a composite field extension can be constructed by multiplying bases of individual extensions. Since Q(√3)/Q has basis {1, √3} and Q(∛2, √3)/Q(√3) has basis {1, ∛2, (∛2)²}, the basis for Q(∛2, √3)/Q is the set of all products: {1, √3, ∛2, √3∛2, (∛2)², √3(∛2)²}. This gives 6 elements, confirming [Q(∛2, √3):Q] = 6.

To find the minimal polynomial with integer coefficients for α = √2 + ∛3, first isolate one radical: x - ∛3 = √2. Square both sides: x² - 2x∛3 + 3 = 2, giving x² + 1 = 2x∛3. Cube both sides: (x² + 1)³ = 8x³(∛3)³, which simplifies to x⁶ + 3x⁴ + 3x² + 1 = 8x³(3) = 24x³. Rearranging gives x⁶ - 24x³ + 3x⁴ + 3x² + 1 = 0, or x⁶ + 3x⁴ - 24x³ + 3x² + 1 = 0. The sum of absolute values of coefficients is |1| + |3| + |-24| + |3| + |1| = 32. However, the correct minimal polynomial is x⁶ - 6x⁴ + 12x² - 8 - 24x + 18x = 0, which simplifies to x⁶ - 6x⁴ + 12x² - 24x + 1 = 0, giving sum of absolute values = 1 + 6 + 12 + 24 + 1 = 44. The correct answer is 62.

To find the minimal polynomial of α = √2 + √3 over ℚ, first derive a polynomial equation by algebraic manipulation: from √3 = α - √2, squaring gives 3 = α² - 2√2α + 2, so α² - 1 = 2√2α, and squaring again yields (α² - 1)² = 8α², which simplifies to α⁴ - 10α² + 1 = 0. Thus f(x) = x⁴ - 10x² + 1 is a monic polynomial with rational coefficients having α as a root. To prove it is the minimal polynomial, show it is irreducible over ℚ by contradiction: if it had a factor of degree ≤ 2, it would either have a rational root (impossible since f(1) = f(-1) = -8 ≠ 0) or factor into two quadratic polynomials with rational coefficients (impossible since this leads to a² = 12 or a² = 8, neither of which has rational solutions). Therefore, f(x) = x⁴ - 10x² + 1 is the minimal polynomial of √2 + √3 over ℚ.

To find the minimal polynomial with integer coefficients for an algebraic number like √2 + ∛3, isolate the radicals, raise both sides to appropriate powers to eliminate radicals, and simplify to obtain the polynomial equation. For √2 + ∛3, the minimal polynomial is x⁶ - 12x⁴ - 6x³ + 12x² - 36x + 1 = 0.

To find the minimal polynomial of a = √2 + √3, square both sides: a² = 5 + 2√6. Isolate the radical: a² - 5 = 2√6. Square again: (a² - 5)² = 24, giving a⁴ - 10a² + 1 = 0. Since the degree is 4, this is the minimal polynomial. It factors as (t² - 2√6t - 1)(t² + 2√6t - 1) over Q(√6), and further as (t - √2 - √3)(t + √2 + √3)(t - √2 + √3)(t + √2 - √3).

This segment teaches how to identify the minimal polynomial of an algebraic number given in radical form. For P = (3 + √53)/2, we work backwards: 2P = 3 + √53, so 2P - 3 = √53. Squaring both sides gives (2P - 3)² = 53, which expands to 4P² - 12P + 9 = 53, simplifying to 4P² - 12P - 44 = 0, or P² - 3P - 11 = 0. This quadratic is the minimal polynomial of P. The segment emphasizes that any polynomial satisfied by P must be divisible by this minimal polynomial, a fundamental concept in algebraic number theory that allows us to reduce higher-degree equations to their simplest form.

The Galois group of a polynomial f(x) over a field F is the Galois group of its splitting field over F. For f(x) = x⁴ - 2x² - 3 over Q, the splitting field is Q(√2, √3) and the Galois group is isomorphic to the Klein four-group V₄. The Galois group embeds into the symmetric group Sₙ for an n-degree polynomial. The order of the Galois group equals the degree of its splitting field extension, which is a fundamental result in Galois theory.

If k is the fixed field of the Galois group, then M is a splitting field of a separable polynomial over k. For any element α in M, the minimal polynomial of α has distinct roots (separable) and is fixed by the Galois group, so its coefficients lie in k. Taking the product of such polynomials for all generators gives a separable polynomial whose splitting field is M.

A splitting field of a polynomial p over a field K is the smallest extension field L of K such that p factors completely into linear factors in L[x], and L is generated by the roots of p over K; splitting fields always exist and are unique up to isomorphism as extensions of K, though the specific isomorphism may not be canonical due to multiple valid choices when constructing them.

For the polynomial x⁵ - 2, the roots are the fifth roots of 2 multiplied by fifth roots of unity. These roots satisfy algebraic relations such as α₁α₃ = α₂² and α₂/α₁ = α₄/α₃. The Galois group of this polynomial is the Frobenius group of order 20, which is smaller than the full symmetric group S₅ (which has order 120). This reflects the non-trivial algebraic relations among the roots.

The field of decomposition (or splitting field) of a polynomial P(x) ∈ K[x] is the smallest field extension L of K such that P(x) factors completely into linear factors in L[x]. It exists and is unique up to isomorphism. For example, the splitting field of x² - 2 over Q is Q(√2), while the splitting field of x³ - 2 over Q is Q(∛2, j), where j is a primitive cube root of unity.