# Thread: Show that E/Q is normal

1. Hey!!

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?

We have that $E/\mathbb{Q}$ is normal iff $E$ is a splitting field for some polynomial in $\mathbb{Q}[x]$.

We have that one root of $x^3-3x-1$ is $a$. Using the euclidean algorithm we have that $x^3-3x-1=(x-a)(x^2+ax+(a^2-3))$.

The roots of the polynomial $x^2+ax+(a^2-3)$ are $x=\frac{-a+\sqrt{12-3a^2}}{2}, \ \ x=\frac{-a-\sqrt{12-3a^2}}{2}$.

To show that $E$ is a splitting field for $x^3-3x-1$, we have to show that all the roots of that polynomial are in $E$, or not?

We have that $a\in E$. But does it hold that $\sqrt{12-3a^2}\in E$ ?
I tried to check if there are $q_0, q_1, q_2\in \mathbb{Q}$ such that $q_0+q_1a+q_2a^2=\sqrt{12-3a^2}$, but then we get the system:
\left\{\begin{align}q_0^2 +2q_1q_2 &= 12\\q_2^2 + 2q_0q_1 + 6q_1q_2 &= 0\\q_1^2 + 2q_0q_2 + 3q_2^2 &= -3\end{align}\right..
Can we solve it?