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Every finite family of nonzero polynomials has a splitting field, obtained from their product

Statement

Let f1,…,fm∈F[x] be nonzero, where m∈N. A splitting field of the product h=∏j=1mfj is a splitting field of the family {f1,…,fm}. Hence every finite family of nonzero polynomials has a splitting field. For m=0, h=1 and the splitting field is F.

Facts & Assumptions

Given: A finite family f1,…,fm of nonzero polynomials over a field F.

[F1]

Nonzero polynomial products over a domain are nonzero (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[F2]

Every nonzero polynomial has a splitting field (Every nonzero polynomial over a field has a splitting field).

[F3]

For every field K, the polynomial ring K[x] is a unique factorisation domain (For every field F, F[x] is a unique factorisation domain).

[F4]

A splitting field is generated by all roots of the polynomial or family that splits there (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · direct
1.1

If m=0, the product is 1, whose empty root set has splitting field F by [F4]. Assume now that m>0. By [F1], h is nonzero, so [F2] gives a splitting field E/F of h.

F1F2F4
2.1

Each fj divides h in E[x]. Since h is a product of linear factors there, unique factorisation in the ring E[x] from [F3] shows that every irreducible factor of fj is linear; hence every fj splits over E.

F3step 1.1
3.1

An element of an extension is a root of h=f1⋯fm exactly when it is a root of at least one fj, because a field has no zero divisors. Thus the roots generating E are precisely the union of the roots of the family, and [F4] makes E its splitting field.

F4step 1.1algebra∎

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