By K.G. Ramanathan

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**Example text**

Then α + λ1 β, α + λ2 β are in k(γ). Thus (λ1 − λ2 ) ∈ k(γ). Hence β ∈ k(γ) because λ1 − λ2 ∈ k. This means that α ∈ k(γ). 2. Algebraic extension fields 38 Therefore k(α, β) ⊂ k(γ) ⊂ k(α, β). Hence every subfield of K, generated by 2 and hence by a finite number of elements is simple. Let K be a maximal subfield of K/k which is simple. (This exists since K/k has only finitely many intermediary fields). Let K0 = k(ω). Let β ∈ K and β K0 . Then k(γ) = k(ω, β) ⊂ K and K0 ⊂ k(γ) contradicting maximality of K0 .

Xm ) = o. ¯ = 0, the Dx ¯ i would have to satisfy the infinity Therefore, since Do of equations m ∂f ¯ Dxi 0 = fD + ∂xi i=1 for every f in Y . Conversely, suppose u1 , . . um are m elements in T satisfying m D f + i=1 ∂f ui = 0 ∂xi for every f in Y . For any ϕ in T define D¯ by m ¯ = ϕD + f D + Dϕ i=1 ∂ϕ ¯ Dxi ∂xi ¯ i = ui . Then clearly D¯ is a derivation of T and it coincides where Dx with D on k.

Zi+1 but in zi+2 , . . , zn+1 . Then ϕ(x1 , . . , zt , . . , xi+1 ) is irreducible over k(x1 , . . , xt−1 , xt+1 , . . , xi+1 ) for every t, 1 ≤ t ≤ i + 1. At least for one t, ϕ(z1 , . . , zt , . . , zi+1 ) is a separable polynomial in zt over k(z1 , . . , zt−1 , . . , zi+1 ). For, if it is inseparable in every zt , then p p ϕ(z1 , . . , zi+1 ) ∈ k[z1 , . . , zi+1 ] and since k is perfect, this will mean that ϕ(z1 , . . , zi+1 ) is the pth power of a polynomial in k[z1 , . . , zi+1 ] which contradicts irreducibility of ϕ.

### Lectures on the Algebraic Theory of Fields by K.G. Ramanathan

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