Integral basis

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

The power basis $1,\theta ,{\theta }^2$ is an integral basis. The defining polynomial is irreducible by Eisenstein's criterion at two and its discriminant is $-140$. The discriminant-index formula for an integral lattice restricts its index in the ring of integers to one or two. An index of two would supply an algebraic integer $\eta =(a+b\theta +c{\theta }^2)/2$ with $a,b,c\in \{0,1\}$ not all zero. Its field trace $(3a-4c)/2$ forces $a=0$. If $b=0$ this gives ${\theta }^2/2$ directly; if $b=1$, the identity $\theta \eta ={\theta }^2/2-c\theta -c$ gives the same integrality. But ${\theta }^2/2=-1-{\theta }^{-1}$, while $N({\theta }^{-1})=-1/2$ excludes integrality by the field norm. Thus the index is one, proving the claim.

For a nonzero squarefree integer $d\equiv 3(\mathrm{mod}4)$, set $\theta =\sqrt[4]{d}$. The power basis $1,\theta ,{\theta }^2,{\theta }^3$ is an integral basis, with discriminant $-256d^3$. The polynomial is Eisenstein at every prime dividing $d$; for $d=-1$, shift $X^4+1$ by one to obtain an Eisenstein polynomial at two. The discriminant-index formula for an integral lattice restricts any power-basis index prime to $2d$. An Eisenstein prime cannot divide the index: the norm of $r{\theta }^i/p$ has valuation $i-4<0$ for $p\nmid r$ and $i<4$, while ${\theta }^4/p$ is integral; multiplying a putative denominator-$p$ integral element successively by powers of $\theta$ forces each numerator coefficient divisible by $p$.

It remains to exclude two. A putative integral element $(a+b\theta +c{\theta }^2+e{\theta }^3)/2$ has relative norm to $\mathbb{Q}(\sqrt{d})$ with coefficientsThe quadratic field has ring of integers $\mathbb{Z}[\sqrt{d}]$, so both coefficients are integers. The first congruence forces $a,c$ to have the same parity. If they are odd, the second requires $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. If they are even, the second forces $b,e$ to have the same parity and the first forces $be$ even, so both are even. Thus no nontrivial denominator-two element exists, proving index one.

Let $f=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0\in \mathbb{Z}[X]$ satisfy Eisenstein's criterion at a prime number $p$, and let $\theta$ be a root. ThenIndeed, ${\theta }^n/p$ is an algebraic integer, whereas for $p\nmid r$ and $0\le i<n$ the field norm of $r{\theta }^i/p$ has $p$-adic valuation $i-n<0$, so that element is not integral. If $\gamma =p^{-1}{\sum }_{i=0}^{n-1}{r}_i{\theta }^i$ is integral, multiply it by ${\theta }^{n-1}$. Every term except $r_0{\theta }^{n-1}/p$ is integral; thus $p\mid r_0$. Subtract $r_0/p$, multiply the remainder by ${\theta }^{n-2}$, and continue to obtain $p\mid r_i$ for every $i$. Hence there is no element of order $p$ in the finite group ${\mathcal{O}}_{\mathbb{Q}(\theta )}/\mathbb{Z}[\theta ]$, proving the claim. The result concerns the Eisenstein prime only: other primes may still divide the index.

Let $m,n$ be coprime square-free integers, $m\equiv 2(\mathrm{mod}4)$ and $n\equiv 1(\mathrm{mod}4)$, with $n\ne 1$. Put ${\alpha }^2=m$, ${\beta }^2=n$, and $\omega =(1+\beta )/2$. Then $1,\alpha ,\omega ,\alpha \omega$ is an integral basis of $\mathbb{Q}(\alpha ,\beta )$. All four entries are algebraic integers. For spanning, field traces to the three quadratic subfields show that an arbitrary algebraic integer has form $(a+b\alpha +c\beta +d\alpha \beta )/2$ with $a,b,c,d\in \mathbb{Z}$. Its field norm to $\mathbb{Q}(\beta )$ is $A/4+(ac-mbd)\beta /2$, where $A=a^2+n{c}^2-m{b}^2-mn{d}^2$. The ring of integers of a quadratic field formula forces $A$ even and $A\equiv 2ac-2mbd(\mathrm{mod}4)$. These give $a\equiv c(\mathrm{mod}2)$ and then $b\equiv d(\mathrm{mod}2)$. The element is consequently $(a-c)/2+((b-d)/2)\alpha +c\omega +d\alpha \omega$. The coefficient determinant $1/4$ proves rational linear independence.

Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.

Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.

Let $m_1,m_2$ be coprime positive square-free integers, with $m_1\not\equiv \pm m_2(\mathrm{mod}9)$. The displayed integral basis has discriminant $-27m_1^2{m}_2^2$. Both nonconstant generators are integral, and their products are integral linear combinations of the three basis elements. At a prime dividing $m_1$, the cubic for $\alpha$ is Eisenstein; at a prime dividing $m_2$, the cubic for ${\alpha }^2/m_2$ is Eisenstein. If $3\nmid m_1{m}_2$, then $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$, and a shift by $1$ or $-1$ gives an Eisenstein polynomial at three. These local uniformizer generators show that the order is maximal at all primes dividing its discriminant; elsewhere the discriminant-index formula for an integral lattice already gives maximality.

For an odd prime $p\equiv 3(\mathrm{mod}4)$, traces to the three quadratic subfields show that an integer has coefficients $(a+b\sqrt{p}+c\sqrt{2}+d\sqrt{2p})/2$ with rational integers $a,b,c,d$. Its norm to $\mathbb{Q}(\sqrt{p})$ forces $a,b$ even and $c\equiv d(\mathrm{mod}2)$. Hence an integral basis is $1,\sqrt{2},\sqrt{p},(1+\sqrt{p})\sqrt{2}/2$. The last generator satisfies the monic polynomial $X^4-(p+1)X^2+(p-1{)}^2/4$, proving integrality as well as spanning. The subfield used for the norm matters: the norm to $\mathbb{Q}(\sqrt{2})$ has different coefficients.

Write $\alpha =\sqrt[4]{2}$ and an algebraic integer as $a+b\alpha +c{\alpha }^2+d{\alpha }^3$ with rational coefficients. Traces to $\mathbb{Q}(\sqrt{2})$ show $2a,2b,2c,4d$ are integers. The norm has coefficients $a^2+2c^2-4bd$ and $2ac-b^2-2d^2$ in the integral basis $1,\sqrt{2}$. Set $A=2a$, $B=2b$, $C=2c$, $D=4d$. Integrality gives $A^2+2C^2-2BD\equiv 0(\mathrm{mod}4)$ and $4AC-2B^2-D^2\equiv 0(\mathrm{mod}8)$. The second forces $D$ even and then $B$ even; the first forces $A,C$ even; returning to the second forces $D$ divisible by four. Thus all original coefficients are integers, proving the integral basis $1,\alpha ,{\alpha }^2,{\alpha }^3$.

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.

Two finite Galois extensions of the rational number field with coprime field discriminants have trivial intersection. Any ramified prime in the intersection would ramify in both fields. The intersection is therefore everywhere unramified at finite primes and has field discriminant of absolute value one. The Minkowski lower bound for a number-field discriminant excludes this in degree greater than one. The Galois correspondence then gives the product degree for their compositum. The normality hypothesis is used in this last intersection-to-degree argument.

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.