Exercise 10.20

Show that if α ∈ 𝔽 q has trace zero, then α = β − β p for some β ∈ 𝔽 q .

Answers

Proof. Here q = p n . Consider first the map

tr { 𝔽 p n → 𝔽 p α ↦ tr ( α ) = α + α p + α p 2 + ⋯ + α p n − 1

This makes sense, since by Proposition 10.3.1(a), tr ( α ) ∈ 𝔽 p for all α ∈ 𝔽 p n . Moreover, parts (b),(c) of this proposition show that tr is 𝔽 p -linear, and by part (d) that tr is surjective (onto): Im ( tr ) = 𝔽 p .

The rank theorem gives

dim 𝔽 p Im ( tr ) = dim 𝔽 p 𝔽 p n − dim 𝔽 p ker ⁡ ( tr ) ,

thus

dim 𝔽 p ker ⁡ ( tr ) = n − 1 .

Consider now

T { 𝔽 p n → 𝔽 p n β ↦ β − β p

T is a 𝔽 p -linear map: for a , b ∈ 𝔽 p , and α , β ∈ 𝔽 p n , using a p = a , b p = b ,

T ( aα + bγ ) = aα + bγ − ( a p α p + b p β p ) = a ( α − α p ) + b ( β − β p ) = aT ( α ) + bT ( β ) .

If γ = T ( β ) = β − β p is in Im ( T ) , then

tr ( γ ) = tr ( β ) − tr ( β p ) = ( β + β p + β p 2 + ⋯ + β p n − 1 ) − ( β p + β p 2 + β p 3 + ⋯ + β p n ) = β − β p n = 0 .

This proves that

Im ( T ) ⊂ ker ⁡ ( tr ) .

Moreover,

β ∈ ker ⁡ ( T ) ⟺ β = β p ⟺ β ∈ 𝔽 p ,

so that ker ⁡ ( T ) = 𝔽 p .

Using newly the rank theorem on T , we obtain

dim 𝔽 p Im ( T ) = dim 𝔽 p 𝔽 p n − dim 𝔽 p ker ⁡ ( T ) = n − 1 .

From Im ( T ) ⊂ ker ⁡ ( tr ) , where dim 𝔽 p Im ( T ) = dim 𝔽 p ker ⁡ ( tr ) = n − 1 , we deduce

Im ( T ) = ker ⁡ ( tr ) .

To conclude, if α ∈ 𝔽 q has trace zero, then α ∈ Im ( T ) , i.e. α = β − β p for some β ∈ 𝔽 q . □

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2022-07-19 00:00
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