I am self studying Galois theory through lecture notes but I am not able to find a reason why an argument should hold.
Please help.
Argument is - Let $K$ be a field and $f(x)$ belonging to $K[x]$ be irreducible, separable polynomial of degree $3$.Then if Galois group of $f$ is $S_3$ , then there are $4$ proper intermediate fields, $K(\delta)$ , $K(a)$ , $K(b)$ , $K(c)$, where $a$, $b$, $c$ are roots of $f$ and $\delta$ is discriminant raised to power $\frac{1}{2}$
Can someone please tell why there must be $4$ intermediate fields.
Let $L$ be the splitting field of $f$ over $K$.
The Galois correspondence provides a one-to-one correspondence between the intermediate fields of $L/K$ and the subgroups of $S_3$.
From group theory, we know that $S_3=\{(),(12),(13),(23),(123),(132) \}$ contains three subgroups of order $2$ (= index 3) (the transpositions $(12),(13)$ and $(23)$) and one subgroup of order 3 (= index 2), which is $A_3$.
It is well-known (see also this thread) that the intermediate field corresponding to $A_3$ is generated by $\sqrt{\Delta}$, where $\Delta$ is the discriminant. Then the other three subgroups are then exactly the ones generated by a root of $f$, since $f$ is irreducible.