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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 different exponent at a prime ideal $\mathfrak{P}$ is its exponent in the prime ideal factorization of the different ideal. It equals the local different exponent of the corresponding extension of completions. The different exponent and tame ramification theorem detects unramified, tame, and wild behavior without assuming a Galois extension.

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.

Let $R={\mathcal{O}}_k$, $S={\mathcal{O}}_K$ and $n=[K:k]$. For each prime ideal $\mathfrak{P}$ of $S$ above $\mathfrak{p}$ of $R$, set $f_{\mathfrak{P}}=[S/\mathfrak{P}:R/\mathfrak{p}]$. The relative norm of a fractional ideal can be specified on the unique prime ideal factorization byHere is why this definition is the intrinsic relative ideal norm from norms of elements, and why the residue weights occur. Localize at $\mathfrak{p}$. The finite torsion-free $R_{\mathfrak{p}}$-module $S_{\mathfrak{p}}$ is free of rank $n$, because $R_{\mathfrak{p}}$ is a discrete valuation ring. Multiplication by a nonzero integral $a$ has determinant $N_{K/k}(a)$. The Smith normal form givesDecompose the quotient at the prime ideals above $\mathfrak{p}$. In $S_{\mathfrak{P}}$ each quotient ${\mathfrak{P}}^j/{\mathfrak{P}}^{j+1}$ is one-dimensional over $S/\mathfrak{P}$, so has length $f_{\mathfrak{P}}$ over $R_{\mathfrak{p}}$. ThereforeFor an integral ideal $I$, all norms of elements of $I$ have valuations at least $\sum f_{\mathfrak{P}}{v}_{\mathfrak{P}}(I)$. These minima can be attained simultaneously at the finitely many prime ideals above a fixed $\mathfrak{p}$. In fact the Chinese remainder theorem for ideals, applied inside the finitely generated module $I$, chooses an $a\in I$ with nonzero image in each $I/\mathfrak{P}I$. Equivalently $v_{\mathfrak{P}}(a)=v_{\mathfrak{P}}(I)$ at all these primes. The generated ideal of the element norms consequently has exactly the displayed valuations. Clearing a denominator extends this description to fractional ideals.

Multiplication of fractional ideals adds their prime valuations. The displayed formula therefore provesIt also gives $N_{K/k}((a))=(N_{K/k}(a))$.

Finally, reduction of the free rank-$n$ module $S_{\mathfrak{p}}$ modulo $\mathfrak{p}$ has dimension $n$ over $R/\mathfrak{p}$. The factorization $\mathfrak{p}S={\prod }_i{\mathfrak{P}}_i^{e_i}$ and the Chinese remainder theorem for ideals identify this reduction with the product of the quotients by ${\mathfrak{P}}_i^{e_i}$. Their filtrations have $e_i$ successive quotients of dimension $f_i$ over $R/\mathfrak{p}$. Adding dimensions proves the fundamental identity for prime decomposition:

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

Since $26\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{26}]$ with discriminant $104$. Write $P=(2,ϵ+1)=(2,\sqrt{26})$ and $Q_{\pm }=(5,ϵ\pm 1)$. Reduction of $X^2-26$ modulo two gives a double root, and modulo five gives the roots $\pm 1$. Thus $N(P)=2$, $N(Q_{\pm })=5$, and $(5)=Q_{+}{Q}_{-}$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot 4=2$. The element $ϵ+1=6+\sqrt{26}$ belongs to $P$ and $Q_{+}$ and has field norm ten, so equality of ideal norms givesThe Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every ideal class contains an integral ideal of ideal norm at most five. Three is inert and gives no prime ideal of ideal norm three; ideals of ideal norm four are $P^2$, principal. Thus the class group is generated by $P,Q_{+},Q_{-}$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have field norm $\pm 2$, but $x^2-26y^2=\pm 2$ would give $x^2\equiv 2$ or $11(\mathrm{mod}13)$, neither a quadratic residue. Hence the class number is two.

The unit $ϵ=5+\sqrt{26}$ has field norm $-1$. To show it is fundamental, suppose $1<u<ϵ$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm 1/u$, so $y=(u-u^{\prime })/(2\sqrt{26})$ is a positive integer less than two. Thus $y=1$, and $x^2-26=\pm 1$ forces $x=\pm 5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of ${ϵ}^{-1}$ puts it in $[1,ϵ)$, proving that all units are $\pm {ϵ}^n$.

An element of field norm $\pm 10$ has principal ideal $P{Q}_{+}$ or $P{Q}_{-}$ by prime ideal factorization. The first is generated by $ϵ+1$; the second by $ϵ-1=4+\sqrt{26}$, of field norm $-10$. Two generators of the same principal ideal differ by a unit. Therefore all, and only, the integer solutions areThe two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two field norm signs.

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.

Let $K=\mathbb{Q}(\alpha )$ with $\alpha$ an algebraic integer, let $m\in \mathbb{Z}[X]$ be its monic minimal polynomial of an algebraic element, and write $A=\mathbb{Z}[\alpha ]$. The Kummer-Dedekind theorem applies to a rational prime $p$ not dividing the finite index $[{\mathcal{O}}_K:A]$. Ifwith distinct monic irreducible polynomials $g_j$, choose monic integer lifts ${\tilde{g}}_j$. Then the distinct prime ideals above $p$ areIn particular, $p$ is unramified precisely when $\overline{m}$ has no repeated irreducible factor. The exclusion of primes dividing the index is essential.

Here is the quotient-ring proof of the Kummer-Dedekind theorem. The finite group ${\mathcal{O}}_K/A$ has order coprime to $p$. Localizing at the integers coprime to $p$ therefore gives $A\otimes {\mathbb{Z}}_{(p)}={\mathcal{O}}_K\otimes {\mathbb{Z}}_{(p)}$, and reduction yieldsThe second isomorphism follows from $A\simeq \mathbb{Z}[X]/(m)$. The Chinese remainder theorem givesEach factor is a local ring with unique maximal ideal generated by $g_j$ and residue field ${\mathbb{F}}_p[X]/(g_j)$. Pulling its maximal ideal back to ${\mathcal{O}}_K$ gives exactly ${\mathfrak{p}}_j$; these exhaust the primes above $p$. This proves the asserted residue degrees and their independence of the chosen lifts.

It remains to prove the exponents rather than simply identify the primes. Unique prime ideal factorization in the Dedekind domain ${\mathcal{O}}_K$ writes $(p)={\prod }_j{\mathfrak{p}}_j^{b_j}$. At ${\mathfrak{p}}_j$, its discrete valuation ring has $p=u{\pi }^{b_j}$, with $u$ a unit. Thus the maximal ideal of ${\mathcal{O}}_{K,{\mathfrak{p}}_j}/p{\mathcal{O}}_{K,{\mathfrak{p}}_j}$ has nilpotence index of an ideal $b_j$: its $b_j$th power is zero and its $(b_j-1)$st is nonzero. In ${\mathbb{F}}_p[X]/(g_j^{e_j})$ this index is $e_j$, since $g_j^{e_j}=0$ but $g_j^{e_j-1}\ne 0$. The displayed product isomorphism identifies these local factors, hence $b_j=e_j$. This completes the theorem's proof. The same local argument works over a base number field whenever ${\mathcal{O}}_K[\alpha ]$ equals the full relative ring of integers after localizing at the chosen base prime.

For the required cyclotomic field, the allowed ring of integers formula makes the index one. Its cyclotomic polynomial isFor $p\notin \{2,5\}$, the polynomial $X^{80}-1$ has derivative $80X^{79}$ coprime to it modulo $p$. Its divisor ${\overline{\Phi }}_{80}$ is therefore squarefree, so the Kummer-Dedekind theorem gives no ramification.

The cyclotomic reduction at a conductor prime is explicit at the remaining two primes:The quartic over ${\mathbb{F}}_2$ has no linear factor. Its only possible irreducible quadratic divisor would be $X^2+X+1$, whose roots have order three, whereas its own roots have order five. Hence it is irreducible. The two quartics over ${\mathbb{F}}_5$ are also irreducible: their roots have multiplicative order sixteen because their fourth powers are $2$ or $3$, both of order four; the finite-field Frobenius automorphism has orbit length four on an element of order sixteen, since $5^4\equiv 1(\mathrm{mod}16)$ and neither $5$ nor $5^2$ is $1$ modulo sixteen.

Consequently the prime ideal factorization above $2$ has one prime with ramification index eight and residue degree four. Above $5$ it has two primes, each with ramification index four and residue degree four. In particular,

Put $w=\sqrt{-14}$. The ring of integers is $\mathbb{Z}[w]$, with discriminant $-56$. The Minkowski bound for ideal classes for an imaginary quadratic field says that every ideal class has an integral representative of norm at most $(2/\pi )\sqrt{56}<5$. Therefore the ideal class group is generated by primes above two and three: norm-four representatives contribute only products of primes above two.

The prime two ramifies: ${\mathfrak{p}}_2=(2,w)$ has norm two and ${\mathfrak{p}}_2^2=(2)$. It is not principal, since an element of norm two would require $a^2+14b^2=2$, which has no integral solution. Thus $[{\mathfrak{p}}_2]$ has order two. Since $X^2+14=(X-1)(X+1)$ modulo three, three splits into ${\mathfrak{p}}_3=(3,w-1)$ and its conjugate, whose classes are inverses.

The element $2+w$ has norm $18$. It belongs to ${\mathfrak{p}}_3$ but not its conjugate, and its norm has exactly one factor two. Unique prime ideal factorization therefore givesHence $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]\ne 1$. Since all generating classes are powers of $[{\mathfrak{p}}_3]$, $\boxed{\mathrm{Cl}(\mathbb{Q}(\sqrt{-14}))\cong \mathbb{Z}/4\mathbb{Z}}$.

Put $s=\sqrt{-17}$. The ring of integers of a number field is $\mathbb{Z}[s]$ and the field discriminant is $-68$. The Minkowski bound for ideal classes says that each ideal class has an integral ideal representative of norm at most $(2/\pi )\sqrt{68}<6$.

The prime two ramifies as $(2)={\mathfrak{p}}_2^2$, where ${\mathfrak{p}}_2=(2,1+s)$. Three splits as $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, choosing ${\mathfrak{p}}_3=(3,1+s)$, whereas five is inert because $-17\equiv 3(\mathrm{mod}5)$ is not a square. The norm bound therefore leaves only the classes of $1$, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\overline{\mathfrak{p}}}_3$; a norm-four ideal is ${\mathfrak{p}}_2^2$ and is principal.

The ideal ${\mathfrak{p}}_2$ is not principal, since an element of norm two would solve $a^2+17b^2=2$. On the other hand $N(1+s)=18$, and $1+s$ is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$. Unique prime ideal factorization and the norm give $(1+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Thus $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2]$, which has order two, and

For the Diophantine equation, $x$ must be positive. If $x$ were even, $y^2\equiv 15(\mathrm{mod}32)$, impossible for a square. Thus $x$ is odd and $y$ even. If $17\mid x$, then $17\mid y$, but $y^2+17$ would have 17-adic valuation one while $x^5$ has valuation divisible by five, also impossible.

The ideals $(y+s)$ and $(y-s)$ are consequently coprime. A common prime divides $2s$ and hence lies over two or seventeen, both excluded above. Their product is $(x{)}^5$, so unique prime ideal factorization gives $(y+s)={\mathfrak{a}}^5$. Since the ideal class group has order four, $[\mathfrak{a}{]}^5=1$ forces $[\mathfrak{a}]=1$. Write $\mathfrak{a}=(a+bs)$ with integers $a,b$. The only units are $\pm 1$ by their norm, and the sign can be absorbed into the fifth power. Hence $y+s=(a+bs{)}^5$.

Comparing the coefficient of $s$ givesFor $b=1$, the bracket would be one, but it is four modulo five. For $b=-1$, the bracket would be minus one, but it is one modulo three for either possible square residue of $a$. Both cases are impossible. Therefore there are no integer solutions.

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.

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.

For $K=\mathbb{Q}(\sqrt{-163})$, the fundamental discriminant is $-163$ and the Minkowski bound for ideal classes isThus every ideal class contains an integral ideal of norm at most eight.

The ring of integers of a quadratic field is $\mathbb{Z}[\theta ]$ for $\theta =(1+\sqrt{-163})/2$, whose minimal polynomial is $X^2-X+41$. Its discriminant $-163$ is a quadratic nonresidue modulo each of $3,5,$ and $7$; directly modulo two, $X^2+X+1$ has no root. Hence these primes are inert by the splitting of rational primes in a quadratic field. The only prime ideal over one of them with norm at most eight is $(2)$, which has norm four and is principal. Unique prime ideal factorization now makes every ideal of norm at most eight principal. The Minkowski bound therefore provesThis is the Ideal class group of Q of square root of minus one hundred sixty-three.

Put $\alpha =\sqrt[n]{m}$. It is an algebraic integer because it satisfies the monic polynomial $X^n-m$. Choose a prime number $q$ dividing the square-free integer $m$, and choose a prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_L$ above $q$. Fromand prime ideal factorization, the $\mathfrak{p}$-adic exponents satisfybecause $q$ occurs to exponent one in $m$. Hence $n$ divides the ramification index of a prime ideal $e_{\mathfrak{p}}$. If $d=[L:\mathbb{Q}]$, then $e_{\mathfrak{p}}\le d$, while the equation ${\alpha }^n=m$ gives $d\le n$. Thereforeso

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

The polynomial of $\alpha$ satisfies the Eisenstein criterion at $p$. Let $\mathfrak{p}$ be any prime ideal of ${\mathcal{O}}_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak{p}}(\mathfrak{p})=1$, and write $e=v_{\mathfrak{p}}(p)$ for its ramification index of a prime ideal. Put $t=v_{\mathfrak{p}}(\alpha )$. Inthe constant term has valuation $e$, while every other $a_j{\alpha }^j$ has valuation at least $e+jt$. The non-Archimedean valuation property then forcesIndeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.

The fundamental inequality ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}}{f}_{\mathfrak{p}}\le n$ now shows that $\mathfrak{p}$ is the unique prime above $p$ and has residue degree one. At this prime,while $(p,\alpha )$ has no factor away from $p$. Unique prime ideal factorization therefore givesFinally $v_P(\alpha )=1$, soThis is total ramification from an Eisenstein polynomial.

The product formula says that every $x\in K^{\times }$ satisfiesFirst suppose that $x$ is an algebraic integer. Its principal ideal has the prime ideal factorizationTaking the ideal norm givesOn the other hand, the field norm is the product over embeddings, soEquating these expressions proves the formula for algebraic integers. Every nonzero element of $K$ is a quotient of two algebraic integers, and multiplicativity completes the proof.

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]