Galois group of $X^3-X+1$ over $\mathbb{Q}$ and $\mathbb{R}$ without discriminant.

Question

Yesterday I had an exam and I had to find the galois group of the polynomial $f = x^3-x+1$. My answer was $A_3$ which is probably wrong. First of all it has no roots by the rational root theorem so it is irreducible and the galois group of this polynomial is a subgroup of $S_3$. Let $\alpha\in\mathbb{\overline{ Q}}$ be a root of f, then the extension $\mathbb{Q}\subset \mathbb{Q}(\alpha)$ is of degree 3. If this field is a splitting field of f then the galois group must be $A_3$. So how do I show that this field is not the splitting field of f and that $S_3$ is the galois group without using the discriminant.

For the second case I think it is $C_2$. We have $f(-1)=-1$ and $f(1)=1$, so by the intermediate value theorem f has at least one root in the interval $(-1,1)$, but f is strictly increasing so it can only be one root. For any $x\in \mathbb{R}$ with $|x|>1$ f is increasing/decreasing. This shows that f has only one real root and two complex which means that a splitting field of is $\mathbb{C}$. It follows that the galois group must be isomorphic to $C_2$.

How can I find those two complex roots in terms of radicals?

You've shown that the polynomial has two complex roots, which therefore must be a pair of complex conjugates since $f$ has real coefficients. This shows that complex conjugation is an element of order $2$ in the Galois group. $A_3$ has no elements of order $2$, so the Galois group must be $S_3$. — Viktor Vaughn, Jun 18 '24 at 15:45
I don't understand why those complex roots must be conjugates. Could you expand on that? Other than that it seems very logical. thx — muhammed gunes, Jun 18 '24 at 15:59

score 4 · Answer 1 · answered Jun 18 '24 at 16:13

4

To be clear, $f$ is not strictly increasing, but its derivative is $f'(x)=3x^2-1$, hence it's strictly increasing over $(-\infty, -\sqrt{1/3}]$ and $[\sqrt{1/3}, \infty)$, but decrease over $[-\sqrt{1/3}, \sqrt{1/3}]$. And since $$f(\sqrt{1/3})>1-\sqrt{1/3}>0 \\ f(-\sqrt{-1/3})=(2/3)\sqrt{1/3}+1>0$$ there is at most one real root whose existence can be confirmed by intermediate value theorem.

There are many different approaches to finish from here, such as the one given in the comment. It's worth to know there is a general result:

if an irreducible polynomial $f(x)\in\mathbb Q[x]$ is of degre $p$ where $p$ is a prime and has precisely one pair of nonreal roots, then its Galois group is isomorphic to $S_p$.

Another one would be let $\alpha$ be the real root of $f$, then $\mathbb Q(\alpha)\subset\mathbb R$ doesn't contain the other two roots, hence not the splitting field of $f$, therefore the degree of the splitting field over $\mathbb Q$ must be $6$ instead of $3$.

answered Jun 18 '24 at 16:13

Just a user

22,048

Why do we take $\alpha $ to be the real root in $\mathbb{Q}(\alpha)$. Would it also for $\alpha$, the complex root. – muhammed gunes Jun 18 '24 at 16:25
3

We are free to choose any one of those roots, but the point here is to break the symmetry. Because if $\alpha\in\mathbb R$ then we know $\mathbb Q(\alpha)$ is a subfield of $\mathbb R$ that cannot contain the other two nonreal roots. If you take an imaginary root $\beta$, you will have as abstract fields $\mathbb Q(\alpha)\simeq\mathbb Q(\beta)$, but without the convenient intermediate field $\mathbb R$, we don't know if $\mathbb Q(\beta)$ contains $\overline{\beta}$ and $\alpha$. – Just a user Jun 18 '24 at 16:28
@muhammedgunes: you can choose any root $\alpha$ and show that the other two roots don't lie in $\mathbb{Q} (\alpha) $ without worrying about real or complex nature of roots. But this requires a bit more algebra. – Paramanand Singh Jan 04 '25 at 14:14

Galois group of $X^3-X+1$ over $\mathbb{Q}$ and $\mathbb{R}$ without discriminant.

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