Adjoin all roots and prove uniqueness up to base-field isomorphism

Try

Adjoining the real cube root α=∛2 gives a degree-three field Q(α), but x³−2 has two more roots: αω and αω², where ω is a nonreal cube root of unity. The real field Q(α) cannot contain them. A field containing one root is not necessarily a field in which the whole polynomial splits.

A splitting field is generated over the base by all roots of the polynomial and makes it a product of linear factors. The generation requirement excludes unnecessary extra elements.

Predict

Predict how to construct a splitting field without assuming the desired roots already exist. An irreducible factor gives a quotient field containing one new root. After removing that linear factor, repeat with the remaining polynomial. Each step lowers the unsplit degree, so the process must terminate.

Experiment

Trace x³−2 over Q.

Field step	New element	Degree
Q to Q(α)	α³=2	3 by Eisenstein
Q(α) to Q(α,ω)	ω²+ω+1=0	2 because the first field is real
Final field	α,αω,αω² all present	Total degree6

The second degree is not merely assumed from a quadratic equation. The polynomial has no root in the real intermediate field, so it stays irreducible there. The tower law then gives six.

Understand

For a nonconstant polynomial over F, choose an irreducible factor m and construct F[x]/(m). The class of x is a root. Divide out its linear factor in the larger field and continue. Each adjunction is finite, and finitely many steps suffice, so a finite splitting extension exists. The subfield generated by all roots is the splitting field.

Splitting fields are unique up to an isomorphism fixing F, not necessarily equal as sets. The key extension lemma sends a root α of an irreducible polynomial to any chosen root β of the same polynomial, identifying F(α) and F(β) through the quotient F[x]/(m). Starting with a base-field embedding, repeat this step using transported minimal polynomials. In another splitting field the needed roots are available. After all roots are adjoined, the embedding maps the finite root set onto the full root set, so its image generates the target splitting field. This proves surjectivity and uniqueness up to base-field isomorphism.

Now prove finite-field existence. In a splitting field of f=x^(pⁿ)−x over F_p, its derivative is −1, so it has pⁿ distinct roots. Those roots form a field: Frobenius identities preserve sums, differences and products; for a nonzero root a, the equation a^(pⁿ)=a also gives (a⁻¹)^(pⁿ)=a⁻¹. Zero and one are roots. Thus the root set is already the splitting field and has exactly pⁿ elements.

Any field with pⁿ elements is a splitting field of the same polynomial, since every element satisfies that equation and there are exactly pⁿ distinct roots. Splitting-field uniqueness therefore proves uniqueness of the finite field up to F_p-isomorphism. This completes the existence and subfield arguments used in the preceding lesson without assuming a field for every integer size.

Apply

Compare the splitting field of x²−7 with that of x³−2. In the quadratic case, adjoining one root automatically supplies its negative; the cubic example requires an additional nonreal element. Explain why this difference is about all roots, not merely polynomial degree.

When proving two root fields isomorphic, verify that the proposed generator images satisfy the same minimal polynomial. Arbitrary assignments of roots of unrelated equations do not automatically extend to a consistent field map.

Continue

Separability next measures whether roots are distinct. The derivative provides an efficient algebraic test, and positive characteristic supplies examples where irreducibility does not prevent repeated roots.