Splitting field

Let $𝐾$ be a field and $𝑓 (𝑥) \in 𝐾 [𝑥]$ be a polynomial of degree $𝑑$ . The splitting field $𝐹$ for $𝑓 (𝑥)$ over $𝐾$ is an extension $𝐹 : 𝐾$ such that

𝑓 (𝑥) = 𝑐 𝑑 \prod 𝑖 = 1 (𝑥 - 𝛼 𝑖)

splits in $𝐹 [𝑥]$ , and $𝐹 = 𝐾 (𝛼 1, \dots, 𝛼 𝑑)$ . It is unique up to isomorphism with¹

[𝐹 : 𝐾] \leq (d e g 𝑓)!

Proof

We construct the splitting field by iterating the process of adjoining a root to a field. Let $𝑓 (𝑥) \in 𝐾 [𝑥]$ Suppose the statement and bound have been proven for polynomials $𝑔 (𝑥) \in 𝐾 [𝑥]$ with $d e g 𝑔 < d e g 𝑓$ . Let $𝑞 (𝑥) \in 𝐾 [𝑥]$ be an irreducible factor of $𝑓 (𝑥)$ , so that

𝐾 (𝛼) : = 𝐾 [𝑥] ⟨ 𝑞 (𝑥) ⟩ : 𝐾

is an extension of degree $d e g 𝑞 \leq d e g 𝑓$ , in which $𝑓 (𝑥)$ has a linear factor $(𝑥 - 𝛼)$ , so letting $𝑔 (𝑥) = 𝑓 (𝑥) / (𝑥 - 𝛼)$ gives $d e g 𝑔 = d e g 𝑓 - 1$ so the splitting field $𝐹$ of $𝑔 (𝑥)$ over $𝐾 (𝛼)$ exists with $[𝐹 : 𝐾 (𝛼)] \leq (d e g 𝑓 - 1)!$ . It follows that $𝐹$ is a splitting field for $𝑓 (𝑥)$ over $𝐾$ and

[𝐹 : 𝐾] = [𝐹 : 𝐾 (𝛼)] [𝐾 (𝛼) : 𝐾] \leq (d e g 𝑓) (d e g 𝑓 - 1)! = (d e g 𝑓)!

as claimed.

Now suppose that $𝜓 : 𝐾' \to 𝐾$ is an isomorphism of fields, and let $ℎ (𝑥) \in 𝐾' [𝑥]$ such that $𝑓 (𝑥) = 𝜓 (ℎ (𝑥))$ , and let $𝐹'$ be a splitting field for $ℎ (𝑥)$ over $𝐾'$ . Consider the composite extension $―― 𝐾 : 𝐾 ≅ 𝐾'$ . Since $𝐹 : 𝐾'$ is algebraic, by Embedding an algebraic extension into an algebraically closed field there exists a morphism

𝑖 \in 𝖥 𝗅 𝖽 𝐾' (𝐹', ―― 𝐾) .

where $𝑖 ↾ 𝐾' = 𝜄 ↾ 𝐾'$ . Since $𝐹' = 𝐾' (𝛼' 𝑖) 𝑑 𝑖 = 1$ where ${𝛼' 𝑖} 𝑑 𝑖 = 1 \subseteq 𝐹'$ are the roots of $ℎ (𝑥)$ , it follows

𝑖 (𝐹') = 𝐾 (𝛼 𝑖) 𝑑 𝑖 = 1 \leq ―― 𝐾

where ${𝛼 𝑖} 𝑑 𝑖 = 1$ are the roots of $𝑖 (𝑔 (𝑥)) = 𝜄 (𝑔 (𝑥)) = 𝑓 (𝑥)$ in $―― 𝐾$ , so $𝑖 (𝐹')$ is independent of the chosen morphism $𝑖$ and the splitting field $𝐹'$ .

Properties

Suppose $𝐹$ is the splitting field of $𝑓 (𝑥) \in 𝐾 [𝑥]$ , and that $𝑔 (𝑥) \in 𝐾 [𝑥]$ is a factor of $𝑓 (𝑥)$ . Then $𝐹$ contains a unique subfield which is the splitting field for $𝑔 (𝑥)$ .

Proof of 1

Let

𝑓 (𝑥) = 𝑐 𝑑 \prod 𝑖 = 1 (𝑥 - 𝛼 𝑖)

as above. Then for some subset of indices $𝐼 \in ℕ 𝑑$ and some $𝐶 \in 𝐾$ ,

ℎ (𝑥) = \prod 𝑖 \in 𝐼 (𝑥 - 𝛼 𝑖) .

Then $𝐾 (𝛼 𝑖) 𝑖 \in 𝐼$ is the splitting field of $ℎ (𝑥)$ , and indeed is the only such field contained in $𝐹$ since $𝛼 𝑖$ are the only roots of $ℎ (𝑥)$ in $𝐹$ , proving ^P1

#state/develop | #lang/en | #SemBr

2009. Algebra: Chapter 0, §VII.4.1, pp. 429–430 ↩

Splitting field

Properties

Footnotes