Consider α = ∛5 (the real cube root of 5). Over ℚ, which of the following is the minimal polynomial of α, and what is [ℚ(α):ℚ]?
Ax³ − 5 (degree 3), so [ℚ(α):ℚ] = 3
Bx − ∛5 (degree 1), because α is a single specific number
Cx⁶ − 25 (degree 6), obtained by eliminating the cube root algebraically
DThere is no minimal polynomial because ∛5 is irrational
The minimal polynomial must lie in ℚ[x] (coefficients in ℚ) and have α as a root. x³ − 5 satisfies this: (∛5)³ − 5 = 0, and it is irreducible over ℚ by Eisenstein's criterion with p = 5. Option B is wrong because x − ∛5 has the irrational number ∛5 as a coefficient — it is not in ℚ[x]. Option D confuses irrational with transcendental: ∛5 is irrational but algebraic over ℚ. The degree of the minimal polynomial is 3, giving [ℚ(∛5):ℚ] = 3, meaning ℚ(∛5) is a 3-dimensional ℚ-vector space with basis {1, ∛5, ∛25}.
Question 2 Multiple Choice
A student argues: 'The minimal polynomial of π over ℚ must exist — just compute it from the decimal expansion.' What is the correct response?
AThe student is right — every real number satisfies some polynomial over ℚ with high enough degree
Bπ is transcendental over ℚ: Lindemann proved in 1882 that no nonzero polynomial with rational coefficients has π as a root, so no minimal polynomial exists
Cπ is irrational, and all irrational numbers are transcendental, so no minimal polynomial exists
DThe minimal polynomial exists but has infinite degree, which is why it cannot be written down
Transcendental over ℚ means exactly that no nonzero polynomial in ℚ[x] vanishes at that element — a fact Lindemann proved in 1882 for π. The decimal expansion is irrelevant: what matters is whether any finite polynomial equation with rational coefficients is satisfied. Option C is wrong: being irrational does not imply transcendental. √2 is irrational but algebraic (satisfies x² − 2 = 0). Transcendence is a stronger, harder-to-prove condition that requires ruling out infinitely many polynomial equations simultaneously.
Question 3 True / False
The minimal polynomial of an algebraic element α over K must be irreducible over K.
TTrue
FFalse
Answer: True
If the minimal polynomial factored as p(x) = q(x)r(x) over K with both factors having strictly smaller degree, then p(α) = 0 implies q(α) = 0 or r(α) = 0. Either factor would be a polynomial in K[x] of smaller degree having α as a root, contradicting the minimality of p. Irreducibility is therefore not an additional assumption — it follows directly from the definition of the minimal polynomial as the lowest-degree polynomial in K[x] that vanishes at α.
Question 4 True / False
If the minimal polynomial of α over K has degree 3, then [K(α):K] = 9, because the extension contains elements up to degree 3 in both α and α².
TTrue
FFalse
Answer: False
[K(α):K] equals the degree of the minimal polynomial — here 3, not 9. K(α) has K-basis {1, α, α²} as a vector space (dimension 3). The element α³ can be expressed in terms of {1, α, α²} using the minimal polynomial relation, so no higher powers of α are needed as independent basis elements. There is no squaring of the degree: [K(α):K] = deg(min_K(α)) is a direct equality, and this dimension is exactly the minimum number of K-linear directions needed to describe the extension.
Question 5 Short Answer
What does it mean to say the degree of the minimal polynomial measures 'how far' α is from K algebraically? Give a concrete example illustrating two different degrees.
Think about your answer, then reveal below.
Model answer: The degree [K(α):K] = deg(min_K(α)) is the dimension of K(α) as a K-vector space — the minimum number of K-linearly independent basis elements needed to reach α from K. A degree-1 minimal polynomial means α ∈ K (no extension). A degree-2 polynomial (e.g., x² − 2 for √2 over ℚ) gives a 2-dimensional extension with basis {1, √2}, so every element of ℚ(√2) looks like a + b√2. A degree-3 polynomial (e.g., x³ − 5 for ∛5 over ℚ) gives a 3-dimensional extension {1, ∛5, ∛25}. Higher degree means more algebraically 'remote' from K.
The degree is the key invariant that carries through into Galois theory and splitting fields: it measures the 'algebraic complexity' of α relative to K with a single integer. Two algebraic elements with the same minimal polynomial degree are, in a precise sense, equidistant from K, even if they look very different.