Integral basis

An integral basis is a $\mathbb{Z}$-basis of the ring of integers of a number field.

If $m>1$ is a square-free integer, $3\nmid m$, and $m\not\equiv \pm 1(\mathrm{mod}9)$, then $1,\sqrt[3]{m},\sqrt[3]{m^2}$ is an integral basis of the pure cubic number field, with field discriminant $-27m^2$. The discriminant-index formula for an integral lattice restricts a possible index to primes dividing $3m$. At primes dividing $m$ the different exponent is $2$ by tame ramification. A shifted Eisenstein polynomial gives total wild ramification at $3$, so its different exponent is at least $3$. These exponents exhaust the polynomial discriminant and force index one.

The discriminant of an integral full-rank lattice is the field discriminant multiplied by the square of its index in the ring of integers. A squarefree lattice discriminant therefore forces index one.

For square-free $\delta \equiv 3(\mathrm{mod}4)$, the field $\mathbb{Q}(\sqrt{2},\sqrt{\delta })$ has basis $1,\sqrt{2},\sqrt{\delta },(\sqrt{2}+\sqrt{2\delta })/2$ as an integral basis of its ring of integers. Relative traces force all rational coefficients to be half-integers; the norm to $\mathbb{Q}(\sqrt{\delta })$ forces the constant and $\sqrt{\delta }$ coefficients to be integers and the two other numerators to have equal parity. The half-sum is integral because it satisfies $X^4-(1+\delta )X^2+(\delta -1{)}^2/4=0$.

For elements ${\alpha }_1,\dots ,{\alpha }_n$ of a degree-$n$ number field with embeddings ${\sigma }_1,\dots ,{\sigma }_n$, their discriminant isIt is nonzero exactly when the elements form a $\mathbb{Q}$-basis.

The field discriminant is the discriminant of any integral basis and controls the index of every suborder.

A ramified prime makes the finite algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ have a nonzero nilpotent element. Multiplication by its product with any other element is nilpotent and has trace zero, so the reduced trace pairing is degenerate. Therefore the field discriminant is divisible by $p$. This proves finiteness of the ramified rational primes without requiring explicit prime factorization in the field.

Let ${\omega }_1,\dots ,{\omega }_n$ be an integral basis and let ${\alpha }_j={\sum }_i{a}_{ij}{\omega }_i$ be a rational basis of algebraic integers, with integer matrix $A=(a_{ij})$. Their trace Gram matrices satisfysoThe absolute discriminant is minimized exactly when $A$ is unimodular, equivalently when the ${\alpha }_i$ form an integral basis.