Splitting Fields

A polynomial dreams of factoring completely. Over the rationals, x^2 - 2 is stuck — it will not break into linear pieces because \sqrt 2 is missing. But climb into the right extension and it relaxes into (x - \sqrt 2)(x + \sqrt 2). The algebraic question this page answers is: what is the smallest field in which a given polynomial factors all the way down into linear factors? That field is its splitting field — the birthplace of every root at once, and the natural stage on which Galois theory is played out.

Let f(x) \in F[x] be a nonconstant polynomial. A splitting field of f over F is a field E \supseteq F such that:

It splits f. Over E, f factors into linear factors f(x) = c(x - \alpha_1)\cdots(x - \alpha_n).
It is smallest. E = F(\alpha_1, \dots, \alpha_n) is generated over F by exactly those roots — nothing to spare.
It exists and is unique. A splitting field exists for every f, and any two are isomorphic (fixing F).
It is not too big. Its degree satisfies [E : F] \le n! for \deg f = n.

The easy cases

For a quadratic you adjoin one root and the other comes free, because the two roots are \pm partners.

x^2 - 2 over \mathbb{Q}: the roots are \pm\sqrt 2, both living in \mathbb{Q}(\sqrt 2). Splitting field \mathbb{Q}(\sqrt 2), degree 2.
x^2 + 1 over \mathbb{Q}: roots \pm i, splitting field \mathbb{Q}(i), degree 2.
x^2 + x + 1 over \mathbb{Q}: roots the two primitive cube roots of unity \omega, \omega^2, splitting field \mathbb{Q}(\omega), degree 2.

The pattern feels reassuring: adjoin a root, get the splitting field. But that pattern breaks the moment the degree exceeds 2, and the break is the whole point of this page.

The cautionary star: x³ − 2

Consider f(x) = x^3 - 2 over \mathbb{Q}. It is tempting to say the splitting field is \mathbb{Q}(\sqrt[3]{2}) — after all, \sqrt[3]{2} is a root. But that field is a subfield of the real numbers, and x^3 - 2 has only one real root. The other two are complex:

x^3 - 2 = (x - \sqrt[3]{2})\,(x - \omega\sqrt[3]{2})\,(x - \omega^2\sqrt[3]{2}), \qquad \omega = e^{2\pi i/3}.

The three roots are \sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2} — three points spaced 120^\circ apart on a circle of radius \sqrt[3]{2} \approx 1.26 in the complex plane. To hold all three you must adjoin both \sqrt[3]{2} and \omega. So the splitting field is \mathbb{Q}(\sqrt[3]{2}, \omega), and by the Tower Law its degree is

[\mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q}(\sqrt[3]{2})] \cdot [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 2 \cdot 3 = 6,

not 3. The figure shows the three roots on their circle — the splitting field is the smallest field containing all of these points, not just the real one on the right.

Existence, uniqueness, and the n! ceiling

Two guarantees make splitting fields dependable. Existence: you can always build one, by repeatedly adjoining a root of an irreducible factor (formally, F[x]/(p(x)) manufactures a root of any irreducible p) until f is fully split. Uniqueness: any two splitting fields of the same f over F are isomorphic by an isomorphism fixing F — so it is legitimate to speak of the splitting field.

The degree bound [E : F] \le n! comes from counting: adjoining the first root costs at most n (its minimal polynomial has degree \le n), the next at most n - 1 (one factor is already removed), and so on, giving n \cdot (n-1) \cdots 1 = n!. For x^3 - 2 the ceiling is 3! = 6, which is exactly the degree we found — this cubic saturates the bound.

The classic trap is to adjoin one root and declare victory. \mathbb{Q}(\sqrt[3]{2}) contains the real cube root but lives entirely inside \mathbb{R}, and the other two roots \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2} are genuinely complex — they are not in there. Over \mathbb{Q}(\sqrt[3]{2}) the polynomial only partly factors, x^3 - 2 = (x - \sqrt[3]{2})(x^2 + \sqrt[3]{2}\,x + \sqrt[3]{4}), leaving an irreducible quadratic behind.

The moral: adjoining a single root is not enough unless the extension happens to be normal — the property that when one root of an irreducible polynomial lands in a field, all its roots do. A splitting field is normal by construction, which is exactly why we must adjoin \omega as well and climb all the way to degree 6.

The n! ceiling is the size of the symmetric group S_n permuting the n roots — and that is no coincidence. The symmetries of a splitting field (its Galois group) act by shuffling the roots, so they form a subgroup of S_n, and the degree of the splitting field equals the size of that group. The bound is achieved exactly when the Galois group is the whole of S_n — the "generic" case. For x^3 - 2 the group is all of S_3 (size 6), so the bound is tight; for x^2 - 2 the group is S_2 (size 2 = 2!), also tight. But a polynomial like x^4 + 1 splits in degree 4 \ll 24 = 4!, its group far smaller than S_4.