## Cubic polynomial

There was a question about a classic exercise: Let $f(x)\in Q[x]$ be a cubic irreducible polynomial, which we take to be monic, and let $a, b, c$ be the roots. Let $F=Q(a)$ and $K=Q(a,b,c)$. Clearly, $[F:Q]=3$, but the exercise is about the degree of $K$.

Fact: $[K:F]$ is 1 or 2.

Proof: We have $f(x)=(x-a)(x-b)(x-c)$, so that we can divide $f$ by $x-a$ in $F[x]$. Therefore, $b$ and $c$ are roots of the quadratic polynomial $g(x)=f(x)/(x-a) \in F[x]$. Since $a+b+c\in Q$, we have $K=F(b,c)=F[b]$. Thus, if $b\in F$, then $[K:F]=1$. If $b\notin F$, then $[K:F]=2$.

By the tower theorem corresponding to these two possibilities, we have that $[K:Q]$ is 3 or 6.

Fact: $[K:Q]=3$ if and only if $\delta=(b-a)(c-b)(c-a)\in Q$.

Proof: Note that $\delta^2=(b-a)^2(c-b)^2(c-a)^2\in Q$, since it is a symmetric polynomial in $a,b,c$. Furthermore, $Q(\delta)\subset K$. If $\delta\notin Q$, then $[Q[\delta):Q]=2$ and this degree must divide $[K:Q]$. So we have $[K:Q]=6$. On the other hand, suppose $\delta\in Q$. The polynomial $h(x)=f(x+a)\in F[x]$ has the roots $0, b'=b-a$, and $c'=c-a$. So $h(x)/x$ has roots $b'$ and $c'$. Therefore,
$b'c'\in F$. But we are assuming that $b'c'(c'-b')=b'c'(c-b)\in Q$. So $c-b\in F$. Since $b+c\in F$ as well, we have $b,c\in F$ and $K=F$, so that $[K:Q]=3$.