Splitting Fields

Central Question

What is the smallest field in which a polynomial reveals all of its roots?

Why This Matters

One root may not reveal the entire polynomial. Splitting fields supply the setting in which the relationships among every root can be studied together.

Mathematical Notes

One root versus all roots

Inside a chosen algebraic closure of $F$, the splitting field of $f\in F[x]$ is the field generated by all its roots. It is the smallest subfield there over which $f$ factors into linear factors. Abstractly, it is unique up to an $F$-isomorphism, rather than literally independent of an ambient choice.

For $x^2-2$, adjoining $\sqrt2$ also provides $-\sqrt2$, so the splitting field is $\mathbb Q(\sqrt2)$.

For $x^3-2$, let $\alpha=\sqrt[3]2\in\mathbb R$ and $\omega=e^{2\pi i/3}$. The roots are $\alpha,\alpha\omega,\alpha\omega^2$. The real field $\mathbb Q(\alpha)$ misses the nonreal roots. The splitting field is

The cubic is Eisenstein at 2, giving degree three. The equation $\omega^2+\omega+1=0$ gives degree at most two over $\mathbb Q(\alpha)$, and nonreality makes that degree exactly two. A rational basis is $1,\alpha,\alpha^2,\omega,\alpha\omega,\alpha^2\omega$.

A complementary example is $x^4-2$. Adjoining the real root $\beta=\sqrt[4]2$ does not supply $i\beta$. Its splitting field $\mathbb Q(\beta,i)$ has degree eight, because the real degree-four field acquires an additional quadratic generator.

The two hypotheses that make symmetries complete

Repeated roots of a polynomial are detected by $\gcd(f,f')\ne1$. In characteristic zero, irreducible polynomials are automatically separable: their derivatives are nonzero of smaller degree, so cannot share a nonconstant factor with them.

In characteristic $p$, this can fail. Over $F=\mathbb F_p(t)$, $x^p-t$ is irreducible (Eisenstein at $t$ in $\mathbb F_p[t]$), but has derivative zero. Adjoining $\alpha$ with $\alpha^p=t$ gives a degree-$p$ extension with only one distinct conjugate. It is normal but not separable, and its automorphism group is trivial. Normality alone is not enough over general fields.

Learning Prompt

Motivate splitting fields rather than beginning with the definition. Given a polynomial over $F$, ask what smallest field we need so that it factors completely into linear factors. Work through $x^2-2$, $x^3-2$, and one carefully chosen example where adjoining one root does not automatically give all roots.

Ideas

Examples

Questions

Connections

• Algebraic Numbers and Minimal Polynomials
• Tower Extensions
• My First Genuine Galois Group
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, §§9.1–9.3; §§17.3–17.6. Publisher and edition details.