By M. Pavaman Murthy, K. G. Ramanathan, C. S. Seshadri, U. Shukla, R. Sridharan
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Extra resources for Galois theory, Edition: version 8 Sep 2011
M be the roots of f . These are called the m-th roots of unity. They form an abelian group under multiplication. Let t be the exponent of this abelian group. 33, Chapter 1). Since X t − 1 has only t roots in its splitting field over K, we see that t = m. This means that the ρi (1 ≤ i ≤ m) form a cyclic group order m. Any generator of this group is called a primitive m-th root of unity. If ρ is primitive m-th root of unity, we have f = Xm − 1 = 0≤i≤m−1 47 (X − ρi ), 48 Chapter 5. Applications of Galois Theory and the field L = K(ρ) is clearly the splitting field of f over K.
Let deg f = n. Since 1, α, . . 1), we note that (k[X]/(f ): k) = deg f . We prove the proposition by induction on n = deg f . If n = 1, f is a linear polynomial and k is obviously a splitting field of f . Let us assume n ≥ 2 and that for any polynomial g of degree n − 1 over any field, there exists a finite extension in which g can be written as a product of linear factors. Let f be any polynomial of degree n. Let f1 be an irreducible factor of f . Let K1 = k(α) be an extension of k such that f1 (α) = 0.
If σ is an element of G(K/k), then σ(K1 ) is again a subfield of K containing k and G(K/σ(K1 )) is the subgroup of σG(K/K1 )σ −1 of G(K/k), as is verified easily. Now K1 /k is normal if and only if σ(K1 ) = K1 for 44 Chapter 4. 15, Chapter 3). Now if σ(K1 ) = K1 , σG(K/K1 )σ −1 = G(K/K1 ) for every σ ∈ G(K/k), which shows that G(K/K1 ) is a normal subgroup of G(K/k). Conversely if σG(K/K1 )σ −1 = G(K/K1 ) for every σ ∈ G(K/k), the fixed field of G(K/K1 ) is the same as that of σG(K/K1 )σ −1 for every σ ∈ G(K/K1 ).