Let $L_f$ be the splitting field of the irreduicble polynomial $f = x^3 − x + 1$ over $\Bbb{Q}[x]$. I want to determine $\operatorname{dim}_{\Bbb{Q}}L_f$.

$f$ has three roots in its splitting field and it has no roots in $\mathbb{Q}$ because it is irreducible (can I deduce that?).

$f$ is irreducible so every polynomial which has a common root with $f$ must be other $f$ or a $f$ divides the polynomial. Hence, $f$ is the minimal polynomial of the roots of $f$. So $\operatorname{dim}_{\Bbb{Q}}L_f = \deg(f) = 3$ ?

Is this correct?

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    $\begingroup$ It has no roots in $\mathbb Q$, hence it is irreducible over $\mathbb Q$. Do you know about the discriminant? It determines whether the degree of the splitting field is $3$ or $6$. Do you see why $6 = \#S_3$ is an upper bound? $\endgroup$ – Dylan Moreland Sep 6 '12 at 17:52
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    $\begingroup$ To be super explicit, $f$ is irreducible because it is degree 3 and has no roots. $\endgroup$ – John Stalfos Sep 6 '12 at 17:58
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    $\begingroup$ A cubic with integer coefficients which is irreducible over $\mathbb Q$ has either one or three real roots (note that a double root would be rational, since it would also be a root of the formal derivative). What can you say about the two cases? Which case does yours belong to? $\endgroup$ – Mark Bennet Sep 6 '12 at 17:59
  • $\begingroup$ Edit: Sorry, I forgot. The irreducible is given. $\endgroup$ – Joachim Sep 6 '12 at 18:01
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    $\begingroup$ Just for, um, culture. Given a prime $p > 3$ with Legendre symbol $(-23|p) = 1,$ then $$x^3 - x + 1 \equiv 0 \pmod p $$ has three distinct roots if and only if there is an integer expression $ p = u^2 + u v + 6 v^2. $ Otherwise we get some expression $ p = 2 u^2 + u v + 3 v^2. $ Not both. If you decide to experiment with this, a good idea, note you need to allow $uv < 0$ sometimes. $\endgroup$ – Will Jagy Sep 6 '12 at 18:35

I'm sure you're looking for a completely algebraic solution, which I'm sure you're very close to, based on the other comments. I'm going to toss in a method that brings in elementary calculus.

I'm beginning at the point where you believe the polynomial is irreducible over $\mathbb{Q}$. Since it is has odd degree and it's in $\mathbb{R}[x]$, it has a real root. Using basic calculus you can check that the graph has exactly one real root, so the rest are complex.

Let $\alpha$ be the real root. Since $f$ has degree 3 and is irreducible and has $\alpha$ as a root, $[\mathbb{Q}[\alpha]:\mathbb{Q}]=3$. This field contains only reals, so you will need to adjoin another (complex) root to finish splitting the polynomial. After factoring out the real root, the complex roots are the remaining zeros of a quadratic factor of $f$.

So, adjoining another root is another extension of degree 2.

Thus the splitting field's degree over $\mathbb{Q}$ is 6.


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