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Algebraic number theory studies integer factorization, divisibility, prime ideals, and units after extending the rational numbers to a number field. Its basic objects are a number field $K$ and its ring of integers ${\mathcal{O}}_K$: the counterparts of $\mathbb{Q}$ and $\mathbb{Z}$. The field norm turns multiplication in $K$ into multiplication of rational numbers, and takes algebraic integers to integers. These connections let linear algebra and Euclidean geometry answer arithmetic questions.^[ref]

A first example is $K=\mathbb{Q}(\sqrt{2})$ with ring of integers $\mathbb{Z}[\sqrt{2}]$. The integer $7$ acquires the factorizationEach factor has field norm $7$. Neither is a unit, and each is an irreducible element: any factorization into two nonunits would factor the prime number $7$ into two integers of absolute value greater than one. Arithmetic in a larger ring thus reveals structure invisible in ordinary integer factorization.

There is also a complication: unique factorization of elements can fail, as in the failure of unique factorization in the ring of integers generated by square root of minus five. The general replacement is prime ideal factorization. The ideal class group measures the obstruction to making every ideal a principal ideal.

The historical motivation is visible in Richard Dedekind's original 1877 exposition: he begins with ordinary integer factorization, passes to Gaussian integers, and then uses a quadratic example to explain why ideals are needed. His later section establishes unique factorization of ideals in a number field. This is a direct record of the development of the theory, rather than a claim that all its ideas originated in one year.^[ref][ref]

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

Every ideal class of this real quadratic number field contains an integral ideal of ideal norm at most $\sqrt{14}<4$, by the diamond form of the Minkowski convex body theorem. The only possible norms are one, two and three. The sole prime ideal above two has norm two and is principal, generated by $4+\sqrt{14}$, whose field norm is two. The prime three is inert, since $X^2-14$ is irreducible modulo three, so there is no ideal of norm three. Every class therefore has a principal small representative, and the ideal class group is trivial.

This illustrates why the small-ideal bound can prove a class number without listing arbitrary ideals: only prime ideals of a few possible norms need to be tested. Real quadratic fields require the absolute field norm; the generator here happens to have positive norm.

The S-unramified power class group of a number field is finite. For $[a]\in K(S,n)$, the fractional ideal of $a$ away from $S$ is an $n$th power ${\mathfrak{a}}^n$. Sending $[a]$ to $[\mathfrak{a}]$ gives an exact sequenceThe S-unit group is finitely generated, and the localized ideal class group is a quotient of the finite ordinary ideal class group. Both end groups are finite. This is the arithmetic finiteness needed in the Kummer-theoretic proof of the weak Mordell-Weil theorem.

The ideal class group of $K=\mathbb{Q}(\sqrt{-6})$ has order two, andThis quadratic extension of $K$ is unramified at every finite prime and therefore has the degree required of the Hilbert class field.

The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

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.

Put $\omega =(1+\sqrt{229})/2$, $P=(3,\omega )$ and $Q=(5,\omega -4)$. The Minkowski bound for ideal classes is $\sqrt{229}/2<8$, so the ideal class group is generated by primes above three and five; two and seven are inert. The prime ideal factorizations $(6+\omega )=PQ$ and $(15+2\omega )=P^3$ reduce it to a cyclic group of order dividing three. It is not trivial: the unit $ϵ=(15+\sqrt{229})/2$ has field norm minus one. If an algebraic integer had norm of absolute value three, multiplying by a power of this unit would make both real absolute values at most $\sqrt{3ϵ}$. For $a=x+y\omega$, this bounds $|y|\sqrt{229}=|a-a^{\prime }|\le 2\sqrt{3ϵ}<\sqrt{229}$, forcing $y=0$, which cannot have norm of absolute value three.

Put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$ and its discriminant is $-120$. The Minkowski bound for ideal classes $4\sqrt{30}/\pi <7$ shows that the ideal class group is generated by the classes of the ramified prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$ and $P_5=(5,\theta )$. There is no additional prime ideal of norm four because two ramifies with residue degree one. The relations $P_i^2=(i)$ and $(\theta )=P_2{P}_3{P}_5$ show that these classes have order dividing two and their product is one. None is a principal ideal: the element field norm $a^2+30b^2$ cannot equal two, three or five. Thus there are exactly four classes, with the displayed group structure. This also supplies the class-number input for the Hilbert class field of Q of square root minus thirty.

Write $\eta =\sqrt{-34}$. The ring of integers is $\mathbb{Z}[\eta ]$ with discriminant $-136$. The Minkowski bound for ideal classes is less than eight, so classes of prime ideals of norm at most seven generate the ideal class group. These are the ramified prime $P_2=(2,\eta )$ and conjugate pairs at five and seven; three is inert with prime norm nine. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Conjugate prime classes are inverse, so these three classes suffice.

The field norm identities $N(4+\eta )=50$ and $N(1+\eta )=35$ give $(4+\eta )=P_2{P}_5^2$ and $(1+\eta )={\bar{P}}_5{P}_7$, by reduction modulo the indicated primes and equality of norms. Thus $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$. Since $P_2^2=(2)$ but $a^2+34b^2=2$ has no integer solution, $[P_2]$ has order two. Therefore $[P_5]$ has order four and generates the entire group, proving the displayed isomorphism.

The ring of integers is $\mathbb{Z}[\sqrt{35}]$ and the field discriminant is $140$. The Minkowski bound for ideal classes is $\sqrt{35}<6$. At three the defining polynomial is irreducible, so the only prime ideals needed to generate the ideal class group are $P_2=(2,\sqrt{35}-1)$ and $P_5=(5,\sqrt{35})$, of ideal norms two and five. They satisfy $P_2^2=(2)$, $P_5^2=(5)$ and $P_2{P}_5=(5+\sqrt{35})$, so one class of order of a group element at most two generates. The field norm equation $a^2-35b^2=\pm 2$ is impossible modulo five, proving $P_2$ is not a principal ideal. Hence the ideal class group has exactly two elements.

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.

The map from an idele to its finite-place fractional ideal is surjective, has kernel the product of the infinite multiplicative groups and finite-place unit groups, and sends diagonal elements to principal fractional ideals. It realizes the ideal class group as the corresponding quotient of the idele group.

For the finite layers of a Zp-extension of a number field, the p-primary ideal class groups eventually satisfy the displayed formula with integer invariants $\mu ,\lambda \ge 0$ and $\nu \in \mathbb{Z}$. The first two are the Iwasawa invariants of the unramified Iwasawa module. The Iwasawa module structure theorem and control of the finite-layer norm and ramification terms give the result. For a cyclotomic extension of an abelian field, the Ferrero-Washington theorem gives $\mu =0$. This growth theorem records two invariants; it is weaker than identifying the full characteristic ideal in the Iwasawa main conjecture.

Iwasawa theory studies arithmetic objects in infinite towers of number fields, often Zp-extensions. Passing to inverse limits turns growing ideal class groups, unit groups and Galois groups into modules over an Iwasawa algebra. Their algebraic ranks and characteristic ideals organize the growth and relate it to p-adic L-functions.

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.

Parity and valuations at 3 and 7 make the ideals $(y+\sqrt{-21})$ and $(y-\sqrt{-21})$ coprime in any hypothetical solution. Their product is a cube, so the first is an ideal cube. The ideal class group has exponent two, forcing its cube root ideal to be principal. Cubing $a+b\sqrt{-21}$ would then give coefficient $3b(a^2-7b^2)$ on $\sqrt{-21}$, which cannot equal one. This proves the nonexistence of integer solutions without assuming unique element factorization.

The ring $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain for the absolute field norm. Approximate both rational coefficients of a quotient by nearest integers. If the errors are $u,w\in [-1/2,1/2]$, then $|u^2-2w^2|\le 1/2<1$, giving a remainder of strictly smaller absolute norm. In particular this ring is a principal ideal domain and has trivial ideal class group.

The norm-one idele group is the kernel of the idelic modulus. Its quotient by the diagonal $K^{\times }$ is compact for a number field and surjects onto the ideal class group.

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.