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For a nonzero integral ideal $\mathfrak{a}$ of the full ring of integers of a number field of degree two,Here conjugation is the nonidentity field automorphism and $N(\mathfrak{a})$ is the additive index of the ideal. A direct proof writes $\mathfrak{a}=s⟨d,e+\omega {⟩}_{\mathbb{Z}}$ in an integral basis $1,\omega$. Put $\theta =e+\omega$, $T=\theta +\bar{\theta }$, $U=\theta \bar{\theta }$. The ideal property gives $d\mid U$. Moreover $\gcd (d,T,U/d)=1$: a prime dividing all three would make $\theta /\ell$ an algebraic integer, contradicting its nonintegral coefficient of $\omega$. The product of $⟨d,\theta ⟩$ with its conjugate contains $d^2,dT,U$, so Bezout identity puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^{2}d)$, and the lattice determinant identifies $s^{2}d=N(\mathfrak{a})$. The full ring of integers is essential to the integrality argument.

Let $j(\mathfrak{b})$ be the Minkowski embedding of a number field applied to a nonzero fractional ideal. Under the metric $|x{|}^2={\sum }_{v\text{ real}}{x}_v^2+2{\sum }_{v\text{ complex}}|x_v{|}^2$, its covolume is $\sqrt{|D_K|}N(\mathfrak{b})$. For ordinary coordinate Lebesgue measure $\prod d{x}_v\prod d\mathrm{Re}{z}_{v}d\mathrm{Im}{z}_v$, its covolume is instead $2^{-r_2}\sqrt{|D_K|}N(\mathfrak{b})$. Here $D_K$ is the field discriminant, and the absolute norm of a fractional ideal is positive and multiplicative. The formula follows by taking the embedding determinant of an integral basis, then using the index of an integral ideal and scaling to handle a fractional ideal.

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 $X^4-(1+\delta )X^2+(\delta -1{)}^2/4=0$.

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.

A place of a number field is an equivalence class of nontrivial absolute values on a field. Its finite places correspond to nonzero prime ideals of the ring of integers of a number field; its infinite places come from real embeddings and conjugate pairs of complex embeddings.

For a embedding of a number field into a p-adic algebraic closure $\iota :K\hookrightarrow {\overline{\mathbb{Q}}}_p$, pull back the p-adic absolute value. This gives a place above $p$. Every element of $\mathrm{Gal}({\overline{\mathbb{Q}}}_p/{\mathbb{Q}}_p)$ preserves that absolute value, by uniqueness of its extension to each finite local extension, so equivalent embeddings give the same place.

Conversely, a place $v$ above $p$ gives a completion of a number field at a prime ideal $K_v$, a finite extension of ${\mathbb{Q}}_p$. Embed it into ${\overline{\mathbb{Q}}}_p$ and restrict to $K$. If two embeddings give the same place, they extend to two ${\mathbb{Q}}_p$-embeddings of this same completion. They are conjugate under the absolute Galois group: their finite separable images lie in a finite normal closure, and an isomorphism of such images extends to an automorphism of the algebraic closure. The completion really is the closure of the embedded $K$, since rational coefficients are dense in the ${\mathbb{Q}}_p$-span of a primitive element. Thus we obtain the embedding orbit description of finite places:

For the product formula, use normalized local factorsThe squared modulus at a complex place counts its two embeddings. Equivalently one can use ordinary complex modulus and put exponent two in the product. These normalized local factors for the product formula must be specified; arbitrary representatives of place classes would not satisfy the unweighted printed formula.

The fractional principal ideal $(x)$ has only finitely many nonzero prime exponents. Its ideal norm is $|N_{K/\mathbb{Q}}(x)|$: for integral $x$, multiplication by $x$ on an integral basis has this determinant in absolute value, equal to the index of $(x)$ in ${\mathcal{O}}_K$; a quotient gives the fractional case. Hence the finite-place product is $|N_{K/\mathbb{Q}}(x){|}^{-1}$. The infinite-place product is $|N_{K/\mathbb{Q}}(x)|$, by the embedding expression for the field norm. Therefore all but finitely many local factors are one, and

Finally, let ${\Delta }_K\ne 0$ be the integer field discriminant, defined by the determinant of the trace pairing on an integral basis. If a rational prime $p$ ramifies, the algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ has a nonzero nilpotent element: use prime ideal factorization and the Chinese remainder theorem to see a nonzero nilpotent in a factor ${\mathcal{O}}_K/{\mathfrak{p}}^e$ with $e>1$. If $y$ is nilpotent, multiplication by $yz$ is nilpotent for every $z$ in this commutative algebra, and therefore has trace zero. Thus $y$ is in the radical of its trace pairing, making the reduced discriminant zero. We have proved the discriminant obstruction to ramification:Only finitely many rational primes divide this nonzero integer, so only finitely many primes ramify.

The cubic has no rational integer root among the divisors of four, so is irreducible. Its polynomial discriminant is $4-27\cdot 16=-428=-4\cdot 107$. Thus the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ is one or two, since its square divides the discriminant.

Set $\beta =({\alpha }^2+\alpha )/2$. Using ${\alpha }^3=\alpha +4$ givesand hence ${\beta }^3-{\beta }^2-3\beta -2=0$. Therefore $\beta$ is integral. The lattice spanned by $1,\alpha ,\beta$ contains $\mathbb{Z}[\alpha ]$ with index two. Changing basis divides the discriminant by four, giving $-107$, which is squarefree. An integral lattice of full rank with squarefree discriminant cannot have any further index in the full ring of integers. Henceis the required integral basis. The relations also give $\alpha ={\beta }^2-\beta -2$, so this ring can equally be generated by $\beta$.

By the preceding argument take a nonzero integer $m\in \mathfrak{a}$. The ideal property gives $m{\mathcal{O}}_L\subseteq \mathfrak{a}$. If $d=[L:\mathbb{Q}]$, an integral basis identifies the additive group of ${\mathcal{O}}_L$ with ${\mathbb{Z}}^d$, soThe natural projection onto ${\mathcal{O}}_L/\mathfrak{a}$ is surjective. Thus the latter is an abelian group of order at most $|m{|}^d$, provingIn particular the ideal is a full-rank integer lattice; this permits the integral-basis argument in the next part.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.

We prove the conjugate product of a quadratic ideal identity $\mathfrak{a}\overline{\mathfrak{a}}=(N(\mathfrak{a}))$, which applies to the given two-generator ideal without assuming its displayed generators form an integral basis.

The quadratic ring of integers of a number field has a basis $1,\omega$ over $\mathbb{Z}$: the vector 1 is primitive, since a rational algebraic integer must be an integer, and extends to an integral basis. Write ${\omega }^2-T\omega +U=0$, with $T,U\in \mathbb{Z}$. Let $B$ be the least positive integer in $\mathfrak{a}$ and let $s>0$ generate the set of coefficients of $\omega$ in elements of $\mathfrak{a}$. Its lattice basis can be chosen as $B,r+s\omega$, with $r\in \mathbb{Z}$. Since $B\omega \in \mathfrak{a}$, comparison of coefficients shows $s\mid B$ and $s\mid r$. Write $B=sd$, $r=se$ and $\theta =e+\omega$. ThenThe scaled ideal $\mathfrak{j}=s^{-1}\mathfrak{a}$ remains an ${\mathcal{O}}_L$-ideal. From ${\theta }^2-T^{\prime }\theta +U^{\prime }=0$ we get $U^{\prime }\in \mathfrak{j}\cap \mathbb{Z}=d\mathbb{Z}$, so $d\mid U^{\prime }$. The product $\mathfrak{j}\bar{\mathfrak{j}}$ is generated as an ideal by $d^2,d\theta ,d\bar{\theta },U^{\prime }$ and is contained in $(d)$.

Moreover $\gcd (d,T^{\prime },U^{\prime }/d)=1$. If a prime $\ell$ divided all three, then $\ell \mid T^{\prime }$ and ${\ell }^2\mid U^{\prime }$, so $\theta /\ell$ would satisfy the monic integer polynomial $z^2-(T^{\prime }/\ell )z+U^{\prime }/{\ell }^2$. It would lie in ${\mathcal{O}}_L$, impossible because its $\omega$ coefficient in the integral basis is $1/\ell$. Bezout identity now expresses one as an integer combination of $d,T^{\prime },U^{\prime }/d$. Multiplying by $d$, the terms are $d^2$, $d{T}^{\prime }=d\theta +d\bar{\theta }$, and $U^{\prime }$, all in the product. Hence $d$ belongs to it, proving $\mathfrak{j}\bar{\mathfrak{j}}=(d)$.

The determinant of the lattice basis $sd,s(e+\omega )$ gives the ideal norm $N(\mathfrak{a})=s^{2}d$. ConsequentlyIn particular $⟨b,\alpha ⟩⟨b,\bar{\alpha }⟩$ is principal. This does not require the original $b$ to be nonzero or the ideal itself to be principal.

Let ${\omega }_1,\dots ,{\omega }_n$ be an integral basis of ${\mathcal{O}}_K$. Since each ${\alpha }_j$ is an algebraic integer, there is an integer matrix $A=(a_{ij})$ such thatThe ${\alpha }_j$ form a $\mathbb{Q}$-basis, so $\det A\ne 0$.

Products of algebraic integers are algebraic integers, and their traces are rational algebraic integers, hence ordinary integers. Therefore every entryis integral. Part (i) makes the trace Gram matrix nonsingular, so $\Delta ({\alpha }_1,\dots ,{\alpha }_n)$ is a nonzero integer.

Let $G_{\omega }$ and $G_{\alpha }$ be the two trace Gram matrices. A change of basis givesTaking determinants yields the discriminant-index formula for an integral latticewhere $d_K=\det G_{\omega }$ is the field discriminant. Since $\det A$ is a nonzero integer,Equality holds exactly when $|\det A|=1$, which is equivalent to $A$ being unimodular and to ${\alpha }_1,\dots ,{\alpha }_n$ being a $\mathbb{Z}$-basis of ${\mathcal{O}}_K$. Thus the minimum is the positive integer