Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 20G (Number Fields)

Use the module criterion: an element $x$ is an algebraic integer if and only if it lies in a nonzero finitely generated $\mathbb{Z}$-module $M$ with $xM\subseteq M$.

Let the coefficients of the monic polynomial be $c_0,\dots ,c_{n-1}$ and put $A=\mathbb{Z}[c_0,\dots ,c_{n-1}]$. A ring generated by finitely many algebraic integers is a finitely generated $\mathbb{Z}$-module. If $\alpha$ is a root, thenis finitely generated, and the monic relationshows that $\alpha M\subseteq M$. The criterion proves that every root is an algebraic integer.

Since $17$ is square-free and $17\equiv 1(\mathrm{mod}4)$, the standard quadratic-field integral-basis theorem givesIndeed, the displayed generator has monic polynomial $x^2-x-4$, and the discriminant of this basis is $17$, the field discriminant; no larger order is possible.

One has ${\alpha }^2=4+\sqrt{17}\in K$ andIf $\alpha$ belonged to $K$, multiplicativity would givewhich is impossible because $N_{K/\mathbb{Q}}(\alpha )\in \mathbb{Q}$. Therefore $\alpha \notin K$ and $[L:K]=2$.

If $\beta \in {\mathcal{O}}_L$, all of its $K$-conjugates are algebraic integers, so their sum and product give

Conversely, if these two quantities lie in ${\mathcal{O}}_K$, then $\beta$ is a root ofa monic polynomial whose coefficients are algebraic integers. Part (a), or transitivity of integral dependence, implies $\beta \in {\mathcal{O}}_L$.

Let $\beta =a+b\alpha \in {\mathcal{O}}_L$. The nontrivial $K$-automorphism sends $\alpha$ to $-\alpha$, soAlso $\alpha$ is integral, hence $\alpha \beta$ is integral, andBut $4+\sqrt{17}$ has norm $-1$, so it is a unit of ${\mathcal{O}}_K$. Multiplying by its inverse proves $2b\in {\mathcal{O}}_K$.