I need to find $[F:\mathbb{Q}]$ where $F$ is the splitting field of the polynomial $f = x^{3} + x^{2} + 1$ over $\mathbb{Q}$.
Now, I know from looking at its derivative that $f$ has only one real root. Moreover, from the rational roots theorem, I know that $f$ has no rational roots. Therefore, $f$ has no roots in $\mathbb{Q}$. Is $[F:\mathbb{Q}] = 3$, then? Do I need to find out what $F$ actually is here? And in order to do that, do I need to find what the actual roots are?
Thanks.
If all you need is $[F:\Bbb{Q}]$, you have everything required to compute it. You've determined that $f$ is irreducible, so let $\alpha$ be a root. Then $[\Bbb{Q}[\alpha]:\Bbb{Q}]=3$ and since all roots are not real, $\Bbb{Q}[\alpha]$ cannot be the splitting field. The minimum polynomial for the other two roots of $f$ is a quadratic with coefficients in $\Bbb{Q}[\alpha]$, in fact $f/(x-\alpha)$. By multiplicativity of degrees the splitting field must have degree $3\cdot 2=6$ over $\Bbb{Q}$.