Past exam of the algebraic number theory course of the University of Oxford / 2023 / 1 / a

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.