Chapter 19

Rigorous

Splitting Fields & Algebraic Closures

00 · Symbol Glossary

$\mathrm{Split}_F(f)$Splitting field of f over F (informal notation)

The smallest extension of $F$ in which $f$ factors completely into linear pieces. This chapter's central construction.

$\overline{F}$Algebraic closure of F

An algebraic extension of $F$ in which every polynomial splits completely — the "largest algebraic extension," in a precise sense, and unique up to isomorphism.

$f'$Formal derivative of f

For $f=\sum a_ix^i$ , $f' := \sum ia_ix^{i-1}$ — defined by the power-rule formula alone, with no limits, so it makes sense over any field, including finite ones.

$\omega$Primitive cube root of unity

A complex number with $\omega^3=1$ , $\omega\neq1$ : concretely $\omega = e^{2\pi i/3} = -\tfrac12+\tfrac{\sqrt3}{2}i$ . Appears constantly in splitting fields of cubics.

01 · Splitting Fields

Definition — Splitting Field

Let $f\in F[x]$ . A field $K\supseteq F$ is a splitting field of $f$ over $F$ if: (a) $f$ factors into linear factors in $K[x]$ ; (b) $K = F(\alpha_1,\ldots,\alpha_n)$ , where $\alpha_1,\ldots,\alpha_n$ are the roots of $f$ in $K$ (no smaller field works).

Example — Splitting field of x²-2 over Q

$x^2-2 = (x-\sqrt2)(x+\sqrt2)$ over $\mathbb{Q}(\sqrt2)$ , and both roots $\pm\sqrt2$ generate exactly $\mathbb{Q}(\sqrt2)$ . $\mathbb{Q}(\sqrt2)$ is the splitting field of $x^2-2$ over $\mathbb{Q}$ .

02 · Existence of Splitting Fields

Theorem 19.1 — Splitting Fields Always Exist

For every field $F$ and every $f\in F[x]$ with $\deg f\geq1$ , a splitting field of $f$ over $F$ exists.

Proof

By strong induction on $\deg f$ . If $f$ already factors into linear pieces over $F$ , then $K=F$ (generated over itself by the roots, which already lie in $F$ ) is a splitting field.

Otherwise, $f$ has an irreducible factor $p$ of degree $\geq2$ . By Theorem 18.4, $F_1 := F[x]/(p)$ is a field extension of $F$ containing a root $\alpha$ of $p$ (hence of $f$ , since $p\mid f$ ). In $F_1[x]$ , the Factor Theorem gives $f=(x-\alpha)g$ for some $g\in F_1[x]$ with $\deg g = \deg f-1$ .

By the induction hypothesis (applied to $g$ over the field $F_1$ ), $g$ has a splitting field $K$ over $F_1$ : $g$ factors into linear pieces over $K$ , and $K=F_1(\beta_1,\ldots,\beta_{n-1})$ where $\beta_i$ are $g$ 's roots. Then $f=(x-\alpha)g$ also factors into linear pieces over $K$ , and $K = F_1(\beta_1,\ldots,\beta_{n-1}) = F(\alpha)(\beta_1,\ldots,\beta_{n-1}) = F(\alpha,\beta_1,\ldots,\beta_{n-1})$ — exactly $F$ adjoined with all of $f$ 's roots. So $K$ is a splitting field of $f$ over $F$ .

03 · Uniqueness of Splitting Fields

Existence alone would be unsatisfying if different constructions gave genuinely different fields. They don't — up to relabeling.

Theorem 19.2 — Isomorphism Extension Theorem

Let $\sigma: F\to F'$ be a field isomorphism, extend it to $\sigma: F[x]\to F'[x]$ by applying $\sigma$ to each coefficient (a ring isomorphism, since $\sigma$ preserves $+,\times$ coefficientwise), and let $f'=\sigma(f)$ . If $K$ is a splitting field of $f$ over $F$ and $K'$ is a splitting field of $f'$ over $F'$ , then $\sigma$ extends to an isomorphism $K\cong K'$ .

Proof

By induction on $[K:F]$ . If $[K:F]=1$ , $K=F$ , so $f$ already splits over $F$ ; applying $\sigma$ termwise to the linear factorization shows $f'$ splits over $F'$ too, so $K'=F'$ (generated by roots already present) and $\sigma$ itself is the required isomorphism.

Otherwise pick a root $\alpha\in K$ of $f$ with $\alpha\notin F$ , and let $p=m_\alpha$ (an irreducible factor of $f$ over $F$ , by Theorem 18.2). Applying $\sigma$ to $p$ 's coefficients gives an irreducible factor $p'=\sigma(p)$ of $f'$ over $F'$ (irreducibility is preserved: any nontrivial factorization $p'=gh$ would pull back via $\sigma^{-1}$ to a nontrivial factorization of $p$ , contradiction). Since $K'$ splits $f'$ completely, $p'$ has a root $\alpha'\in K'$ .

By Theorem 18.4 applied on both sides, $F(\alpha)\cong F[x]/(p)$ and $F'(\alpha')\cong F'[x]/(p')$ ; since $\sigma$ identifies $F[x]/(p)$ with $F'[x]/(p')$ (it's an isomorphism sending the ideal $(p)$ to $(p')$ ), composing gives an isomorphism $\tau: F(\alpha)\to F'(\alpha')$ extending $\sigma$ , with $\tau(\alpha)=\alpha'$ .

Now $K$ is a splitting field of $f$ over the larger base field $F(\alpha)$ (same field $K$ , same roots, just a bigger base), and $K'$ is a splitting field of $f'=\tau(f)$ over $F'(\alpha')$ . Since $[K:F(\alpha)] = [K:F]/[F(\alpha):F] < [K:F]$ (as $[F(\alpha):F]=\deg p>1$ , Theorem 18.4), the induction hypothesis applies to $\tau$ in place of $\sigma$ : $\tau$ (hence $\sigma$ ) extends to an isomorphism $K\cong K'$ .

Corollary 19.3 — Splitting Fields Are Unique

Any two splitting fields of $f\in F[x]$ over $F$ are isomorphic via a map fixing $F$ pointwise.

Proof

Apply Theorem 19.2 with $F'=F$ , $\sigma=\mathrm{id}_F$ .

Why uniqueness matters for what's coming

Corollary 19.3 licenses the phrase "the splitting field of $f$ ," treated as a single well-defined object up to relabeling — exactly the setting Chapter 20 needs to define the Galois group of $f$ as "the automorphisms of the splitting field fixing $F$ ," without ambiguity about which splitting field was chosen.

04 · Algebraic Closures

Definition — Algebraically Closed, Algebraic Closure

A field $K$ is algebraically closed if every nonconstant $f\in K[x]$ has a root in $K$ (equivalently, splits completely into linear factors over $K$ ). An algebraic closure of $F$ , written $\overline{F}$ , is an algebraic extension of $F$ that is algebraically closed.

Existence and uniqueness, taken as given

Every field has an algebraic closure, unique up to isomorphism — a genuine theorem, but its general proof requires Zorn's Lemma (to handle possibly infinite towers of extensions) rather than the finite inductive techniques of this chapter. We take existence and uniqueness as given, the same deferral used for free-group associativity (Chapter 10) and the field of fractions (Chapter 17): the construction is real, but its technical machinery is a detour from this course's algebraic focus.

Example — C is algebraically closed

The Fundamental Theorem of Algebra (an analysis result, proved via complex analysis or topology rather than pure algebra, and not reproved here) states every nonconstant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$ . So $\mathbb{C}$ is algebraically closed, and since $\mathbb{C}/\mathbb{R}$ is algebraic ( $[\mathbb{C}:\mathbb{R}]=2$ is finite, so Theorem 18.5 applies), $\mathbb{C} = \overline{\mathbb{R}}$ .

Example — The algebraic closure of Q is smaller than C

$\overline{\mathbb{Q}}$ (the algebraic numbers: all complex numbers that are roots of some polynomial in $\mathbb{Q}[x]$ ) is a proper subfield of $\mathbb{C}$ — it's countable (each polynomial has finitely many roots, and there are only countably many polynomials with rational coefficients), while $\mathbb{C}$ is uncountable. So $\overline{\mathbb{Q}} \subsetneq \mathbb{C}$ , even though both are algebraically closed: algebraic closures are only unique relative to the base field, not universally.

05 · Multiple Roots and Separability

Definition — Formal Derivative

For $f=\sum_{i=0}^n a_ix^i \in F[x]$ , define $f' := \sum_{i=1}^n ia_ix^{i-1}$ (where $ia_i$ means $a_i+a_i+\cdots+a_i$ , $i$ times — makes sense in any field, no limits needed). The product rule $(fg)'=f'g+fg'$ holds by direct expansion, exactly as in calculus.

Theorem 19.4 — Detecting Multiple Roots

Let $K$ be a splitting field of $f\in F[x]$ and $\alpha\in K$ a root of $f$ . Then $\alpha$ is a multiple root (i.e. $(x-\alpha)^2 \mid f$ in $K[x]$ ) if and only if $f'(\alpha)=0$ .

Proof

By the Factor Theorem, $f=(x-\alpha)g$ for some $g\in K[x]$ . By the product rule, $f' = (x-\alpha)'g + (x-\alpha)g' = g + (x-\alpha)g'$ . Evaluating at $\alpha$ : $f'(\alpha) = g(\alpha)$ .

$(x-\alpha)^2\mid f \iff (x-\alpha)\mid g \iff g(\alpha)=0$ (Factor Theorem again) $\iff f'(\alpha)=0$ .

Definition — Separable Polynomial

$f\in F[x]$ is separable if it has no multiple roots in its splitting field.

Example — Irreducible polynomials are separable in characteristic 0

Suppose $F$ has characteristic $0$ and $f\in F[x]$ is irreducible with $\deg f=n\geq1$ . Its leading term $a_nx^n$ has derivative $na_nx^{n-1}$ , and $na_n\neq0$ in $F$ (since $\mathrm{char}(F)=0$ means no positive integer multiple of a nonzero element vanishes). So $f'\neq0$ and $\deg f' = n-1 < n = \deg f$ . Any common root $\alpha$ of $f$ and $f'$ would make $\gcd(f,f')$ (computable via the Euclidean algorithm in $F[x]$ , Chapter 15) a nonconstant common divisor — but $f$ is irreducible, so its only divisors are units and associates of itself; since $\deg f' < \deg f$ , $f'$ cannot be an associate of $f$ , forcing $\gcd(f,f')$ to be a unit (constant), meaning no shared root exists. By Theorem 19.4, $f$ has no multiple roots: every irreducible polynomial over a characteristic-0 field is separable.

Why this matters going forward

Chapters 20–21 work with extensions of $\mathbb{Q}$ (characteristic 0) throughout, so separability is automatic there by the Example above — every irreducible polynomial behaves as "nicely" as possible, with no repeated roots to complicate the Galois correspondence. (Characteristic- $p$ fields can have genuinely inseparable irreducible polynomials, but that phenomenon is outside this course's scope.)

06 · A Worked Splitting Field

Step-by-step — The splitting field of $x^3-2$ over $\mathbb{Q}$

Find the roots in $\mathbb{C}$ :

x^3=2

has solutions

\sqrt[3]2,\ \sqrt[3]2\,\omega,\ \sqrt[3]2\,\omega^2

, where

\omega=e^{2\pi i/3}

is a primitive cube root of unity (satisfying

\omega^3=1

\omega\neq1

Adjoin the real root first:

\mathbb{Q}(\sqrt[3]2)

has degree 3 over

\mathbb{Q}

(Exercise 18.2), but contains only one of the three roots — the other two,

\sqrt[3]2\,\omega

and

\sqrt[3]2\,\omega^2

, are not real, while

\mathbb{Q}(\sqrt[3]2)\subset\mathbb{R}

entirely. So

\mathbb{Q}(\sqrt[3]2)

is not yet a splitting field.

Adjoin $\omega$ as well:

K=\mathbb{Q}(\sqrt[3]2,\omega)

now contains all three roots (

\sqrt[3]2

\sqrt[3]2\,\omega

, and

\sqrt[3]2\,\omega^2 = \sqrt[3]2\,\omega\cdot\omega

, both products of elements already present).

Compute the degree:

\omega

satisfies

x^2+x+1

(dividing

x^3-1=(x-1)(x^2+x+1)

, and

\omega\neq1

), irreducible over

\mathbb{Q}(\sqrt[3]2)

(a degree-2 polynomial with no root in

\mathbb{Q}(\sqrt[3]2) \subset \mathbb{R}

, since

\omega

is nonreal). So

[\mathbb{Q}(\sqrt[3]2,\omega):\mathbb{Q}(\sqrt[3]2)]=2

, and by the Tower Law,

[K:\mathbb{Q}] = 3\times2=6

Conclusion:

K=\mathbb{Q}(\sqrt[3]2,\omega)

, with

[K:\mathbb{Q}]=6

, is the splitting field of

x^3-2

over

\mathbb{Q}

— the exact example Chapter 20 revisits to compute a Galois group of order 6.

07 · Exercises

EXERCISE 19.1

Find all roots of $x^4-1$ in $\mathbb{C}$ first, then determine the smallest field over $\mathbb{Q}$ containing all of them.

Roots of $x^4-1$ : $1,-1,i,-i$ . The smallest field over $\mathbb{Q}$ containing all four is $\mathbb{Q}(i)$ (since $1,-1\in\mathbb{Q}$ already, and $-i=-1\cdot i\in\mathbb{Q}(i)$ once $i$ is adjoined). $x^4-1=(x-1)(x+1)(x-i)(x+i)$ splits completely over $\mathbb{Q}(i)$ .

The splitting field of $x^4-1$ over $\mathbb{Q}$ is $\mathbb{Q}(i)$ , with degree 2.

Find the splitting field of $x^4-1$ over $\mathbb{Q}$ , and its degree.

EXERCISE 19.2

Apply Theorem 19.4: compute $f'$ and check for a common root with $f$ .

$f=x^3-3x+2$ , $f'=3x^2-3$ . Check $x=1$ : $f(1)=1-3+2=0$ ✓; $f'(1)=3-3=0$ ✓. By Theorem 19.4, $x=1$ is a multiple root. (Confirmed: $x^3-3x+2 = (x-1)^2(x+2)$ .)

Using Theorem 19.4, determine whether $x=1$ is a multiple root of $f=x^3-3x+2$ .

EXERCISE 19.3

Recall that $\mathbb{Q}$ has characteristic 0, and apply the Example from Section 05.

$\mathbb{Q}$ has characteristic 0. By the Example in Section 05, every irreducible polynomial over $\mathbb{Q}$ is automatically separable — including $x^2-2$ (irreducible, Section 04's earlier chapters) with no need to check $\gcd(f,f')$ by hand.

Is $x^2-2$ separable over $\mathbb{Q}$ ? Justify using the characteristic-0 fact from Section 05, without computing a gcd directly.

08 · Chapter Summary

Concept	Statement
Splitting field	Smallest extension where $f$ factors into linear pieces
Existence	Every $f\in F[x]$ has a splitting field (Thm 19.1)
Isomorphism Extension Theorem	Field isomorphisms extend across matching splitting fields (Thm 19.2)
Uniqueness	Splitting fields of $f$ over $F$ are unique up to $F$ -isomorphism (Cor 19.3)
Algebraic closure $\overline{F}$	Algebraic + algebraically closed; exists, unique up to isomorphism
$\overline{\mathbb{R}}=\mathbb{C}$	Fundamental Theorem of Algebra
Formal derivative	$f'=\sum ia_ix^{i-1}$ ; product rule holds in any field
Multiple root test	$\alpha$ multiple root $\iff f(\alpha)=f'(\alpha)=0$ (Thm 19.4)
Separable	No multiple roots; automatic for irreducibles in characteristic 0

Next: Chapter 20 — Galois Theory: Fundamentals defines the Galois group of a splitting field and proves the Fundamental Theorem of Galois Theory, translating field-theoretic questions about a polynomial into purely group-theoretic ones.