Let $f(x) \in \mathbb Q[x]$ and $$f(x)= x^3 + px +q=(x-r_1)(x-r_2)(x-r_3),$$ and $r_1, r_2,r_3\in K$ for the splitting field $K$ of $f(x)$.
Let the $D = (r_3 - r_1)^2(r_2- r_1)^2(r_3 - r_2)^2$, which is the discriminant of $f(x)$.
The square root of the discriminant $D$ is clearly in the splitting field because $\sqrt D= |(r_3 - r_1)(r_2- r_1)(r_3 - r_2)|$.
How to find $[\mathbb Q(r_1)(\sqrt{D}):\mathbb Q]$, i.e. the degree of the extension field of $\mathbb Q$ adjoining $\sqrt D$ and $r_1$?
I tried to find the degree of $\sqrt D$ over $\mathbb Q(r_1)$ and did not find the minimal polynomial of it. The degree is surely at most $2$. It is just not clear to me whether it is $1$ or $2$, or it depends on $f(x)$.
related: Splitting field of a degree 3 polynomial is generated by a root and the square root of the discriminant Since proposition points out this extension field is simply the splitting field of $f(x)$ over $\Bbb Q$.
In short, my question:
- what is the degree of $\sqrt D$ (over $\Bbb Q(r_1)$ and $\Bbb Q$) ? What are all possibilities and what condition does these depend on?
- Equivalently, what is the degree of the splitting field (which is equal to $\Bbb Q(r_1)(\sqrt D)$ according to the linked question) of $f(x)$ given its discriminant? (Is there even relations between them?)
The degree of $\sqrt D$ over the rationals is one if $D=-27q^2-4p^3$ is a square, two otherwise. The degree of the splitting field is three if $D$ is a square, six otherwise.
The full story is best explained via Galois Theory, which is a bit of a long story. Copying from my comment on the question,
"very loosely, the idea is that any bijection $\sigma$ from the splitting field to itself that preserves the field operations must act as a permutation on the roots of $f$, from which it follows that $\sigma(D)=D$; but the only elements of the splitting field that are fixed by all the operation-preserving bijections of the splitting field are the rationals."