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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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    What about my answer to one of your questions (this should be a counter example) ? http://math.stackexchange.com/questions/191371/dimension-of-a-splitting-field/191375#1913752012-09-06

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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.