What is the splitting field of f(x) = x⁴ − 5 over ℚ?
Aℚ(⁴√5), since it contains one real root of f
Bℚ(⁴√5, i), since all four roots are ±⁴√5 and ±i·⁴√5, requiring both ⁴√5 and i
Cℂ, because all polynomials ultimately split over ℂ
Dℚ(√5), since f factors as (x²−√5)(x²+√5) over that extension
The four roots of x⁴ − 5 are ⁴√5, −⁴√5, i·⁴√5, and −i·⁴√5. Adjoining ⁴√5 to ℚ gives the real roots but not the complex ones. You also need i to get i·⁴√5 and −i·⁴√5. So the splitting field is ℚ(⁴√5, i) — the smallest field containing all four roots. Option A is wrong because ℚ(⁴√5) ⊂ ℝ cannot contain i·⁴√5. Option C is too large: ℂ is not the *smallest* field where f splits. Option D gives ℚ(√5), which is degree 2 over ℚ and does not contain ⁴√5.
Question 2 Multiple Choice
Two students construct the splitting field of f(x) = x³ − 2 over ℚ by adjoining roots in different orders. Student A adjoins ∛2 first, then ω (a primitive cube root of unity). Student B adjoins ω first, then ∛2. Why must their resulting fields be the same?
ABecause ℚ(∛2, ω) = ℚ(ω, ∛2) trivially as sets
BBecause the splitting field is unique up to isomorphism: any two splitting fields of f over F are isomorphic by a map fixing F
CBecause all finite extensions of ℚ of the same degree are isomorphic
DBecause the order of adjunction does not matter for polynomials of degree 3 specifically
The uniqueness up to isomorphism is not just set-equality (though in this case the fields happen to literally be equal). The deeper point is that no matter how you construct a splitting field — which root you adjoin first, which tower you climb — you always arrive at a field that is isomorphic to any other splitting field of f over F by an isomorphism fixing F. This makes the splitting field a canonical, well-defined object, not an artifact of construction. This uniqueness is what allows the Galois group to be defined: you need the splitting field to be canonical before you can count its automorphisms.
Question 3 True / False
Adjoining one root of x² − 2 to ℚ automatically provides the other root as well, since −√2 is already in ℚ(√2).
TTrue
FFalse
Answer: True
ℚ(√2) = {a + b√2 : a, b ∈ ℚ}. The other root of x² − 2 is −√2 = 0 + (−1)·√2, which is clearly in this set with a = 0, b = −1. So the splitting field of x² − 2 over ℚ is just ℚ(√2), a degree-2 extension. This does not always happen: for x³ − 2 over ℚ, adjoining ∛2 does not give the complex cube roots, so a further extension is required.
Question 4 True / False
The degree of the splitting field of a degree-n polynomial over F usually equals n! (n factorial).
TTrue
FFalse
Answer: False
The degree of the splitting field *divides* n!, but it is often much smaller. The factorial bound arises because the first root creates an extension of degree at most n, the second at most n−1, and so on. But when roots are algebraically related — for instance, when adjoining one root gives you others for free — the actual degree is smaller. For x² − 2, the splitting field has degree 2 over ℚ, not 2! = 2 (coincidentally equal here), but for x⁴ − 1 the splitting field has degree 2 over ℚ even though 4! = 24.
Question 5 Short Answer
What does 'smallest' mean in the definition of a splitting field, and why is this minimality condition important?
Think about your answer, then reveal below.
Model answer: The splitting field of f over F is the smallest field extension of F containing all roots of f — meaning it is contained in every other field extension of F in which f splits completely. Minimality means you adjoin exactly what is needed (the roots of f) and nothing extra. The condition matters because without it 'a splitting field' would be non-unique: any field containing a splitting field would also qualify. With minimality, the splitting field is the canonical choice, uniquely determined up to isomorphism, and the right object for defining the Galois group as its automorphism group over F.
ℂ contains all roots of every polynomial over ℚ, but ℂ is not the splitting field of x² − 2: ℚ(√2) is much smaller and still splits f. Minimality pins down which extension we care about. The uniqueness up to isomorphism then ensures that even though we could in principle choose different embeddings, the resulting structure is always 'the same' in the relevant sense.