Ideal class group

The ideal class group is the group of nonzero fractional ideals modulo principal fractional ideals.

The narrow ideal class group of a number field is the group of nonzero fractional ideals modulo principal fractional ideals generated by totally positive elements. It is the ray class group for the modulus consisting of all real Archimedean places and trivial finite part. The forgetful map to the ordinary ideal class group is surjective, with kernel $\{\pm 1{\}}^{r_1}/\mathrm{sgn}({\mathcal{O}}_K^{\times })$ under the unit signature map.

The ideal-form correspondence for imaginary quadratic fields and the reduction bound $a\le \sqrt{31/3}$ give exactly the forms $[1,1,8]$, $[2,1,4]$ and $[2,-1,4]$. Thus the class number is three and the ideal class group is cyclic of order three.

For a negative fundamental discriminant $D$, primitive positive definite integral binary quadratic forms of discriminant $D$, modulo proper equivalence, correspond bijectively to the ideal class group of $\mathbb{Q}(\sqrt{D})$. The displayed ideal gives one conventional choice of this correspondence. Reduced positive definite binary quadratic forms give unique class representatives with the usual boundary sign convention, making class-number computations finite.

In $K=\mathbb{Q}(\sqrt{-14})$, the Minkowski bound for ideal classes is less than five. The ideals $\mathfrak{p}=(2,\sqrt{-14})$ and $\mathfrak{q}=(3,1+\sqrt{-14})$ generate all classes, with ${\mathfrak{p}}^2=(2)$ and $(2-\sqrt{-14})=\mathfrak{p}{\mathfrak{q}}^2$. No element has field norm two, so $[\mathfrak{p}]$ has order two and $[\mathfrak{q}]$ has order four.

An ideal class is an element of the ideal class group, represented by a nonzero fractional ideal. Two fractional ideals represent the same class exactly when their quotient is a principal ideal.

The class number of a number field is the order $h_K=|\mathrm{Cl}(K)|$ of its ideal class group.

If a number field of degree $n$ and signature $(r_1,r_2)$ has discriminant $d_K$, every ideal class contains an integral ideal $I$ with

Since every nonzero integral ideal has norm at least one, the Minkowski bound for ideal classes impliesfor every number field of degree $n$ and signature $(r_1,r_2)$.

For $K=\mathbb{Q}(\sqrt[3]{3})$ with ${\mathcal{O}}_K=\mathbb{Z}[\sqrt[3]{3}]$, the field discriminant is $-243$ and the Minkowski bound is $8\sqrt{3}/\pi <5$. The prime ideals above $2$ and $3$ are principal, so every ideal class has a principal representative and

The imaginary quadratic field $K=\mathbb{Q}(\sqrt{-163})$ has discriminant $-163$, and its Minkowski bound for ideal classes isThe polynomial $X^2-X+41$ defining its ring of integers of a quadratic field has no root modulo $2,3,5,$ or $7$, so these rational primes are inert. Factoring an integral ideal of norm at most eight into prime ideals then shows that its only possible nontrivial factor is the principal ideal $(2)$, of norm four. Every ideal class is therefore principal, and

The field has ring of integers $\mathbb{Z}[\sqrt{-33}]$, discriminant $-132$, and Minkowski bound below eight. The nonprincipal ramified ideals above two and three generate four classes, and the split primes above seven introduce no new class. Hence

The ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, and its Minkowski bound for ideal classes is $4\sqrt{22}/\pi <6$. The primes three and five are inert, whileThe ideal ${\mathfrak{p}}_2$ is nonprincipal because $a^2+22b^2=2$ has no integer solution. Every class is represented by an ideal of norm at most five, so $[{\mathfrak{p}}_2]$ is the only possible nontrivial class and the class group is $C_2$.

The equation $y^3=x^2+22$ has no integer solutions. Factoring it in $\mathbb{Z}[\sqrt{-22}]$, the two conjugate factors are coprime. Unique factorization of ideals in a number field and the absence of three-torsion in the class group force $x+\sqrt{-22}$ to be a cube up to a unit. Writing it as $(a+b\sqrt{-22}{)}^3$ makes its $\sqrt{-22}$ coefficientwhich has no integer solution.

Let $m\ge 2$, let $p=4m-1$ be prime, and suppose $\mathbb{Q}(\sqrt{-p})$ has class number one. If $n\ge 0$ and $n^2+n+m<m^2$, then $n^2+n+m$ is prime. The key identity is the algebraic normA hypothetical prime divisor below $m$ would split and yield an algebraic integer of that norm, but the positive norm form $(a^2+p{b}^2)/4$ represents no prime below $m$.

The order of the class of a fractional ideal $I$ is the least positive integer $n$ for which $I^n$ is principal.

Let $p\equiv 11(\mathrm{mod}12)$, $K=\mathbb{Q}(\sqrt{-p})$, and let $\mathfrak{p}$ be a prime-ideal factor of $(3)$. If $p>3^{n+2}$ and $n$ is odd, then $[\mathfrak{p}]$ does not have order $n$. Otherwise ${\mathfrak{p}}^n=(\alpha )$ would givefor $\alpha =(a+b\sqrt{-p})/2$. The bound forces $b=0$, after which $3^n=N(\alpha )$ is an integer square, impossible for odd $n$.

If every integral ideal $I$ contains nonzero $\alpha$ with $|N(\alpha )|\le CN(I)$, then $(\alpha )I^{-1}$ is an integral representative of $[I{]}^{-1}$ with norm at most $C$.

Bounded-norm representatives prove class-group finiteness because only finitely many integral ideals have norm at most a fixed constant.

Every element of a finite ideal class group has finite order. Therefore every nonzero ideal has a positive power that is principal.