Q:

Finding all subfields of splitting field for x^3+x+1.

I have the polynomial $f(x)=x^3+x+1 \in \mathbb{Q}[x]$ and have to find all the subfields of its splitting field $K$ over $\mathbb{Q}$. I'm not supposed to find them explicitly. Heres my try:
Denote the roots of $f$ by $r_1, r_2, r_3$. These roots are obviously not in $\mathbb{Q}$, and so we have $\mathbb{Q}$, $K$, and $\mathbb{Q}(r_i)$, as subfields.
I've been told there one more, but i can't figure out how to find it, and doing the calculations explicitly yields too ugly roots, to be the smartest way. Maybe the composite of some of the $\mathbb{Q}(r_i)$, but i can't figure out which ones.

A:

The last one is given by
$$ \mathbb Q(\sqrt{\Delta}),$$
where $\Delta$ is the discriminant of the polynomial $f(x)$.
For a monic polynomial 
$$p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0=\prod_{i=1}^n(x-r_i),$$
the discriminant of the polynomial $p$ is defined by
$$ \Delta=\prod_{i\neq j}(r_i-r_j)^2.$$
You can observe that $\sqrt{\Delta}$ belongs to the splitting field of $p$.
In terms of Galois correspondence, the subfield $\mathbb Q(\sqrt{\Delta})$ correspond to the subgroup 
$$ G\cap A_n,$$
where $G$ is the Galois group of the splitting field of $p$ (seen as a subgroup of $S_n$) and $A_n$ is the alternating group.
The discriminant is also given by 
$$ \Delta=Res_x(p(x),p'(x)),$$
where $Res$ denotes the resultant.
For a cubic polynomial  of the form
$$p(x)=x^3+cx+d$$
we have $$\Delta=-4c^3-27d^2.$$
So the extension that you are looking for is 
$$\mathbb Q(\sqrt{\Delta})=\mathbb Q(\sqrt{-31})$$

