# Download e-book for kindle: An Introduction to Galois Theory by Andrew Baker

By Andrew Baker

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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.

Example. 57. 68. Definition. Let L/K be a ﬁnite extension. The separable degree of L over K is (L : K) = | MonoK (L, K)|. 69. Lemma. For a finite simple extension K(u)/K, (K(u) : K) = | Roots(minpolyK,u , K)|. If K(u)/K is separable, then [K(u) : K] = (K(u) : K). 50 Proof. 34 applied to the case L = K. 1) K(u1 )/K, K(u1 , u2 )/K(u1 ), · · · , L = K(u1 , . . , uk )/K(u1 , . . , uk−1 ). So we can use the following to compute (L : K) = (K(u1 , . . , uk ) : K). 70. Proposition. Let L/K and M/L be finite extensions.

Ur , . . in F and denote by K(u1 , . . , ur , . ) F the smallest extension ﬁeld of K containing all the elements ur . 1) K(u1 , . . , ur ) = { } f (u1 , . . , ur ) ∈ F : f (X1 , . . , Xr ), g(X1 , . . , Xr ) ∈ K[X1 , . . , Xr ], g(u1 , . . , ur ) ̸= 0 . g(u1 , . . , ur ) Reordering the ui does not change K(u1 , . . , un ). 8. Proposition. Let K(u)/K and K(u, v)/K(u) be simple extensions. Then K(u, v) = K(u)(v) = K(v)(u). More generally, K(u1 , . . , un ) = K(u1 , . . , un−1 )(un ) and this is independent of the order of the sequence u1 , .