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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.

An atomic domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements. Every unique factorization domain is atomic, but atomicity alone does not imply uniqueness.

The ascending chain condition supplies existence of factorizations into irreducible elements. The Bezout identity makes every irreducible element prime, which supplies uniqueness up to order and multiplication by units.

This integral domain embeds in $\mathbb{C}$ but has the inequivalent factorizations $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$. The multiplicative norm $N(a+b\sqrt{-5})=a^2+5b^2$ takes neither the value two nor three, proving that the four displayed factors are irreducible elements. Its only units are $\pm 1$, so those two factorizations cannot be related by units and reordering.

Let $\pi$ be an irreducible element of a principal ideal domain $R$, and suppose $\pi \nmid a$. The ideal $(\pi ,a)=(d)$ has generator $d$ dividing $\pi$. Irreducibility makes $d$ a unit or an associate of $\pi$; the latter would imply $\pi \mid a$. Hence $(\pi ,a)=R$, giving $u\pi +va=1$ for some $u,v\in R$. If $\pi \mid ab$, multiply this identity by $b$: both $u\pi b$ and $vab$ are divisible by $\pi$, so $\pi \mid b$. This is precisely the prime element property.

In this localization of a ring, a nonzero fraction $a/b$ with odd $b$ is a unit exactly when $a$ is odd: its reciprocal then has odd denominator; if $a$ is even, an inverse would equate an even integer with an odd one after clearing denominators. Consequently every nonzero element has a unique formHere $r$ is the exponent of $2$ in the numerator of the reduced fraction, its nonnegative 2-adic valuation.

Every nonzero ideal has the form $dR$ by part (c). Write the nonzero integer $d=2^{r}v$ with $v$ odd; since $v$ is a unit, $dR=2^{r}R$. Hence the nonzero proper ideals are exactlyThe zero ideal and $R=2^{0}R$ complete the full list. These ideals are distinct: $2^r$ does not belong to $2^{r+1}R$, since $1/2\notin R$.

The reduction map $a/b\mapsto a\mathrm{mod}2$ is a well-defined surjective ring homomorphism $R\to {\mathbb{F}}_2$ with kernel of a ring homomorphism $2R$; odd denominators reduce to $1$. Thus $R/2R\cong {\mathbb{F}}_2$ is a field, so $2R$ is a prime ideal and $2$ is a prime element. Every nonzero nonunit is associated to $2^r$ with $r\ge 1$. If $r\ge 2$, the factorization $2^r=2\cdot 2^{r-1}$ is into two nonunits, so it is not an irreducible element and therefore not a prime element. We concludeThe failed step in the argument of part (b) is the guarantee that the product plus one is a nonunit and hence has an irreducible factor. Here every prime element is an even fraction, so a product of prime elements plus $1$ is an odd fraction, therefore a unit. Already $2+1=3$ has no prime divisor in this ring. In $k[t]$, positive degree prevented that failure.

An element $a$ of an integral domain is an irreducible element when it is nonzero and not a unit, and any factorization $a=bc$ has at least one unit factor. The exclusions of zero and units are part of the definition.

An element $a$ is a prime element when it is nonzero and not a unit, and $a\mid bc$ implies $a\mid b$ or $a\mid c$. If $a=bc$ and $a$ is a prime element, assume for example $b=ad$. Cancellation in $a=adc$ gives $dc=1$, so $c$ is a unit. This proves every prime element is an irreducible element.

In a Euclidean domain, each nonzero ideal has a generator of least Euclidean value: division of another member by that generator leaves a remainder in the ideal of smaller value, so the remainder must vanish. Let $a$ be an irreducible element and suppose $a\mid bc$ but $a\nmid b$. Write $(a,b)=(d)$. Since $d\mid a$, irreducibility says $d$ is a unit or is associate to $a$. The second case would give $a\mid b$, so $d$ is a unit. Hence $1=ra+sb$ for some $r,s$. Multiplying by $c$ shows $a\mid c$. Thus irreducibles in a Euclidean domain are prime.

For $R=\mathbb{Z}[i\sqrt{7}]$, use the multiplicative integer field norm $N(m+ni\sqrt{7})=m^2+7n^2$. Units have field norm one and are exactly $\pm 1$. If $2=uv$ were a product of nonunits, their norms would be integers greater than one multiplying to four, hence both would have field norm two. But $m^2+7n^2=2$ has no integer solution: $n\ne 0$ makes the left side at least seven, and $n=0$ would give $m^2=2$. Therefore 2 is an irreducible element.

On the other hand,is divisible by 2, while neither factor is, because divisibility by 2 requires both integer coefficients to be even. Thus 2 is not a prime element in this ring. It is also not a prime element in $\mathbb{Z}[i]$: $2\mid (1+i{)}^2=2i$, but $2\nmid 1+i$. In fact $2=(1+i)(1-i)$ is already a nontrivial factorization there.

A unit in a ring is an element $u$ with a multiplicative inverse in $R$, so $uv=1$ for some $v\in R$.

An irreducible element is a nonzero nonunit $a$ such that every factorization $a=bc$ has at least one of $b,c$ a unit in a ring.

A prime element is a nonzero nonunit $p$ such that $p\mid bc$ implies $p\mid b$ or $p\mid c$. Equivalently its principal ideal $(p)$ is a nonzero prime ideal. These are definitions in the given commutative ring; the equivalence between irreducibility and primality requires additional hypotheses on the ring.

A Euclidean domain is an integral domain $R$ with a function $d:R\setminus \{0\}\to {\mathbb{N}}_0$ such that, for $a\in R$ and nonzero $b\in R$, there exist $q,r\in R$ with $a=qb+r$ and either $r=0$ or $d(r)<d(b)$. This is the Euclidean-ring convention used here.

If an ideal $I$ is zero, it is generated by zero. Otherwise choose nonzero $a\in I$ with $d(a)$ minimal. For any $b\in I$, Euclidean division gives $b=qa+r$, with $r\in I$. A nonzero remainder would have $d(r)<d(a)$, contradicting the choice. Hence $r=0$ and $b\in aR$. The reverse inclusion follows from $a\in I$, provingNow suppose $a$ is an irreducible element. Its principal ideal is proper because $a$ is not a unit. If $aR\subseteq J\subseteq R$, write $J=bR$ by the result just proved. Then $a=bc$ for some $c\in R$. Irreducibility gives either $b$ a unit, so $J=R$, or $c$ a unit, so $bR=aR$. There is no intermediate proper ideal, and therefore $aR$ is a maximal ideal.

For a field $F$, the polynomial ring $F[X]$ is an integral domain and the polynomial division algorithm gives $A=QB+R$ with $R=0$ or $\mathrm{deg}R<\mathrm{deg}B$. Thus the polynomial degree is a Euclidean function, proving $F[X]$ is a Euclidean domain.

Take the integral domain $R=\mathbb{Z}[\sqrt{-5}]$, embedded in $\mathbb{C}$. It exhibits failure of unique factorization in the ring of integers generated by square root of minus five:Here is why these are genuinely distinct factorizations into irreducible elements. The multiplicative algebraic norm $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm 1$, which are therefore the only units. Neither two nor three is a value of $N$: if $b\ne 0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm \sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be associate elements of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is an integral domain which is not a unique factorization domain.

In a principal ideal domain, an ascending chain of ideals stabilizes: its union is an ideal $(d)$, and its generator belongs to one member, which then contains the union. This proves the needed ascending chain condition. If a nonzero nonunit had no finite irreducible factorization, split it into two nonunits and select a factor that still has no such factorization. Continuing would give a strictly ascending chain of principal ideals, because $a=bc$ with $c$ a nonunit implies $(a)⊊(b)$. This contradicts the chain condition, so factorization exists.

An irreducible element $p$ generates a maximal ideal: any intermediate ideal is $(d)$ with $p=dc$, and irreducibility forces $d$ or $c$ to be a unit. Therefore $p$ is a prime element. In any two irreducible factorizations, the first prime factor divides a factor of the other, hence is associate to it. Cancel and continue. This proves every principal ideal domain is a unique factorization domain.

In $\mathbb{Z}[\sqrt{-3}]$, the multiplicative field norm is $N(a+b\sqrt{-3})=a^2+3b^2$. The only units are $\pm 1$. There is no element of norm two, so every element of norm four is irreducible: a factorization into nonunits would require norms two and two. Consequentlyis a pair of genuinely different irreducible factorizations. Neither $1+\sqrt{-3}$ nor $1-\sqrt{-3}$ is associate to $2$.

Take the Eisenstein integers $R=\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. Since $\sqrt{-3}=1+2\omega$, the embedded subring is $\mathbb{Z}+2\mathbb{Z}\omega$, whose additive index in $R$ is two. Its larger ring has Eisenstein-integer norm $N(a+b\omega )=a^2-ab+b^2$ and is Euclidean. Indeed, write a complex quotient in the real basis $1,\omega$ and round both coordinates to integers. The coordinate errors $u,v$ satisfy $|u|,|v|\le 1/2$, giving $|u+v\omega {|}^2=u^2-uv+v^2\le 3/4<1$. Thus division leaves a remainder of smaller norm. A Euclidean domain is a principal ideal domain, because division by a nonzero element of least norm in an ideal leaves zero remainder. This gives the requested index-two embedding into a principal ideal domain.

With the usual commutative, unital conventions, the chain isA field is an integral domain, and every nonzero ideal contains an invertible element and is consequently the whole ring. Its only ideals are $(0)$ and $(1)$, so it is a principal ideal domain.

To prove the middle implication, first show that a principal ideal domain has the ascending chain condition on ideals. The union of an ascending chain is an ideal, hence is $(a)$; its generator belongs to one member of the chain, and all later members equal $(a)$. Now suppose a nonzero nonunit admitted no factorization into irreducible elements. It would factor into two nonunits, at least one of which again has no such factorization. Repeating would produce a strictly increasing chain of principal ideals: if $a=bc$ with $c$ a nonunit, $(a)⊊(b)$, since equality and cancellation would make $c$ invertible. This contradicts the ascending chain condition. Existence of factorization follows. By the permitted fact that every irreducible element is a prime element in a principal ideal domain, any irreducible factor in one factorization divides a factor in the other; cancellation then proves uniqueness up to units and order. This is a unique factorization domain. The final implication is part of the definition of a unique factorization domain.

None of the adjacent implications reverses. The principal ideal domain $\mathbb{Z}$ is not a field, since $2$ is not invertible. The polynomial ring $\mathbb{Q}[x,y]$ is a unique factorization domain, by applying the polynomial extension theorem for unique factorization domains twice, but its ideal $(x,y)$ is not principal: a generator would divide both $x$ and $y$, so would be a unit, whereas every member of this ideal vanishes at the origin. Finally, $\mathbb{Z}[\sqrt{-5}]$ is an integral domain as a subring of $\mathbb{C}$, but is not a unique factorization domain. Its multiplicative norm is $N(a+b\sqrt{-5})=a^2+5b^2$. There is no element of norm $2$ or $3$. Hence $2,3,1+\sqrt{-5},1-\sqrt{-5}$, of norms $4,9,6,6$, are all irreducible: a nontrivial factorization would require one of those missing norms. The only units have norm one and are $\pm 1$. The two factorizationsare therefore not related by associates and reordering.

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

Suppose $X=f_1\cdots f_s$ were a product of irreducible elements of $R$. Since degree of a polynomial is additive under products, exactly one factor has positive degree and all the others are constant. By the preceding classification, the positive-degree factor has constant term $\pm 1$, while every constant irreducible is $\pm p$ for a prime $p$. Their product therefore has a nonzero constant term, contradicting $X(0)=0$.

Thus $X$ is not a product of irreducibles, so this integral domain is not an atomic domain and in particular is not a unique factorization domain.

The algebraic norm on $\mathbb{Z}[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies multiplicativity of the algebraic norm. The calculation in part (a) therefore gives all elements of norm $9$:If one of these factored into two nonunits, multiplicativity would force both factors to have norm $3$. But the Diophantine equation $a^2+5b^2=3$ has no solution. Hence all six displayed elements are irreducible. This is a standard example in which an irreducible element need not have prime norm.