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Extra info for Honors Algebra 4, Edition: version 13 Apr 2009
If 1, θ, · · · , θn−1 were not linearly independent, then b0 + b1 θ + · + bn−1 θn−1 = 0 in K with b0 , b1 , ·, bn−1 , not all 0. This in turn implies p(x) | b0 + b1 x + · · · + bn−1 xn−1 which is impossible, since deg p(x) = n. This proves that 1, θ, θ2 , · · · , θn−1 are a basis for K over F . Let a(θ), b(θ) ∈ K be two polynomials of degree < n in θ. Then addition in K is defined simply by usual polynomial addition and multiplication in K is defined by a(θ)b(θ) = r(θ) where r(θ) is the residue (degree less than n) of a(θ)b(θ) modulo (p(x)).
1. A polynomial over F is called separable if it has no multiple roots (in its splitting field) in its splitting field. A polynomial which is not separable is called inseparable. Example 37. x2 − 2 is separable over Q and (x2 − 3)5 is not separable over Q. 58 Example 38. x2 − t ∈ F2 (t)[x] over the field F2 (t) of rational functions in t is irreducible, but is not separable. (x − √ t)2 = x2 − t since F2 (t) is a field of characteristic 2. So it is not separable. 2. The derivative of the polynomial f (x) = an xn + an−1 xn−1 · · · + a1 x + a0 ∈ F [x] is defined to be Dx f (x) = nan xn−1 + (n − 1)an−1 xn−1 · · · + 2a2 x + a1 ∈ F [x].
From K1 K2 = K1 (β1 , β2 , · · · , βm ), we also see β1 , · · · , βm span K1 K2 over K1 . Thus [K1 K2 : K1 ] ≤ m = [K2 : F ] with equality iff these elements are linearly independent over K1 . Since [K1 K2 : F ] = [K1 K2 : K1 ] · [K1 : F ] this proves the proposition. 16. If we assume gcd(n, m) = (n, m) = 1 in the above proposition, then [K1 K2 : F ] = [K1 : F ][K2 : F ]. Proof. 8 we know [K1 K2 : F ] is divisible by both n and m, and so divisible by nm because (n, m) = 1. By the proposition the equality follows.