Show that $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is a root of the irreducible polynomial $x^3-3x-1$ We have that $E=\mathbb{Q}(a)$, where $a\in \mathbb{C}$ is a root of the irreducible polynomial $x^3-3x-1\in \mathbb{Q}[x]$. 
I want to show that $E/\mathbb{Q}$ is normal. 
I have done the following: 
Let $b\in E$. 
A basis of the extension is $1, a, a^2$. So, $b$ can be written as $$b=q_0+q_1a+q_2a^2$$ 
We have to find the minimal irreducible polynomial of $b$ over $\mathbb{Q}$ and compute the other roots to check if they are in $E$, or not? 
Since $[E:\mathbb{Q}]=3$, we have that $\deg m(b,\mathbb{Q})\leq 3$. 
So, the general form of that polynomial is $Ax^3+Bx^2+Cx+D$. 
So, do we have to replace $x$ with $b=q_0+q_1a+q_2a^2$, compute that polynomial, knowing that $a^3=3a+1$, and find the other roots?  
Or is there an other way to show that?  
 A: The discriminant (Wikipedia) of the polynomial $p(x)=x^3-3x+1\in\mathbb{Q}[x]$ is equal to $81$, which is a square in the field $\mathbb{Q}$. Letting $L$ be the splitting field of $p(x)$ over $\mathbb{Q}$, there is a theorem that since the discriminant is a square, we have $[L:\mathbb{Q}]=3$. Since $\mathbb{Q}(a)\subseteq L$ and $[\mathbb{Q}(a):\mathbb{Q}]=3$ also, we must have $L=\mathbb{Q}(a)$, i.e. the field $\mathbb{Q}(a)$ is the splitting field of the polynomial over $\mathbb{Q}$. Therefore it is normal over $\mathbb{Q}$.

(from Galois Theory by Jean-Pierre Escofier)


A: Hint: If $a$ is one of the roots, show that $a^2-2$ is another.

As Zev (+1) explained, observing that the discriminant is a square (of a rational number) implies that the Galois group of the splitting field is cyclic of order three. Therefore:


*

*$E=\Bbb{Q}(a)$, and

*the other zeros are at most quadratic polynomials in $a$.



This is one of the two cubic polynomials, where $a\mapsto a^2-2$ permutes the three zeros cyclically. The other is discussed here.
A few words about figuring out that the zeros $x_k=2\cos(2^k\pi/9), k=1,2,3,$ come up here, or rather, why the splitting field more or less "must" be the real subfield of the ninth cyclotomic field. This is not necessarily conclusive but pointing heavily in that direction:


*

*By the famous Kronecker-Weber theorem any abelian extension of $\Bbb{Q}$ is a subfield of $\Bbb{Q}(e^{2\pi i/m})$ for some $m$.

*For any prime $p$, the discriminant of $\Bbb{Q}(e^{2\pi i/p})$ is a power of $p$ (up to sign?). 

*Because the prime factors of the discriminant of a subfield are also factors of the discriminant of the bigger, and Zev calculated that the discriminant is $81$, we can conclude that in our case $m$ must be divisible by $p=3$, but not necessarily divisible by any other prime.

*$m=9=3^2$ is the smallest power of three such that $Q(e^{2\pi i/m})$ has a cubic subfield.

