Splitting Fields

Splitting fields are the entry point to Galois theory, where the group of automorphisms of a splitting field fixing the base field controls whether a polynomial's roots can be expressed by radicals — this is how one proves the classical impossibility of solving general quintics by radicals. In algebraic number theory, the splitting field of $x^n-1$ over $\mathbb{Q}$ is the cyclotomic field, central to class field theory and the study of prime factorization in number rings. In finite field theory, every finite field $\mathbb{F}_{p^n}$ is literally the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$, which underlies the classification of finite fields used throughout coding theory and cryptography (e.g., Reed–Solomon codes, AES's field arithmetic, and elliptic-curve cryptography over finite fields). In computer algebra systems, algorithms for polynomial factorization and root-finding over algebraic number fields explicitly construct splitting-field towers analogous to the inductive proof above. In physics, the splitting field construction appears implicitly whenever one needs to adjoin all eigenvalues of an operator's characteristic polynomial to guarantee a full eigenbasis over an algebraically closed extension.

:::quiz[What are the two defining conditions of a splitting field $E$ for $f\in F[x]$?]

• $E$ is algebraically closed and contains $F$
• $f$ has at least one root in $E$, and $E/F$ is finite
• $f$ factors into linear factors in $E[x]$, and $E$ is generated over $F$ by the roots of $f$ ✓
• $E$ is the smallest field containing $F$, period

:::quiz[For $f(x)=x^3-2$ over $\mathbb{Q}$, what is $[E:\mathbb{Q}]$ for the splitting field $E$?]

• 3
• 6 ✓
• 2
• 9

:::quiz[In the inductive existence proof, what is the key algebraic fact making $F[x]/(p(x))$ a field?]

• $p(x)$ has degree at least 2
• $p(x)$ is irreducible, so $(p(x))$ is a maximal ideal in the PID $F[x]$ ✓
• $F$ is algebraically closed
• $p(x)$ has no repeated roots

:::quiz[What general upper bound does the theorem give for $[E:F]$ where $E$ is a splitting field of a degree-$n$ polynomial?]

• $n$
• $2^n$
• $n!$ ✓
• $n^2$

:::faq[What goes wrong if $F[x]$ is replaced by a non-UFD ring in this construction?] The entire inductive proof leans on $F[x]$ being a Euclidean domain, hence a PID, hence a UFD, which guarantees every nonconstant polynomial factors into irreducibles and that $(p(x))$ is maximal exactly when $p$ is irreducible. If one tried the same construction over a non-field coefficient ring (say $\mathbb{Z}[x]/(p(x))$ for non-irreducible or non-prime settings), quotients by principal ideals generated by "irreducible-looking" elements need not be fields or even domains, and the entire simple-extension machinery from Simple Extensions and F(alpha) breaks down. This is why the theorem is stated strictly for $F$ a field.

Ask the tutor: How does the uniqueness of splitting fields get proved, and why does it require extending field isomorphisms rather than just comparing degrees?

Think of $F$ as a stage that is too small to hold all the actors a polynomial's factorization demands. Over $\mathbb{Q}$, the polynomial $x^2+1$ has no roots at all — the stage has zero actors present. Adjoining a single root $i$ enlarges the stage to $\mathbb{Q}(i)$, and remarkably the other root $-i$ is already present there too, because $x^2+1 = (x-i)(x+i)$ automatically once $i$ is available. This is a coincidence of the specific polynomial (its roots differ by a rational multiple), not a general law. For $x^3-2$, the situation is more elaborate. Adjoining the real root $\sqrt[3]{2}$ to $\mathbb{Q}$ gives a degree-3 extension, but it is not a splitting field: the other two roots $\sqrt[3]{2}\,\omega$ and $\sqrt[3]{2}\,\omega^2$ (where $\omega = e^{2\pi i/3}$) are not real numbers, so they cannot lie in $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. We must adjoin $\omega$ as well, producing a further degree-2 step. The splitting field is $\mathbb{Q}(\sqrt[3]{2}, \omega)$, of total degree 6 over $\mathbb{Q}$ — matching the bound $3!=6$ exactly. The picture below shows this tower of adjunctions and the corresponding degrees multiplying at each stage, an instance of the Finitely Generated Algebraic Extensions tower law.

The moral: to split a polynomial you may need several successive simple extensions, each governed by an irreducible factor of what remains, and the final degree is the product of the individual step degrees — always dividing $n!$.

Example 1: $f(x) = x^4 - 5x^2 + 6$ over $\mathbb{Q}$. First factor $f$. Substituting $y = x^2$: $y^2 - 5y + 6 = (y-2)(y-3)$, so

Both $x^2-2$ and $x^2-3$ are irreducible over $\mathbb{Q}$ (neither $2$ nor $3$ is a perfect square in $\mathbb{Q}$, so by the rational root theorem these quadratics have no rational roots and, being degree 2, irreducibility over $\mathbb{Q}$ follows immediately from having no roots there). Following the inductive construction: adjoin a root of the first irreducible factor, say $\alpha = \sqrt{2}$, giving $K = \mathbb{Q}(\sqrt2)$ with $[K:\mathbb{Q}]=2$. In $K[x]$, $x^2 - 2 = (x-\sqrt2)(x+\sqrt2)$ splits completely already — both roots $\pm\sqrt2$ lie in $K$. The remaining factor to split is $g(x) = x^2 - 3$, still irreducible over $K = \mathbb{Q}(\sqrt2)$: if $\sqrt3 \in \mathbb{Q}(\sqrt2)$, then writing $\sqrt3 = a+b\sqrt2$ with $a,b\in\mathbb{Q}$ and squaring gives $3 = a^2+2b^2 + 2ab\sqrt2$; since $\sqrt2\notin\mathbb{Q}$, the coefficient of $\sqrt2$ must vanish, so $ab=0$. If $b=0$ then $\sqrt3 = a\in\mathbb{Q}$, false; if $a=0$ then $3=2b^2$, impossible since $3/2$ is not a rational square. So $x^2-3$ stays irreducible over $K$, and adjoining $\beta=\sqrt3$ gives

In $E$, $f(x) = (x-\sqrt2)(x+\sqrt2)(x-\sqrt3)(x+\sqrt3)$ splits completely, and $E$ is generated over $\mathbb{Q}$ by the four roots $\pm\sqrt2,\pm\sqrt3$, so $E$ is indeed the splitting field. Note $[E:\mathbb{Q}] = 4 < 4! = 24$: the bound $n!$ is far from tight here, because each irreducible factor is only quadratic rather than the full quartic being irreducible.

Example 2: $f(x) = x^3 + x + 1$ over $\mathbb{F}_2$. Working in $\mathbb{F}_2 = \{0,1\}$: check for roots. $f(0) = 0+0+1 = 1 \ne 0$, $f(1) = 1+1+1 = 1 \ne 0$ (all arithmetic mod 2). So $f$ has no root in $\mathbb{F}_2$, and since $\deg f = 3$, having no root forces $f$ to be irreducible over $\mathbb{F}_2$ (a cubic with no linear factor cannot factor as (linear)(quadratic), and it cannot factor into three linear terms without having a root). Adjoin a root $\alpha$ of $f$: form $K = \mathbb{F}_2[x]/(f(x))$, a field since $f$ is irreducible, with $[K:\mathbb{F}_2] = 3$, so $|K| = 2^3 = 8$. We must check whether $f$ splits completely already in $K$, i.e. whether the other two roots of $f$ lie in $K = \mathbb{F}_2(\alpha)$. Perform polynomial division of $f(x)$ by $(x-\alpha) = (x+\alpha)$ (characteristic 2) in $K[x]$:

Let's verify by direct expansion instead: since $\alpha^3 + \alpha + 1 = 0$ in $K$, we have $\alpha^3 = -\alpha - 1 = \alpha+1$ (char 2). We seek $g(x)$ with $f(x) = (x-\alpha)g(x)$; write $g(x) = x^2+bx+c$. Expanding:

Matching with $x^3 + 0\cdot x^2 + x + 1$ (coefficients in char 2, so $-$ is $+$):

Consistency requires $\alpha^{-1} = 1+\alpha^2$, i.e. $\alpha(1+\alpha^2) = 1$, i.e. $\alpha + \alpha^3 = 1$. Since $\alpha^3 = \alpha+1$, this gives $\alpha + \alpha + 1 = 1$, true (char 2, $\alpha+\alpha=0$). So consistent, and

Now check whether $g$ has roots in $K$ itself (it might factor further there, or need a bigger field). Since $|K|=8$, every nonzero element of $K$ satisfies $\beta^7=1$, and $K^\times$ is cyclic of order 7. The remaining two roots of $f$ are $\alpha^2$ and $\alpha^4$ — this is a general fact for irreducible polynomials over finite fields: the roots of an irreducible polynomial of degree $d$ over $\mathbb{F}_p$ are $\alpha, \alpha^p, \alpha^{p^2}, \dots, \alpha^{p^{d-1}}$ (the Frobenius orbit). Here $p=2$, $d=3$, so the roots are $\alpha, \alpha^2, \alpha^4$, all of which already lie in $K$ since $K$ is closed under multiplication. Indeed substituting confirms $g(\alpha^2) = \alpha^4 + \alpha\cdot\alpha^2 + 1+\alpha^2 = \alpha^4+\alpha^3+\alpha^2+1$; using $\alpha^3=\alpha+1$: $\alpha^4 = \alpha\cdot\alpha^3 = \alpha(\alpha+1) = \alpha^2+\alpha$, so

in characteristic 2, confirming $\alpha^2$ is a root of $g$, hence of $f$. So $f$ splits completely already inside $K$:

This matches the general theorem: $3 = 3! /2$, well below the crude bound $3! = 6$, since for finite fields all roots of an irreducible polynomial are automatically Frobenius conjugates living in the same extension — a phenomenon special to finite fields that will be explained fully once Normal Extensions are introduced.