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.

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