Past exam of the number theory course of the University of Oxford / 2008 / A / 1

For an irreducible polynomial $f\in K[X]$, take $L=K[X]/(f)$ and let $\alpha$ be the class of $X$. The quotient is a field, is generated by $\alpha$ over $K$, and satisfies $f(\alpha )=0$. Its degree is $\boxed{[L:K]=\mathrm{deg}f}$, with basis $1,\alpha ,\dots ,{\alpha }^{\mathrm{deg}f-1}$.

A splitting field is a field generated over $K$ by all roots of the polynomial and in which the polynomial is a product of linear factors. Prove existence and the factorial bound together by induction on the degree. Constant nonzero polynomials and linear polynomials require no extension. For a reducible degree-$n$ polynomial write $g=hk$ with positive degrees $r,n-r$. Split $h$ by induction, then split $k$ over that field. The resulting field is generated by all roots of $g$, and its degree over $K$ is at most $r!(n-r)!\le n!$. If $g$ is irreducible, adjoin one root in degree $n$, divide out its linear factor, and split the degree-$(n-1)$ quotient over that extension. The tower formula gives degree at most $n(n-1)!=n!$. This also works in inseparable characteristic and proves the required bound without assuming Galoisness.

For the sextic with constant term three, Eisenstein's criterion at three gives degree six. Choose a root $\alpha$, so ${\alpha }^3$ is a square root of $-3$. Thensatisfies ${\zeta }^2-\zeta +1=0$, and is a primitive sixth root of unity: ${\zeta }^3=-1$ and $\zeta \ne -1$. The six distinct roots are $\alpha {\zeta }^j$ for $0\le j<6$, all in $\mathbb{Q}(\alpha )$. Hence the simple extension is itself the splitting field and $\boxed{[L:\mathbb{Q}]=6}$.

For the constant term two, Eisenstein at two gives $[\mathbb{Q}(\gamma ):\mathbb{Q}]=6$ for a root $\gamma$. Its splitting field is $\mathbb{Q}(\gamma ,{\zeta }_6)$. The supplied noncontainment of a root of $X^2+3$ means that adjoining ${\zeta }_6=(1+\sqrt{-3})/2$ has degree two. Thus