Past exam of the number theory course of the University of Oxford / 2011 / B / 4

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.

Let $\theta =\sqrt[3]{3}$. The power-basis discriminant isIf $\mathbb{Z}[\theta ]$ were a proper sublattice of ${\mathcal{O}}_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb{Z}[\theta ]$ already belongs to $\mathbb{Z}[\theta ]$. This is a contradiction. Hence