The extension 𝔽(α)

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 02 February 2012

The extension 𝔽(α)

Let 𝔽 be a field and let 𝔼 be an extension of 𝔽.

Let 𝔼 be an extension of 𝔽 and let α𝔼. Then 𝔽(α) = 𝔽[α] 𝔽[x] ( mα,𝔽(x) ) .

(Theorem of the primitive element) If 𝔼 over 𝔽 is a finite separable extension of 𝔽 then there exists an element θ𝔼 such that 𝔼=𝔽(θ).

Let 𝔼 be a finite extension of 𝔽. Then 𝔼=𝔽(θ) for an element θ𝔼 if and only if there are only a finite number of intermediate fields 𝔼𝕂𝔽.

  1. All roots of mα,𝔽 (x) have the same multiplicity.
  2. If char(𝔽)=0 or if 𝔽 is finite then all elements of 𝔽_ are separable.

Proof.
  1. Let α and β be roots of mα,𝔽 (x). The isomorphism σ: 𝔽(α) 𝔽[x] ( mα ,𝔽 (x)) 𝔽(β) α x β extends to an automorphism σ: 𝔽_ 𝔽_ α β which is the identity on 𝔽. Thus σ( mα,𝔽 (x)) = mα,𝔽 (x) and σ( x-α k )= x-β k . So if α is an root of mα,𝔽 (x) of multiplicity k then x-β k is a factor of mα,𝔽 (x) in 𝔽_[x]. So β has multiplicity k.
  2. Let mα,𝔽 be the minimal polynomial of α𝔽_ over 𝔽. Then mα,𝔽 (x) = (x-α)n(x) and d dx mα,𝔽 (x) = (x-α) dn dx + n(x). So mα,𝔽 (x) has α as a multiple root if and only if mα,𝔽' (α)=0. Since mα,𝔽 (x) is the minimal polynomial which has α as a root, mα,𝔽 (x) has a multiple root if and only if mα,𝔽' (α)=0. If mα,𝔽 (x) = i=0 r aixi then mα,𝔽' (x) = i=0 r iai xi-1. So mα,𝔽' (x)=0 if and only if p divides iai for all 1ir, where p=char(𝔽) . Since p divides iai if and only if p divides ai it follows that mα,𝔽 (x) has no multiple root when char(𝔽)=0. If char(𝔽)=p then mα,𝔽 (x) has a multiple root if and only if mα,𝔽 (x) = g(xp) for some polynomial g(x)𝔽[x].

(Theorem of the primitive element) If 𝔼 over 𝔽 is a finite separable extension of 𝔽 then there exists an element θ𝔼 such that 𝔼=𝔽(θ).

Proof.
Case 2: 𝔽 is infinite.
Since 𝔼 is a finite extension of 𝔽, 𝔼=𝔽 γ1γ2 ...γk for some α1 ... αk in 𝔼. The proof is by induction on k. It is sufficient to show that if 𝔼=𝔽αβ then there is an element θ𝔼 such that 𝔼=𝔽(θ).
Let α= α1 α2 ... αn , and β= β1 β2 ... βm be the roots of mα,𝔽 (x) and mβ,𝔽 (x), respectively. Since 𝔽 is infinite there exists a𝔽 such that a αi-α βj-β , for any   j   such that   βjβ. Let θ=α+aβ. Let p(x) = mα,𝔽 ( θ-ax ) 𝔽(θ) [x]. Then p(β) = mα,𝔽 ( α+aβ -aβ )=0, and pβj = mα,𝔽 ( α+aβ -aβj )0, for all   j1, since α+a (β-βj) αi for any i. Now mβ,𝔽(θ) (x) divides mβ,𝔽 (x) and mβ,𝔽(θ) (x) divides p(x). Since the only factor in common between p(x) and mβ,𝔽 (x) is (x-β), mβ,𝔽(θ) (x) = (x-β). So β𝔽(θ) and α=θ-aβ 𝔽(θ). So 𝔼=𝔽 αβ = 𝔽(θ).

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