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Part CCSIR NET June 2025irreducibility-is-what-makes-one-root-generate-the-whole-splitting-field

Irreducibility is what makes one root generate the whole splitting field

Let p > 2 be a prime number. Let 𝔽_p denote the field with p elements and 𝔽̄_p an algebraic closure of 𝔽_p. Which of the following statements are true?

  1. A.Let f(X) ∈ 𝔽_p[X] and Ξ±\alpha be a root of f in 𝔽̄_p. Then 𝔽p(Ξ±)_p(\alpha) is the splitting field of f in 𝔽̄_p.
  2. B.Let f, g ∈ 𝔽_p[X] be irreducible polynomials of same degree and Ξ±\alpha be a root of f in 𝔽̄_p. Then 𝔽p(Ξ±)_p(\alpha) is the splitting field of g in 𝔽̄_p.βœ“ Correct
  3. C.𝔽_p[X] has infinitely many irreducible polynomials.βœ“ Correct
  4. D.The set {a + b | a, b ∈ 𝔽_p} is contained in {a2+b2a^{2} + b^{2} | a, b ∈ 𝔽_p}.βœ“ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times β€” the specific reasoning error each wrong option was built to reward, and why this one tests hypothesis dropped.

See pricing

50 are analysed free β€” try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Related counterexample: An algebraic extension is a finite extension

More on this topic

The chapter behind this: Field extensions, degrees and splitting fields β€” free to read

From Rings and Fields β€Ί Field extensions, splitting fields, finite fields

Last revised . Found a mistake? Tell us β€” corrections are the fastest thing we act on.

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