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The unique factorization of ideals in a Dedekind domain states that every nonzero proper ideal $I$ of a Dedekind domain $D$ has a factorizationwhere the $P_j$ are distinct nonzero prime ideals and the $n_j$ are positive integers. The prime ideals and their exponents are unique up to reordering. The unit ideal $D$ is the empty product; the zero ideal is excluded.

Write $v_P(I)$ for the prime-ideal valuation in a Dedekind domain, namely the exponent of $P$ in this factorization, with zero assigned when $P$ is absent. The associated containment rule isLarger exponents give smaller ideals, just as higher divisibility gives smaller principal ideals in $\mathbb{Z}$. The theorem concerns unique factorization of ideals; it does not assert unique factorization of elements or that these ideals are principal.

Let $D$ be a Dedekind domain. The zero ideal is generated by $0$, and $D$ is generated by $1$, so consider a nonzero proper ideal $I$. Choose $0\ne a\in I$. We use these consequences of the unique factorization of ideals in a Dedekind domain: every nonzero ideal has finitely many nonzero prime-ideal valuations in a Dedekind domain; containment reverses the valuations as in part (b); andThe last formula follows from the containment rule: $J+K$ is the smallest ideal containing both $J$ and $K$, and the product with the indicated minimum exponents is exactly that ideal.

Let $T$ be the finite set of prime ideals dividing $(a)$. For each $P\in T$, set $r_P=v_P(I)$. Since $(a)\subseteq I$, we have $r_P\le v_P((a))$. The containment rule also gives $I⊈P^{r_P+1}$, so chooseDistinct nonzero prime ideals of $D$ are distinct maximal ideals and hence comaximal ideals; their positive powers are also comaximal. Indeed, if $u+v=1$ with $u\in P$, $v\in Q$, for positive integers $m,n$ expanding $(u+v{)}^{m+n-1}$ shows $1\in P^m+Q^n$, since each term contains either at least $m$ factors of $u$ or at least $n$ factors of $v$. We use the Chinese remainder theorem in its ideal form: for pairwise comaximal ideals $J_1,\dots ,J_s$, the map $D\to {\prod }_{j}D/J_j$ is surjective. Apply it to the powers $P^{r_P+1}$ to choose $e_P\in D$ withThen $b={\sum }_{P\in T}{e}_P{b}_P$ belongs to $I$ and satisfies $b\equiv b_P(\mathrm{mod}{P}^{r_P+1})$ for every $P\in T$. Thus $b\notin P^{r_P+1}$, whereas $b\in I\subseteq P^{r_P}$, proving $v_P((b))=r_P$.

For $P\in T$, therefore,For $P\notin T$, $v_P((a))=0$, and $(a)\subseteq I$ forces $v_P(I)=0$ as well. Hence $(a,b)$ and $I$ have the same valuation at every nonzero prime ideal. Uniqueness of their factorizations givesThis proves two generators for an ideal in a Dedekind domain; in fact the first nonzero generator $a\in I$ can be chosen in advance.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.

The original PDF asks about a discrete valuation ring; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.

Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

The first isomorphism theorem for rings states that a ring homomorphism $\varphi :R\to S$ induces an isomorphismBecause $J$ is an ideal, sums and products of elements of $R+J$ remain in $R+J$, so it is a subring. The surjective homomorphismhas kernel $R\cap J$. The theorem givesEvaluation at $-2$ givesso this quotient is a field of characteristic zero. The other quotient isIt has characteristic $3$ but is not a field, since the nonzero class of $X-1$ is nilpotent.

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.

Assume (iii). Because $R$ is a Noetherian ring, every ideal $I\subseteq R$ is finitely generated. Regard $I$ as an $R$-submodule of $F$. By (iii), $I=Rx$ for some $x\in F$; since $x\in I\subseteq R$ unless $I=0$, this is a principal ideal of $R$. Thus every ideal is principal and (i) holds.

Together with the previous two implications, this proves (i), (ii), and (iii) equivalent.

Assume (i). For $a,b\in R$, write the ideal $(a,b)=(d)$. Then $d=ra+sb$ for some $r,s\in R$. Since $a,b\in (d)$, the element $d$ divides both; and every common divisor of $a,b$ divides the displayed linear combination $d$. Thus $d$ is a greatest common divisor and (ii) holds.