By Andrew Baker
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Extra info for An Introduction to Galois Theory
34). Since θ permutes the roots vj , we have θE = θK(v1 , . . , vk ) = K(θ(v1 ), . . , θ(vk )) = K(v1 , . . , vk ) = E. 81. Corollary. Let E/L and L/K be finite extensions. If E/K is normal then E/L is normal. 54 Proof. If E is the splitting ﬁeld of a polynomial f (X) ∈ K[X] over K, then E is the splitting ﬁeld of f (X) over L. These result makes it easy to recognize a normal extension since it is suﬃcient to describe it as a splitting ﬁeld for some polynomial over K. In Chapter 4 we will see that separable normal extensions play a central rˆole in Galois Theory, indeed these are known as Galois extensions.
For n 1, show that each of the roots of f (X) = X n − 1 in C is simple. Solution. We have f ′ (X) = ∂(X n − 1) = nX n−1 , so for any root ζ of f (X), f ′ (ζ) = nζ n−1 ̸= 0. 59. Example. Show that 2i is a multiple root of f (X) = X 4 + 8X 2 + 16. Solution. We have f ′ (X) = 4X 3 +16X. Using Long Division and the Euclidean Algorithm we ﬁnd that gcd(f (X), f ′ (X)) = X 2 + 4, where 2i is also a root of X 2 + 4. Hence 2i is a multiple root of f (X). In fact, X 4 + 8X 2 + 16 = (X 2 + 4)2 , so this is obvious.
So we have [Q( 3 + i) : Q] = 4 and Q( 3 + i) = Q( 3, i). 7. Normal extensions and splitting fields Let K be an algebraic closure for the ﬁeld K and let E/K K/K be a ﬁnite extension. 32, φE = E if and only if φE E. 78. Definition. E/K is normal if φE = E for every φ ∈ MonoK (E, K). 79. Remark. 52 each pair of roots of p(X) is conjugate over K and one can be mapped to the other by a monomorphism K −→ K which must map E into itself. 80. Theorem. A finite extension E/K is normal if and only if it is a splitting field over K for some polynomial f (X) ∈ K[X].