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A normal extension generated by finitely many elements is the splitting field of the product of their minimal polynomials

Statement

Let $E / F$ be normal and suppose $E = F (α_{1}, \dots, α_{m})$ for some $m \in N$ . If $p_{j}$ is the minimal polynomial of $α_{j}$ over $F$ , then $E$ is a splitting field over $F$ of $p_{1} p_{2} \dots p_{m} .$ For $m = 0$ , the product is $1$ and the assertion reads $E = F$ .

Facts & Assumptions

Given: A normal extension $E / F$ with the displayed finite generating family.

[F1]

Normality makes the minimal polynomial over $F$ of every element of $E$ split over $E$ (A normal algebraic extension is one in which every minimal polynomial with a root in the extension splits there).

[F2]

Each $α_{j}$ is a root of its minimal polynomial $p_{j}$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

A splitting field is generated over $F$ by all roots of a polynomial 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 generating hypothesis says $E = F$ , and the nonzero constant $1$ has empty root set, so [F3] makes $F$ its splitting field.

1.2

Suppose $m > 0$ . By [F1], every $p_{j}$ splits over $E$ , so their product splits over $E$ . Let $K$ be the subfield of $E$ generated over $F$ by all roots of that product. Then $K \subseteq E$ .

F1F3

2.1

Each generator $α_{j}$ is one of those roots by [F2], so $E = F (α_{1}, \dots, α_{m}) \subseteq K$ . Thus $E = K$ , and [F3] says exactly that $E$ is the splitting field of the product.

F2F3step 1.2∎

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 32 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, Proposition 2.12 (standard reference, not scraped)