Mathematics.

field theory

Splitting Fields

Abstract Algebra II40 minDifficulty8 out of 10

You should know: field extensions

Overview

The splitting field of a polynomial f(x) over a field K is the smallest field extension of K in which f factors completely into linear factors. Splitting fields exist for every polynomial over every field and are unique up to isomorphism, and they are the natural home for Galois theory, since the Galois group of f is defined as the automorphism group of its splitting field fixing K. Constructing a splitting field is an iterative process: adjoin one root at a time via a quotient K[x]/(irreducible factor), and repeat until the polynomial fully factors.

Intuition

Not every polynomial factors completely over its original field — x²+1 has no real roots — so the splitting field is the minimal 'upgrade' of the field that supplies exactly the roots needed, and nothing more. You build it by repeatedly adjoining one irreducible factor's root at a time: factor f over the current field, if there's an irreducible factor of degree >1, adjoin one of its roots (via K[x]/(irreducible)), and the polynomial gains a linear factor; repeat until everything splits. The final degree [L:K] measures how 'entangled' the roots are — full degree n! happens when there are no hidden algebraic relations among the roots, as for a 'generic' polynomial.

Formal Definition

Definition

Let f(x) ∈ K[x] have degree n. A splitting field L of f over K satisfies:

f(x)=c(x−α1)(x−α2)⋯(x−αn) in L[x]f(x) = c(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n) \text{ in } L[x]
f splits completely (factors into linear terms) over L
L=K(α1,…,αn)L = K(\alpha_1, \ldots, \alpha_n)
L is generated over K by all the roots
[L:K]≤n![L:K] \leq n!
Degree bound: at most n! (achieved when the Galois group is the full symmetric group)
Splitting fields exist and are unique up to K-isomorphism\text{Splitting fields exist and are unique up to } K\text{-isomorphism}
Existence and uniqueness

Worked Examples

  1. x²+1 is irreducible over ℚ (no real, hence no rational, roots).

    x2+1 irreducible over Qx^2+1 \text{ irreducible over } \mathbb{Q}

  2. Adjoining one root i gives ℚ(i); check if x²+1 splits completely there: x²+1=(x-i)(x+i), and both roots i, −i lie in ℚ(i).

    x2+1=(x−i)(x+i)∈Q(i)[x]x^2+1 = (x-i)(x+i) \in \mathbb{Q}(i)[x]

  3. Since both roots are already in ℚ(i), no further extension is needed.

    [Q(i):Q]=2[\mathbb{Q}(i):\mathbb{Q}] = 2

Answer: The splitting field of x²+1 over ℚ is ℚ(i), with [ℚ(i):ℚ] = 2.

Practice Problems

Difficulty 5/10

Find the splitting field of x²−2 over ℚ and its degree.

Difficulty 7/10

Explain why the splitting field of xⁿ−1 over ℚ (the cyclotomic field) requires adjoining a primitive nth root of unity, not just one root.

Difficulty 7/10

Verify that 𝔽₄ (constructed via 𝔽₂[x]/(x²+x+1)) is the splitting field of x⁴−x over 𝔽₂.

Quiz

The splitting field of f(x) over K is:
For a degree-n polynomial, the degree of its splitting field over K is at most:
The splitting field of x³−2 over ℚ is:

Summary

  • The splitting field of f(x) over K is the minimal extension where f factors completely into linear factors; it always exists and is unique up to isomorphism.
  • Splitting fields are built by iteratively adjoining roots of irreducible factors; [L:K] ≤ n! for a degree-n polynomial.
  • x³−2 over ℚ needs ℚ(∛2, ω) (degree 6), illustrating that a single real root isn't enough when complex roots are also needed.

References

  1. WebsiteWikipedia — Splitting field