Field theory MOC

Splitting field

Let 𝐾 be a field and 𝑓(𝑥) ∈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)=𝑐𝑑∏𝑖=1(𝑥−𝛼𝑖)

splits in 𝐹[𝑥], and 𝐹 =𝐾(𝛼1,…,𝛼𝑑). It is unique up to isomorphism with¹

[𝐹:𝐾]≤(deg⁡𝑓)!

Proof

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

𝐾(𝛼):=𝐾[𝑥]⟨𝑞(𝑥)⟩:𝐾

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

[𝐹:𝐾]=[𝐹:𝐾(𝛼)][𝐾(𝛼):𝐾]≤(deg⁡𝑓)(deg⁡𝑓−1)!=(deg⁡𝑓)!

as claimed.

Now suppose that 𝜓 :𝐾′ →𝐾 is an isomorphism of fields, and let ℎ(𝑥) ∈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∈𝖥𝗅𝖽𝐾′(𝐹′,――𝐾).

where 𝑖 ↾𝐾′ =𝜄 ↾𝐾′. Since 𝐹′ =𝐾′(𝛼′𝑖)𝑑𝑖=1 where {𝛼′𝑖}𝑑𝑖=1 ⊆𝐹′ are the roots of ℎ(𝑥), it follows

𝑖(𝐹′)=𝐾(𝛼𝑖)𝑑𝑖=1≤――𝐾

where {𝛼𝑖}𝑑𝑖=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

1. Suppose 𝐹 is the splitting field of 𝑓(𝑥) ∈𝐾[𝑥], and that 𝑔(𝑥) ∈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)=𝑐𝑑∏𝑖=1(𝑥−𝛼𝑖)

as above. Then for some subset of indices 𝐼 ∈ℕ𝑑 and some 𝐶 ∈𝐾,

ℎ(𝑥)=∏𝑖∈𝐼(𝑥−𝛼𝑖).

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

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Footnotes

1. 2009. Algebra: Chapter 0, §VII.4.1, pp. 429–430 ↩