Prove $\mathbb Q(t)/\mathbb Q(\frac{1-3t+t^3}{t(t-1)})$ is Galois with $G=\mathbb Z/3\mathbb Z$ and $\sigma t = \frac{1}{1-t} $

Question

I just had this question from my algebra finals:

Prove $\mathbb Q(t)/\mathbb Q(\frac{1-3t+t^3}{t(t-1)})$ is Galois with $G = \mathbb Q/3\mathbb Z$ and $\sigma t = \frac{1}{1-t} $ for a generator $\sigma \in G$.

I failed to solve it despite the many attempts and computations. My idea is to assume that $Q(t)$ is the splitting field of a cubic $f$ in $Q(\frac{1-3t+t^3}{t(t-1)})$. Then $G \subset S_3$ must be the $3$-cycle that permutes the three roots of the cubics. Now $Q(t) = Q(\frac{1-3t+t^3}{t(t-1)})/f(x) = Q(\alpha_i)$ so $t$ must be one of the roots $\alpha_1$. And since $\sigma$ permutes the three roots, I can get that $\alpha_2 = \sigma t = \frac{1}{1-t}$ and $\alpha_3 = \sigma^2 t = \sigma \frac{1}{1-t} = \frac{1}{1-\frac{1}{1-t}} $. And thus $f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$. But after factoring out and doing the computation, I realized $f$ doesn't have the coefficients in form of $\frac{1-3t+t^3}{t(t-1)}$. So I am wondering if there is any part of my reason deeply flawed that leads to the mistake. Any helps will be much appreciated.

Answer 1

Let $$u=\frac{1-3t+t^3} {t^2-t}$$ so that $$t^3-ut^2+(u-3)t+1=0$$ and hence $t$ is a root of the polynomial $$f(x) =x^3-ux^2+(u-3)x+1\in\mathbb {Q} (u)[x] $$ and $$[\mathbb {Q} (t) :\mathbb{Q} (u)] \leq \operatorname {deg} (f(x)) =3$$ Next note that if $t$ is a root of $f(x) $ then $1/(1-t)$ is also a root of $f(x)$. We have $$f(1/(1-t))=\frac{1-u(1-t)+(u-3)(1-t)^2+(1-t)^3}{(1-t)^3}=-\frac{1+(u-3)t-ut^2+t^3}{(1-t)^3}=0$$

The other remaining root is then $$u-t-\frac {1}{1-t}=\frac{t-1}{t}$$ Thus all the roots of $f$ are distinct and are rational functions of $t$. Then $\mathbb {Q} (t) $ is the splitting field of separable polynomial $f(x) \in\mathbb {Q} (u) [x] $ and therefore the extension $\mathbb{Q} (t) /\mathbb{Q} (u) $ is Galois.

Next we consider the automorphism $\sigma $ given by $\sigma(t) =1/(1-t)$. We have $$\sigma^2(t)=\sigma(1/(1-t))=\dfrac{1}{1-\dfrac{1}{1-t}}=\frac{t-1}{t} $$ and $$\sigma^3(t)=\sigma((t-1)/t)=\dfrac{1}{1-\dfrac{t-1}{t}}=t$$ It follows that group $G=\{I=\sigma^3,\sigma,\sigma^2\}$ where $I$ is identity map on $\mathbb {Q} (t) $ has generator $\sigma$.

Let $K$ be the fixed field of $G$. Since $u$ is sum of roots of $f(x)$ it is fixed by $\sigma$ and thus by $G$ and hence $\mathbb{Q } (u)\subseteq K\subseteq\mathbb{Q} (t) $. Since $[\mathbb {Q} (t) :K] =|G|=3$ it follows that $[\mathbb {Q} (t) :\mathbb {Q} (u)] \geq 3$. We had remarked earlier that degree of this extension does not exceed $3$ and hence we get $[\mathbb {Q} (t) :\mathbb {Q} (u)] =3$ and $K=\mathbb {Q}(u)$.

It now follows that $G=\text{Gal} (\mathbb{Q} (t) /\mathbb {Q} (u)) $.