By K.G. Ramanathan

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Then (L : k) is a divisor of (K : k) = n. If d = (L : k) then L has qd elements. Also since K/k has a cyclic galois group, there is one and only one subgroup of a given order d. Hence 4) The number of intermediary fields of K/k is equal to the number of divisors of n. K. Schmidt’s theorem Let K/k be an extension field and x1 , . . , xn+1 any n + 1 elements of 58 K. Let R = k[z1 , . . , zn+1 ] be the ring of polynomials in n + 1 variables over k. Let Y be the ideal in R of polynomials f (z1 , .

Thus φ(x) = f (x p ) = {ψ(x)} p Thus if f (x p ) is reducible, it is the pth power of an irreducible polyp nomial. In this case ai = bi , bi ∈ k, i = 1, . . n. p Conversely if ai = bi , bi ∈ k, i = 1, . . n. then f (x p ) is reducible. Hence f (x p ) is reducible f (x) ∈ k p [x]. Let k now be a field of characteristic p > 2 given by k = Γ(x, y), the field of rational functions in two variables x, y over the prime field Γ if p elements. Consider the polynomial f (z) = z2p + xz p + y, in k[z]. Since x1/p , y1/p do not lie in k, f (z) is irreducible in k.

This is evident. We call L the separable closure of k in K. We had already defined an algebraic element ω to be inseparable if its minimum polynomial has repeated roots. Let us study the nature of irreducible polynomials. Let f (x) = ao + a1 x + · · · + an xn be an irreducible polynomial in k[x]. If it has a root ω ∈ Ω which is repeated, then ω is a root of f 1 (x) = a1 + 2a2 x + · · · + nan xn−1 . Thus f (x)| f 1 (x) which can happen only if iai = o, i = 1, . . , n. 31 Let k have characteristic zero.