Questions: Algebraic and Transcendental Elements

Question 1

Let α = ⁴√2 (the real fourth root of 2). The polynomial x⁴ − 2 ∈ ℚ[x] vanishes at α and is irreducible over ℚ. What can you conclude?

• A. α is transcendental over ℚ because no rational number equals ⁴√2
• B. α is algebraic over ℚ with minimal polynomial x⁴ − 2, so [ℚ(α):ℚ] = 4
• C. α is algebraic over ℚ, but the extension degree depends on how many roots x⁴ − 2 has in ℝ
• D. α is algebraic, but since ⁴√2 is irrational, the extension ℚ(α)/ℚ has infinite degree

Answer: (B)

An element is algebraic over F if it satisfies any nonzero polynomial in F[x]. Since x⁴ − 2 is irreducible over ℚ and has α as a root, it is the minimal polynomial of α over ℚ. The extension degree [ℚ(α):ℚ] equals the degree of the minimal polynomial, which is 4. Being irrational has nothing to do with transcendence — an element is transcendental only if *no* polynomial with F-coefficients vanishes at it, not merely if it fails to be rational.

Question 2

An element α is transcendental over F. A student claims that F(α) ≅ F(x) as fields, where x is a formal indeterminate. Which assessment is correct?

• A. The student is wrong — transcendental elements generate degree-2 extensions by definition
• B. The student is correct: [F(α):F] is infinite and F(α) ≅ F(x) as fields
• C. The student is partially right: the degree is infinite, but F(α) is not isomorphic to F(x) because α has a specific numerical value
• D. The isomorphism fails because F(x) contains polynomials while F(α) contains only field elements

Answer: (B)

Both claims are correct. Because α satisfies no polynomial over F, the set {1, α, α², …} is linearly independent over F (any finite vanishing linear combination would be a polynomial with α as a root), so [F(α):F] is infinite. Moreover, the map x ↦ α induces an isomorphism F(x) → F(α): every algebraic relation that would hold in F(α) would correspond to a polynomial vanishing at α, and by assumption there are none. The 'specific numerical value' objection misunderstands field isomorphism — it is a structure-preserving bijection, not a numerical equality.

Question 3

The minimal polynomial of an algebraic element over F is always irreducible over F.

• T. True
• F. False

Answer: True

This is a theorem, not just a definition. Among all nonzero polynomials in F[x] vanishing at α, the minimal polynomial has smallest degree. If it factored as p(x) = g(x)h(x) with g, h of smaller degree, then since p(α) = 0 we would have g(α) = 0 or h(α) = 0 (F[x] is an integral domain). But then g or h would be a lower-degree polynomial vanishing at α, contradicting minimality. So the minimal polynomial is irreducible.

Question 4

If an element α ∈ K satisfies a polynomial of degree 5 over F, then [F(α):F] = 5.

• T. True
• F. False

Answer: False

This confuses 'satisfying some degree-5 polynomial' with 'having a degree-5 minimal polynomial.' The extension degree equals the degree of the *minimal polynomial* — the unique monic irreducible polynomial of smallest degree vanishing at α. If α satisfies x⁵ − 1 = 0 and that polynomial factors over F, then α's minimal polynomial may have smaller degree. For instance, a primitive 5th root of unity satisfies x⁵ − 1 = (x − 1)(x⁴ + x³ + x² + x + 1), so its minimal polynomial over ℚ has degree 4, giving extension degree 4, not 5.

Question 5

Explain why an algebraic element α of degree n over F gives a finite extension [F(α):F] = n, while a transcendental element gives an infinite extension.

Model answer: An algebraic element α of degree n has a minimal polynomial m(x) of degree n. In F(α), the relation m(α) = 0 lets us rewrite αⁿ as a linear combination of lower powers of α, so every element of F(α) can be expressed in terms of {1, α, …, αⁿ⁻¹}. These n elements span F(α) and are linearly independent (if a nontrivial linear combination were zero, that would be a polynomial of degree < n vanishing at α, contradicting minimality). So {1, α, …, αⁿ⁻¹} is a basis, giving [F(α):F] = n. A transcendental element satisfies no polynomial over F, so no finite set of powers suffices: if {1, α, …, αᵏ} spanned F(α), some linear combination would vanish at α, contradicting transcendence. Hence the extension has infinite degree.

The minimal polynomial provides a 'reduction rule' capping the dimension of F(α) at n. Transcendental elements lack any such rule, so no finite basis exists.