Questions: Finite Fields

Question 1

Why does the polynomial x^{pⁿ} − x have exactly pⁿ distinct roots over 𝔽_{pⁿ}?

• A. Its degree is pⁿ, and every polynomial over a field has exactly degree-many roots
• B. Its formal derivative is −1 (a nonzero constant), so it has no repeated roots, and every element α of 𝔽_{pⁿ} satisfies α^{pⁿ} = α
• C. The field 𝔽_{pⁿ} is algebraically closed, guaranteeing that all polynomials split completely
• D. The polynomial is irreducible over 𝔽_p, which ensures all its roots are distinct

Answer: (B)

Two facts combine: (1) since the derivative is −1 ≠ 0, the polynomial is separable — it has no repeated roots. (2) Every element α ∈ 𝔽_{pⁿ} satisfies α^{pⁿ} = α, which follows from the multiplicative group having order pⁿ − 1 (so α^{pⁿ−1} = 1 for α ≠ 0, and 0^{pⁿ} = 0). These pⁿ elements are exactly the pⁿ distinct roots. Option A is wrong: a polynomial of degree d over a field has AT MOST d roots, not exactly d. Option C is wrong: 𝔽_{pⁿ} is not algebraically closed.

Question 2

Two students construct 'the field with 16 elements' independently — one by adjoining a root of x⁴ + x + 1 over 𝔽₂, the other by adjoining a root of x⁴ + x³ + 1. What does the uniqueness theorem for finite fields say about their results?

• A. Only one construction is valid — only one of these polynomials generates a field with 16 elements
• B. Both constructions yield isomorphic fields — there is essentially one 𝔽₁₆ regardless of which irreducible polynomial is used
• C. The two fields are non-isomorphic because they arise from different irreducible polynomials
• D. Whether the fields are isomorphic depends on whether both polynomials are irreducible over 𝔽₂

Answer: (B)

The uniqueness theorem says: for each prime power q, there is a unique field of order q up to isomorphism. Both polynomials are irreducible over 𝔽₂ (this can be verified), so both generate valid degree-4 extensions of 𝔽₂, both with 2⁴ = 16 elements. By uniqueness, they must be isomorphic. This contrasts sharply with rings, where non-isomorphic structures of the same order are common. The field axioms impose enough rigidity that arithmetic determines structure completely.

Question 3

The Frobenius automorphism φ: x ↦ xᵖ generates a subgroup of the Galois group Gal(𝔽_{pⁿ}/𝔽_p), but generally does not generate the whole group — additional automorphisms may exist beyond powers of φ.

• T. True
• F. False

Answer: False

In fact, φ generates the ENTIRE Galois group. Applying φ repeatedly gives φ^k: x ↦ x^{p^k}, and after n applications, φⁿ is the identity (since x^{pⁿ} = x for all elements). Moreover, φ has order exactly n (no smaller power is the identity on all of 𝔽_{pⁿ}). So the cyclic group ⟨φ⟩ has order n, matching |Gal(𝔽_{pⁿ}/𝔽_p)| = n. The Galois group of a finite field extension is always cyclic of order n, generated by Frobenius — a remarkable rigidity.

Question 4

The Galois group of any finite field extension 𝔽_{pⁿ} over 𝔽_p is cyclic of order n.

• T. True
• F. False

Answer: True

This is one of the most striking rigidity results in algebra. Unlike general field extensions, where Galois groups can be any finite group, finite field extensions always have cyclic Galois groups. The generator is always the Frobenius automorphism x ↦ xᵖ, which has order exactly n. The subfields of 𝔽_{pⁿ} correspond to divisors of n by the Galois correspondence, making the lattice of subfields of 𝔽_{pⁿ} isomorphic to the divisor lattice of n.

Question 5

Why is the uniqueness statement 'there is essentially one finite field of each prime-power order' surprising, and what does the proof actually establish?

Model answer: It is surprising because uniqueness is rare in algebra: there are many non-isomorphic rings, groups, or algebras of the same size. For fields, the axioms are strong enough to force uniqueness. The proof shows that any field of order pⁿ must have characteristic p (so it contains 𝔽_p), and all its elements must satisfy x^{pⁿ} = x (from the order of the multiplicative group). This means the field is the splitting field of x^{pⁿ} − x over 𝔽_p, which is unique up to isomorphism by the theory of splitting fields. So any two fields of order pⁿ are both splitting fields of the same polynomial over the same base field, forcing an isomorphism between them.

The uniqueness contrasts with the situation for rings: ℤ/4ℤ and 𝔽₂ × 𝔽₂ are both rings of order 4 but are non-isomorphic. Among fields, however, the additional multiplicative structure eliminates this ambiguity. The proof also shows existence: for every prime power, x^{pⁿ} − x always has a splitting field, so 𝔽_{pⁿ} always exists.