# An Introduction to Galois Theory by Andrew Baker

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By Andrew Baker

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2-3. 12 by induction on n. 2-4. Let K a ﬁeld with char K ̸= 2 and suppose that L/K is an extension. If a, b ∈ K are distinct, suppose that u, v ∈ L satisfy u2 = a and v 2 = b. Show that K(u, v) = K(u + v). ] 2-5. Show that [Q(i) : Q] = 2. √ √ 2-6. Show that [Q( 3, i) : Q] = 4. Find the three subﬁelds L Q( 3, i) with [L : Q] = 2 and display their relationship in a diagram, indicating which ones are subﬁelds of R. 2-7. Let ζ5 = e2πi/5 ∈ C. (a) Explain why [Q(ζ5 ) : Q] = 4. (b) Show that cos(2π/5), sin(2π/5) i ∈ Q(ζ5 ).

An extension of the form K(u)/K is called a simple extension of K with generator u. We can extend this to the case of an inﬁnite sequence u1 , . . , 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.

1. 3. 31. Definition. For extensions F/K and L/K, let MonoK (L, F ) denote the set of all monomorphisms L −→ F which ﬁx the elements of K. 32. Remark. We always have AutK (F ) ⊆ MonoK (F, F ) and MonoK (F, F ) is closed under composition but is not always a group since elements are not necessarily invertible. If F/K is ﬁnite, then we do have MonoK (F, F ) = AutK (F ) since every injective K-linear transformation is surjective and so invertible. We will also use the following notation. 33. Definition.

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