Show that Q[x]/⟨x3−2⟩=[a+bα+cα2:a,b,c∈Q,α3=2] is not a Galois extension of Q. In particular, show that every automorphism of Q[x]/⟨x3−2⟩ that fixes Q also fixes all of Q[x]/⟨x3−2⟩. To do this, we’ll use the following approach. Let φ be an automorphism of Q[x]/⟨x3−2⟩ that fixes Q, i.e. that satisfies φ(a)=a ∀a∈Q. We want to show that φ is necessarily the identity automorphism, namely the function e, given by e(β)=β ∀β∈Q[x]/⟨x3−2⟩.

Part a. We want to show that φ=e, namely that φ(β)=β for all β of the form a+bα+cα2 where a,b,c∈Q and α3=2. Show that to do this, it suffices to show that φ(α)=α.

Thoughts: So this is what I have so far, and I wasn’t sure if I had to actually show that φ(α)=α. Let K = Q[x]/⟨x3−2⟩ and Let F = Q. Let φ be any automorphism of K that fixes F. φ is defined to be φ:K→K such that ∀β, β=a+bα+cα2, where a,b,c∈F and α3=2. Show φ(β)=β. φ(β)=φ(a+bα+cα2)=φ(a)+φ(bα)+φ(cα2)=a+bφ(α)+cφ(α2) (since a,b,c are in F, which is the fixed field). I’m not sure if I actually have to show that φ(α)=α, because then φ(β)=β, which for this part of the problem is what we’re trying to show, I think. Any help would be greatly appreciated.

Part b: Show that [φ(α)]3=2. Once we have this then we can conclude that β3=2 has only one solution β∈K.

Thoughts: I already showed that [φ(α)]3=2 by using that fact that since it’s an automorphism, so then we can bring the power of 3 into the automorphism so that it’s actually: φ(α3)=φ(2)=2 since 2∈F, which is the fixed field. Then this part is done.

Part c: To show that there is only one cube root of 2 in K, factor β3−2=β3−α3=(β−α)(β2+αβ+α2). Conclude that it suffices to show that β2+αβ+α2≠0 for all β∈K.

Thoughts: So I wrote the factoring out, and concluded that the only way for that to happen was if β−α=0, implying that β=α or β2+αβ+α2=0 for all β∈K. So then this part is done.

Part d: By way of contradiction, show that if ∃β∈K such that β2+αβ+α2=0, then writing β as β=a+bα+cα2, we necessarily have

0=a2+2c(1+2b),0=a(1+2b)+2c2,0=1+b+b2+2ac

Show that there are no rational numbers a, b, and c which satisfy these equations. (which this ends the entire proof cuz it shows that φ=e)

Thoughts: I understand what we’re supposed to do for this part of the problem, but I’m not sure how to start showing that there are no rational numbers which satisfy these equations.

Thanks, in advance, for the help. Sorry those three equations aren’t centered, I’m not entirely sure how to do that in Latex quite yet, and what I did try didn’t work.

**Answer**

I think it’d be way simpler to show that quadratic has no roots in Q(3√2) by analyzing its discriminant:

Δ=α2−4α2=−3α2

But α=3√2 , so 0>−3α2∈R and from here the quadratic has no real roots, and this is enough as K is a real field…

**Attribution***Source : Link , Question Author : Community , Answer Author : DonAntonio*