# Galois Theory [Lecture notes] by Sergey Shpectorov By Sergey Shpectorov

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Extra resources for Galois Theory [Lecture notes]

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Now suppose that n > 1. 9 that L is the splitting field for some f ∈ K[x]. Since L > K and since L is generated by the roots of f , there exists some 44 root u with u ∈ K. Let p ∈ K[x] be the minimal polynomial of u over K. 4), the set of roots of p, say R, coincides with the orbit of u under the action of G = Gal(L/K). Notice that p divides f , since f (u) = 0. Thus, the roots of p are also roots of f . Since f splits in L (that is, it has no new roots in any extension of L), so also does p. By separability, p = minu,K has no multiple roots, which means that p has exactly deg (f ) distinct roots; that is, |R| = deg (p).

5) is trivial. Hence every automorphism of L stabilizes every √ subfield of L. In particular, the three intermediate subfields Q( 2), √ Q(i), and Q( 2i) are Galois extensions of Q. Furthermore, in each of the three cases Gal(M/Q) ∼ = G/Φ(M ) is of order two. √ 2. Let now L = Q( 3 2, ζ) be the splitting field of x3 − 2. In this case G = Gal(L/Q) ∼ = S3 is nonabelian, hence conjugation is a nontrivial operation. The subgroup β has index two in G and hence, by a wellknown fact from the group theory, it is normal in G.

Also, α √ fixes all of Q( 2), hence the latter field is Ψ( α ). Similarly, Q(i) = √ Ψ( β ). It remains to determine Ψ( αβ ). Notice that 2i ∈ L and √ √ √ √ √ αβ( 2i) = α(− 2i) = − 2(−i) = 2i. Clearly, 2i ∈ Q and so √ Q( 2i) = Ψ( αβ ). 2. Here is a slightly more complicated example. Let L be the splitting field of x3 − 2. 5 (2) we determined that Gal(L/Q) √ √ √ 2πi induces on the set R = { 3 2, 3 2ζ, 3 2ζ 2 }, where ζ = e 3 , the full group Sym(R) ∼ = S3 . √ The group Sym(R) can be generated by two elements: α fixes 3 2 and interchanges the other two elements of R (α is induced by the √ complex conjugation); β induces a 3-cycle on R, namely, β( 3 2) = √ √ √ √ √ 3 2ζ, β( 3 2ζ) = 3 2ζ 2 , and β( 3 2ζ 2 ) = 3 2.