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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 multiplicative field norm $N(m+n\sqrt{-7})=m^2+7n^2$ is an integer. Units have field norm one and are exactly $\pm 1$. A nontrivial factorization of 2 would force both factors to have field norm two, which the displayed quadratic form cannot represent. Thus 2 is an irreducible element. It is not a prime element, since it divides $(1+\sqrt{-7})(1-\sqrt{-7})=8$ but neither factor. This is an example in a nonmaximal quadratic order, rather than the full ring of integers of $\mathbb{Q}(\sqrt{-7})$; the precise ring matters for factorization.

The Brauer class of a central simple algebra is its equivalence class under adjoining matrix-algebra factors. Two algebras have the same class exactly when their central division-algebra representatives are isomorphic. The zero class consists of split matrix algebras; the inverse of $[A]$ is the class of the opposite algebra. For a cyclic algebra $(L/K,\sigma ,a)$, the class is zero exactly when $a$ is a field norm from $L$.

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.

Let $q\nmid m$ be outside the excluded primes, let $\zeta$ have order $m$, and let $\rho$ be a primitive qth root of unity. Galois equivariance identifies the conjugates with $\Phi ({\rho }^i\zeta )$ for $1\le i\le q-1$. The distribution axiom includes the missing $i=0$ factor, so the field norm is $\Phi ({\zeta }^q)/\Phi (\zeta )$. Arithmetic Frobenius sends $\zeta$ to ${\zeta }^q$, giving the displayed factor. At higher powers of an already ramified prime all q translates occur instead. These are different norm relations; norm compatibility in just a p-power tower does not supply the auxiliary-prime relation by itself.

The diagonal multiplicative group of a number field is discrete in its idele group. Require local units at every finite place and an archimedean neighbourhood of one. A diagonal element $x$ meeting these restrictions has integral $x-1$. If $x\ne 1$, its nonzero integer field norm cannot have absolute value less than one, while sufficiently small archimedean differences would force exactly that.

Let $L$ be a fixed number field and let $V\subset L$ be a fixed rational linear subspace for which the associated norm form has no unit-family degeneracy. There are effective constants $A,C>0$ such that every $z\in V\cap {\mathcal{O}}_L$ with nonzero field norm $m$ satisfiesTo prove this, factor the principal ideal $(z)$, write its generators as a bounded factor times powers of fundamental units, use the linear relations defining $V$, and apply the Baker lower bound for a homogeneous linear form in logarithms to bound the unit exponents by $O(\log |m|)$. For $L=\mathbb{Q}(\alpha )K$ with $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$, $[K:\mathbb{Q}]=2$, and $V={\mathrm{span}}_{\mathbb{Q}}\{1,\alpha ,\omega \}$ for a basis $1,\omega$ of $K$, the coprime degrees rule out the degeneracy.

For odd $p$ and p-adic units $u,v$, the quadratic Hilbert symbol isThe Legendre symbols record the two unit square classes. For $p=2$, put $ϵ(u)=(u-1)/2\mathrm{mod}2$ and $\omega (u)=(u^2-1)/8\mathrm{mod}2$. ThenThe basis $-1,2,5$ of the square-class group of the 2-adic numbers gives this formula: its only negative basis pairings are $(-1,-1{)}_2$ and $(2,5{)}_2=(5,2{)}_2$. The first is obstructed by sums of squares modulo four, the second by $x^2-2y^2$ modulo eight; the remaining pairings have explicit field norm representatives.

For a finite field extension $L/K$, multiplication by $\alpha \in L$ is a $K$-linear map $m_{\alpha }:L\to L$. Its determinant is the field norm:This does not depend on the chosen basis. If $L/K$ is a separable field extension, thenwhere the product runs over all $[L:K]$ field embeddings into an algebraic closure of $K$. Embeddings are counted even when several give the same value on $\alpha$: for $r\in K$, the result is $r^{[L:K]}$. This avoids ambiguity in the phrase “product of algebraic conjugates”.^[ref]

For $L=\mathbb{Q}(\sqrt{2})$ over $K=\mathbb{Q}$, write $\alpha =a+b\sqrt{2}$. In the basis $1,\sqrt{2}$, multiplication by $\alpha$ has columns $(a,b)$ and $(2b,a)$, henceThe determinant calculation and the product of the two field embeddings give the same answer. In particular,A field norm is an arithmetic invariant; it is not a nonnegative norm measuring length, and need not be positive on nonzero elements.

The field norm is multiplicative, $N(\alpha \beta )=N(\alpha )N(\beta )$, because $m_{\alpha \beta }=m_{\alpha }{m}_{\beta }$ and determinants are multiplicative. It vanishes exactly at zero. For a number field over $\mathbb{Q}$, it takes algebraic integers to integers, and an algebraic integer is a unit exactly when its field norm is $1$ or $-1$.^[ref][ref]

This makes the field norm a useful test for divisibility. If nonzero $\alpha ,\beta \in {\mathcal{O}}_L$ and $\alpha$ divides $\beta$, then the integer $N(\alpha )$ divides $N(\beta )$. For example, $3+\sqrt{2}$ has field norm $7$, so it cannot divide $2$, whose field norm is $4$. It does divide $7$, with quotient $3-\sqrt{2}$. This necessary condition can reject a proposed factor without solving for its quotient.

For a degree-$m$ extension $L/K$, substitute a $K$-basis of $L$ into a homogeneous degree-$d$ polynomial in $N>d$ variables over $L$, then take its field norm. The resulting degree-$md$ polynomial has $mN>md$ variables over $K$. A nonzero zero over a C1 field gives a nonzero zero of the original form because the norm of a field element vanishes only at zero. Separability is unnecessary.

For a cyclic extension of number fields, an element of the base field is a global field norm exactly when it is a norm at every finite and infinite completion. The splitting criterion for a cyclic algebra encodes the norm obstruction as a Brauer group class. The Albert-Brauer-Hasse-Noether theorem makes a class that splits at every completion split globally, giving the result. Cyclicity is essential; arbitrary noncyclic extensions can have additional local-global norm obstructions.

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.

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.

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.

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.

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.

The product formula uses finite-place factors given by ideal norms, the ordinary modulus at real embeddings, and the modulus squared at complex places. The complex square accounts for the pair of conjugate embeddings. Equivalently ordinary complex modulus can be used with weight two in the product. These normalizations cancel the absolute global field norm.