Past exam of the options II course of the University of Oxford / 2004 / B / B2

Take $\boxed{K=F[x]/(f)}$ and let $\alpha$ be the residue class of $x$. Division with remainder represents every class by a polynomial of degree less than $n$, so every element is a polynomial in $\alpha$ and $f(\alpha )=0$. A nonzero class represented by $u$ has $\gcd (u,f)=1$ because $f$ is irreducible. Bézout's identity gives $au+bf=1$, making the class of $a$ its inverse. Thus $K$ is a field, containing $F$ through its constant classes.

A splitting field of a nonconstant polynomial is an extension in which it factors into linear factors and which is generated over the original field by its roots. To construct one, choose an irreducible factor and use the preceding quotient construction to adjoin a root. Divide off a linear factor in that extension and repeat on the remaining polynomial. Each stage lowers the remaining degree, so finitely many stages suffice. The field generated by all roots in the resulting field is the required splitting field. This induction works with repeated roots and does not assume separability. The two roman clauses in the source specify properties of this one construction; they are not separate questions.

The roots are $\pm \sqrt{-5}$, so one root already supplies both. The polynomial has no rational root and is irreducible of degree two. Thus its splitting field is $\mathbb{Q}(\sqrt{-5})$, with $\boxed{[K:\mathbb{Q}]=2}$.

Put $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The three roots are $\alpha ,\alpha \zeta ,\alpha {\zeta }^2$, so the splitting field is $K=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein's criterion at two gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This intermediate field is real, while $\zeta$ is not real, and $\zeta$ satisfies $x^2+x+1=0$. Its degree over the intermediate field is therefore exactly two. By the tower law, $\boxed{[K:\mathbb{Q}]=6}$.

Over ${\mathbb{F}}_5$,The quadratic discriminant is $9-16\equiv 3$, while the squares modulo five are $0,1,4$. Thus that factor is irreducible. Adjoining either quadratic root creates a degree-two extension and supplies the other root by the sum-of-roots relation. Consequently the splitting field is ${\mathbb{F}}_{25}$ and $\boxed{[K:{\mathbb{F}}_5]=2}$.