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For a negative discriminant of a binary quadratic form $D\equiv 0$ or $1(\mathrm{mod}4)$, the class number of a negative discriminant $h(D)$ counts proper equivalence classes of primitive binary quadratic forms that are positive definite and have discriminant $D$.

Uniqueness of the reduced positive definite binary quadratic form in each class makes this the number of reduced primitive forms. Enumeration of reduced binary quadratic forms proves finiteness and gives an explicit procedure. For example, $h(-4)=1$, represented by $[1,0,1]$, while $h(-39)=4$, with the four forms listed under reduced positive definite binary quadratic forms of discriminant minus thirty-nine.

Imprimitive forms do not enter $h(D)$. At $D=-32$, the reduced form $[2,0,4]$ has content two, so the three reduced forms listed under reduced forms of discriminant minus thirty-two contribute only two primitive classes. If $D$ is a fundamental discriminant, this form class number agrees with the class number of the imaginary quadratic field of discriminant $D$. For nonfundamental $D$, the associated order in a number field need not be the full ring of integers, so that distinction must be retained.

A class number problem asks which number field or order in a number field has a prescribed class number. The family is part of the question: fixing the value alone does not specify the degree, discriminant or type of field.

For imaginary quadratic fields, it becomes a problem about negative fundamental discriminants $D$ and the class number of a negative discriminant $h(D)$. Allowing nonfundamental discriminants instead includes an order in a number field smaller than the full ring of integers. A reduced positive definite binary quadratic form and the enumeration of reduced binary quadratic forms make each fixed negative-discriminant case an explicit finite calculation.

For example, class number one for discriminant minus sixty-seven proves $h(-67)=1$ by exhausting the reduced coefficient possibilities. In a number field, class number one means every ideal of the ring of integers is principal, so the ring is a principal ideal domain. This connects a numerical classification with the arithmetic question of whether ideal obstructions to unique factorization remain. A computation for one discriminant does not by itself classify all discriminants of that class number.

The polynomial discriminant satisfiesA polynomial has no repeated roots over the algebraic closure exactly when it is coprime to its derivative. Consequently the assumed squarefree reduction has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. On the other hand the power basis is a basis of the order in a number field $\mathbb{Z}[\alpha ]$, and the discriminant-index formula for an integral lattice givesHere the field discriminant $d_L$ is an integer. ThereforeThis is squarefree reduction excludes a prime from the power-basis index; its argument establishes the index condition before applying the Kummer-Dedekind theorem.

Let $L=\mathbb{Q}(\alpha )$, where $\alpha$ is an algebraic integer with monic minimal polynomial $f\in \mathbb{Z}[X]$, and put $I=[{\mathcal{O}}_L:\mathbb{Z}[\alpha ]]$. The Kummer-Dedekind theorem applies to a prime number $p$ with $p\nmid I$. Write its reduction as a product of polynomial factors over ${\mathbb{F}}_p$:where the $g_j$ are distinct monic irreducible polynomials, choose monic integral lifts $G_j$. Then the distinct prime ideals above $p$ are ${\mathfrak{p}}_j=(p,G_j(\alpha ))$, andThus the multiplicity of a factor is its ramification index of a prime ideal, and its degree is the residue degree. The hypothesis on the order in a number field is essential: one cannot infer prime ideal factorization from an arbitrary defining polynomial at a prime dividing its index. There is also a relative version over a base number field $K$: if ${\mathcal{O}}_L$ and ${\mathcal{O}}_K[\alpha ]$ agree after localization at $\mathfrak{p}$, factor the minimal polynomial over ${\mathcal{O}}_K/\mathfrak{p}$ and use $(\mathfrak{p},G_j(\alpha ))$ with the same conclusions.

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

Let $d\ne 0,1$ be a square-free integer, and let $K=\mathbb{Q}(\sqrt{d})$ be the associated quadratic field. ThenThe pair $1,{\omega }_d$ is an integral basis. The field discriminant is $d$ in the first case and $4d$ in the second. This statement requires $d$ to be a square-free integer: for example, $\mathbb{Q}(\sqrt{8})=\mathbb{Q}(\sqrt{2})$ but $\mathbb{Z}[\sqrt{8}]$ omits $\sqrt{2}$.^[ref]

For $d=2$, the integral basis is $1,\sqrt{2}$, and the field discriminant is $8$. For $d=5$, it is $1,\varphi$ with $\varphi =(1+\sqrt{5})/2$, and the field discriminant is $5$. The identity ${\varphi }^2=\varphi +1$ makes $\mathbb{Z}[\varphi ]$ closed under multiplication. Since $\sqrt{5}=2\varphi -1$, the smaller order in a number field $\mathbb{Z}[\sqrt{5}]$ consists precisely of $a+b\varphi$ with even $b$. This computes its index two and shows concretely why adjoining a square root does not always supply every algebraic integer.

In this second integral basis, the field norm isIn particular, $N(\varphi )=-1$ and ${\varphi }^{-1}=\varphi -1$: the extra algebraic integer is also a unit. The units of the quadratic field Q square root of five describe all its powers.