Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 18H (Galois Theory)

The extension $L/K$ is finite when $L$ is finite-dimensional as a vector space over $K$, and its degree is $[L:K]={\dim }_{K}L$.

The extension $L/K$ is separable when it is algebraic and the minimal polynomial over $K$ of every element of $L$ has no repeated root in an algebraic closure.

The extension $L/K$ is simple when there is an element $\alpha \in L$ such that $L=K(\alpha )$.

Finite and separable imply simple by the primitive element theorem. Separable and simple imply finite: if $L=K(\alpha )$ and the extension is separable, then $\alpha$ is algebraic, so $[L:K]=\mathrm{deg}{m}_{\alpha ,K}<\infty$.

Finite and simple do not imply separable. In characteristic $p$, takeThen $L/K$ has degree $p$ and is simple, but the generator has inseparable minimal polynomial $X^p-t$. These are the finite separable simple extension implications.

The rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant iswhich is not a square in $\mathbb{Q}$. The Galois group of an irreducible cubic therefore gives

Quadratic subfields correspond to order-three subgroups. The unique such subgroup is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is

Let the roots be $\alpha ,\beta ,\gamma$ and choose
$\delta =(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$, so ${\delta }^2=-23$. Put $\theta =\alpha +\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta -\alpha =2\delta$. Cancelling the factor $\beta -\alpha$ in the expression for $\delta$ would giveBut $f^{\prime }(\gamma )=3{\gamma }^2-1$, so this would make ${\gamma }^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal polynomial. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so