Show that the field generated by a root of $f(x)$ where $f(x)=x^3-x-1$ over $\mathbb{Q}$ is not normal over $\mathbb{Q}$.
First in order to mention some extension I showed $f(x)=x^3-x-1$ is irreducible over $\mathbb{Q}$.
$$\text{Theorem 1: Suppose } f(x)\in \text{F[x] such that } \deg(f)=\lbrace2,3\rbrace. \; f(x) \text{ is irreducible over F if and only if } f(x)\text{ has no zero in F}$$
$$\text{Theorem 2: Let } f(x) \text{be a monic polynomial in } \mathbb{Z}[x]. \text{ Suppose there exist } r\in \mathbb{Q} \text{ such that } f(r)=0 \text{. Then } r \in \mathbb{Z} \text{ and } r|a_0, \text{ where } a_0 \text{is constant coefficient of} f(x)$$
$$\text{Theorem 3: Suppose E is a finite extension of F. This extension is a normal extension if and only if E is is the splitting field of a polynomial } f(x)\in \text{F}$$
I have used these three theorem to prove what is asked in the question.
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Attemption:
By Theorem 1, Theorem 2 and $f(1)\neq0,\;f(-1)\neq 0$ we can say $f(x)$ is irreducible over $\mathbb{Q}$. Now, we can construct extension field let $\alpha$ be a root of $f(x)$. It follows that $x-\alpha | f(x)$. After some calculation I left with: $$(x-\alpha)(x^2+\alpha x+(\alpha^2-1)) = f(x)\;(*)$$
Before saying $\mathbb{Q}(\alpha)$ is or is not normal extension we have investigate the quadratic factor, say $q(x)=x^2+\alpha x+(\alpha^2-1)$
. Let's calculate its $\Delta$:
$$\Delta = 4 – 3\alpha^2$$
It is clear that we have three cases. I examined case by case:
1-) If $\Delta = 0$. Then:
$$\alpha = \pm \dfrac{2}{\sqrt{3}}$$ But, $\left(x-\dfrac{2}{\sqrt{3}}\right)\left(x^2+\dfrac{2}{\sqrt{3}}x+\dfrac{1}{3}\right) \neq f(x)$
Therefore $\Delta \neq 0$
2-) If $\Delta > 0$. Then:
$$\dfrac{4}{3}>\alpha^2>0$$ We can say $$x_{2,3}=\dfrac{\alpha}{2}\pm\dfrac{\sqrt{4-3\alpha^2}}{2}$$ and it is clear that $\alpha \neq \sqrt{4-3\alpha^2}$ since $\alpha \neq \pm 1$. Thus we can say $\mathbb{Q}(\alpha)$ is not a splitting field of f(x). And by Theorem 3 we can say $\mathbb{Q}(\alpha)$ is not normal.
3-) If $\Delta < 0$. Same idea applicable. Aware that we get two complex roots. Again by Theorem 3 we can say $\mathbb{Q}(\alpha)$ is not normal.
All in all I conclude that the root of the polynomial does not generate normal extension over $\mathbb{Q}$
Is that right? I think it is but I am not sure the part that I use determinants? Thanks in advance!
Best Answer
$x^3-x-1$ has only one real root, so it doesn't generate either of the complex roots.