Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 18H (Galois Theory)

For roots $r_1,r_2,r_3$, the discriminant isand for $f=X^3+pX+q$,If $f$ is irreducible, its Galois group is transitive in $S_3$: it is $A_3\cong C_3$ when $\Delta$ is a square in $K$, and $S_3$ otherwise. If $f$ has one root in $K$ and an irreducible quadratic factor, the group is $C_2$; if it splits, it is trivial. A repeated-root depressed cubic already has all roots in $K$, so adds no case.

For $X^3+3X+1$, the rational-root test proves irreducibility andis not a rational square. Thus $G\cong S_3$. By the fundamental theorem of Galois theory, the complete subfield list iswhere the three cubic fields fix the three order-two subgroups and the quadratic field fixes $A_3$.

If $L/K$ is Galois, form the norm polynomialThe Galois group permutes the factors, so $F\in K[X]$. Since the identity factor is $f$, taking $g=F/f\in L[X]$ proves the claim.

If $L/K$ is merely finite separable, place it in a finite Galois closure $M/K$ and use the same product over $\mathrm{Gal}(M/K)$. It gives $F\in K[X]$ divisible by $f$ in $M[X]$. Polynomial division of $F$ by $f$ uses only coefficients in $L$, and the remainder is zero, so the quotient $g=F/f$ actually lies in $L[X]$. It is nonzero and satisfies $fg=F\in K[X]$.