Wednesday, 21 May 2014

Prove that the Galois group of xn1 is abelian over the rationals




If p(x)=xn1, prove that the Galois group of p(x) over the field of rational numbers is abelian.





Here's what I have so far.



Denote the Galois group G(K,Q), where K is the splitting field for p(x) over Q.



By setting xn1=0, we find the nth roots of unity ω,ω2,,ωn=1, where ω=e2πi/n. Then, the splitting field K=Q(ω).



By a theorem, K is a normal extension of Q. We now wish to examine G(K,Q)=G(Q(ω),Q). By defintion, this is the group of automorphisms of Q(ω) that keep every element of Q fixed. In other words, if aQ, σ(a)=a for all σG(Q(ω),Q).



Suppose σ,τG(Q(ω),Q). We know the group structure is given by composing automorphisms. To show that this group is abelian, we need to show that (στ)(b)=(τσ)(b) for all bQ(ω).




We know that all elements of Q are fixed. That is, if aQ,



(στ)(a)=σ(τ(a))=σ(a)=a



(τσ)(a)=τ(σ(a))=τ(a)=a



Now, consider σ(ω)=σ(e2πi/n). We have (σ(e2πi/n))n=σ(e2πi)=σ(1)=1. This implies that σ(e2πi/n)= an nth root of unity. Thus, σ just permutes roots of unity.



This is where I am confused. If the automorphism permutes roots of unity, it doesn't seem to necessarily be the case that (στ)(ω)=(τσ)(ω).




Please let me know where to go from here (or where I've gone wrong in my argument). Thanks.


Answer



Your argument up to the point where you're stuck seems completely correct to me.



To finish it, suppose that σ(ω)=ωj and that τ(ω)=ωk for some j,kN. Then just observe that
(στ)(ω)=σ(τ(ω))=σ(ωk)=σ(ω)k=(ωj)k=ωjk
and similarly
(τσ)(ω)=τ(σ(ω))=σ(ωj)=τ(ω)j=(ωk)j=ωjk


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