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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.

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.

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence

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.

The quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus

Let $\pi$ be a prime element of the Gaussian integers. Its norm $\pi \overline{\pi }$ is a positive integer greater than one, so $\pi$ divides some ordinary positive prime number in its integer prime factorization. It cannot divide two distinct such primes $p,q$: an integer Bézout identity $up+vq=1$ would imply $\pi \mid 1$, contrary to its being a nonunit. This proves existence and uniqueness of the rational prime lying below $\pi$.

Up to associates, the complete list of Gaussian prime elements is: $1+i$; the positive rational primes $p\equiv 3(\mathrm{mod}4)$; and, for every rational prime $p\equiv 1(\mathrm{mod}4)$, the two nonassociate elements $a+bi$ and $a-bi$, where $a,b>0$ and $p=a^2+b^2$. Representatives related by multiplication by one of $\pm 1,\pm i$ are associates. The last pair accounts for both factors of a split prime.

Every residue class modulo $p$ has a unique representative $a+bi$ with $0\le a,b<p$, since $p\mid (a-a^{\prime })+(b-b^{\prime })i$ means both coordinate differences are divisible by $p$. Hence $\boxed{|R/pR|=p^2}$ andFor $p=2$, the class of $1+i$ is nonzero but its square is $2i=0$; thus the quotient ring is not a field. For $p\equiv 3(\mathrm{mod}4)$, a root $r^2=-1$ would imply $r^{p-1}=(-1{)}^{(p-1)/2}=-1$, contradicting Fermat's little theorem. The quadratic $X^2+1$ is therefore irreducible and its quotient is a field with $p^2$ elements.

For $p\equiv 1(\mathrm{mod}4)$, choose a generator $g$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Then $r=g^{(p-1)/4}$ satisfies $r^2=-1$. The two nonzero classes $i-r$ and $i+r$ have product zero in $R/pR$. ThusFor $p\equiv 1(\mathrm{mod}4)$ it is instead isomorphic to ${\mathbb{F}}_p\times {\mathbb{F}}_p$, by the distinct linear factors and the Chinese remainder theorem.

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.

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 $B$.

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.

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore

Assume that $p$ is a prime element. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the integral domain gives $1=bc$, so $b$ is a unit. The other case is symmetric. Hence every prime element is irreducible.

Suppose first that $A$ is a unique factorization domain, and let $\mathfrak{p}$ be a minimal nonzero prime ideal. Choose $0\ne a\in \mathfrak{p}$ and factor it into irreducibles. In a UFD each irreducible is a prime element, so one factor $\pi$ belongs to $\mathfrak{p}$. The nonzero prime ideal $(\pi )\subseteq \mathfrak{p}$ must equal $\mathfrak{p}$ by minimality.

Conversely, the ascending chain condition in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak{p}$ minimal over $(\pi )$. The Krull principal ideal theorem gives $\mathrm{ht}\mathfrak{p}\le 1$. Since $A$ is a domain and $\mathfrak{p}\ne (0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak{p}=(q)$. The divisibility $q\mid \pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi )=\mathfrak{p}$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the Minimal-prime criterion for a Noetherian unique factorization domain.

Prime elements are first-order definable in the ring first-order language. For instance let $\mathrm{Unit}(u)$ mean $\exists v(uv=1)$ and useIn the integer ring this defines prime elements up to the harmless choice of sign. Expand the first-order language by names for every integer and a new constant $c$. Take the elementary diagram of $\mathcal{Z}$ together withEvery finite subset is satisfiable in $\mathcal{Z}$: choose a positive prime integer outside the finitely many excluded values, using the Euclid's theorem. Compactness theorem gives a first-order model of the entire set. The complete-diagram argument embeds $\mathcal{Z}$ elementarily into its reduct; identify that copy with the original integer ring.

The resulting elementary extension contains $c$, a prime element unequal to every named standard integer. ThereforeThe construction rules out all standard integers, not merely the finite list appearing in any one approximation.

Since ${\omega }^2=-1-\omega$, products of integer linear combinations of $1,\omega$ remain of the same form. Thus the Eisenstein integers $E=\mathbb{Z}[\omega ]$ form a subring of $\mathbb{C}$, and hence an integral domain. Their Eisenstein norm isIt is a positive integer on nonzero elements and is multiplicative, because the complex modulus is multiplicative.

For $\alpha ,\beta \in E$ with $\beta \ne 0$, write $\alpha /\beta =u+v\omega$ with real $u,v$. Choose integers $m,n$ with $|u-m|,|v-n|\le 1/2$, and let $q=m+n\omega$. With $s=u-m$ and $t=v-n$,Consequently $r=\alpha -q\beta$ belongs to $E$ and satisfies $N(r)\le \frac{3}{4}N(\beta )<N(\beta )$ when $r\ne 0$. This proves $E$ is a Euclidean domain with Euclidean function $N$, the Euclidean norm on the Eisenstein integers.

For the representation of the prime $p$, first note that $3=N(1-\omega )$. Now suppose $p>3$ and $p\mid n^2+3$. The factorizationshows that the Eisenstein integer $p$ divides a product. It divides neither factor: an element $a+b\omega$ lies in $pE$ exactly when both integer coefficients are divisible by $p$, and the second coefficients here are $2$ and $-2$. Hence $p$ is not a prime element of $E$.

To justify the necessary factorization step, in any principal ideal domain an irreducible element $\pi$ is prime. If $\pi \nmid a$, write $(\pi ,a)=(d)$. Since $d\mid \pi$, irreducibility makes $d$ a unit or an associate of $\pi$. The latter would give $\pi \mid a$, so $d$ is a unit and $u\pi +va=1$ for some $u,v$. If $\pi \mid ab$, multiplication of this identity by $b$ gives $\pi \mid b$. This proves irreducible elements of a principal ideal domain are prime without assuming unique factorization.

By part (a), $E$ is a principal ideal domain. The integer $p$ has norm $p^2>1$, so it is a nonzero nonunit. Since it is not prime, the result just proved implies that it is reducible: $p=uv$ with $u,v$ both nonunits. A nonzero element of $E$ is a unit exactly when its norm is one, since $N(z)=1$ gives inverse $\overline{z}$ and multiplicativity proves the converse. ThusThe only possible positive integer factors exceeding one are $N(u)=N(v)=p$. Writing $u=a+b\omega$ givesThis is the prime representation by the Eisenstein norm argument.

Regard $\mathbb{R}[X,Y]$ as the polynomial ring $(\mathbb{R}[X])[Y]$. The coefficient ring $\mathbb{R}[X]$ is a Euclidean domain, hence a principal ideal domain. The element $X$ is a prime element: the quotient ring $\mathbb{R}[X]/(X)\cong \mathbb{R}$ is an integral domain.

The monic polynomial $Y^2+X$ has its other coefficients divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the indeterminate, proves that $Y^2+X$ is an irreducible polynomial.

Now use the automorphism of a polynomial ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving