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Write $R={\mathcal{O}}_K$. We prove unique factorization of ideals in a number field, rather than assuming this conclusion as a property of a Dedekind domain. Three preliminary facts about $R$ will be useful. Its additive group is a finite-rank free abelian group, so every ideal is finitely generated and the ascending chain condition holds. It is an integrally closed domain, by integral dependence is transitive. Finally, every nonzero prime ideal is a maximal ideal. For this last fact, choose $0\ne a\in P$. Multiplication by $a$ is an injective endomorphism of the finite-rank free abelian group $R$ with nonzero determinant, so $R/(a)$ is finite. Consequently $R/P$ is a finite integral domain and hence a field.

First, every nonzero proper ideal contains a product of nonzero prime ideals. Otherwise choose a maximal counterexample $J$ under inclusion, using the ascending chain condition. It cannot be a prime ideal. Choose $x,y\notin J$ with $xy\in J$. The strictly larger ideals $J+(x)$ and $J+(y)$ each contain a product of nonzero prime ideals, allowing the empty product $R$ when necessary. Multiplying those products gives a product contained ina contradiction.

Next fix a nonzero prime ideal $P$, choose $0\ne a\in P$, and choose a product of nonzero prime idealswith $r$ minimal. Because $(a)\subseteq P$, primality and maximality imply $P_j=P$ for some $j$; reorder so that $P_1=P$. Minimality provides $b\in P_2\cdots P_r$ with $b\notin (a)$, taking the remaining product to be $R$ if $r=1$. ThusDefine the fractional idealIt contains $R$ and $x$. The ideal $P{P}^{-1}$ lies between $P$ and $R$, so it is $P$ or $R$. If it were $P$, then $xP\subseteq P$. Applying the determinant trick to a finite generating set of an ideal for $P$ gives a monic polynomial over $R$ satisfied by $x$. Since $R$ is an integrally closed domain, this would put $x$ in $R$, a contradiction. ThereforeThis proves that each nonzero prime ideal is an invertible fractional ideal, independently of the desired factorization theorem.

Now take a nonzero proper ideal $A$ and a maximal ideal $P$ containing it. The ideal $B=P^{-1}A$ satisfiesThe first inclusion is strict: if $A=B$, the same $x\in P^{-1}\setminus R$ would stabilize $A$, and the determinant trick would again force $x\in R$. Repeating gives a strictly ascending chain until it reaches $R$; the ascending chain condition ensures this happens after finitely many steps. The identities $A=PB$ then express $A$ as a product of nonzero prime ideals.

For uniqueness, suppose $P_1\cdots P_r=Q_1\cdots Q_s$. Since the right-hand side is contained in $P_1$, some $Q_j$ is contained in $P_1$. Both are nonzero maximal ideals, so $Q_j=P_1$. Multiplication by $P_1^{-1}$ cancels this factor. Repeat. A nonempty product of proper prime ideals cannot equal $R$, so both products finish together. This proves uniqueness up to reordering, including repeated factors.

A commutative ring $R$ is a Noetherian ring if every ideal has a finite generating set of an ideal. Equivalently, it satisfies the ascending chain condition on ideals. Indeed, the union of an ascending chain is an ideal, and finitely many generators all belong to one term, after which the chain stabilizes. Conversely, an ideal without a finite generating set of an ideal would allow us to choose successively $a_{n+1}\notin (a_1,\dots ,a_n)$, producing a strictly ascending chain of ideals.

Let $\varphi :R\to S$ be a surjective ring homomorphism, where $R$ is a Noetherian ring. For any ideal $J$ of $S$, its inverse image is an ideal of $R$, say ${\varphi }^{-1}(J)=(a_1,\dots ,a_r)$. Taking images givesThus homomorphic images of Noetherian rings are Noetherian. This also applies to the image of an arbitrary ring homomorphism, by taking its image as the codomain.

The Hilbert basis theorem states that if $R$ is a Noetherian ring, then the polynomial ring $R[t]$ is a Noetherian ring. Repeating this gives the result for finitely many variables.

To prove it, fix an ideal $K\subseteq R[t]$. For $n\ge 0$, define the coefficient ideals of a polynomial idealHere the zero polynomial is allowed. Each $A_n$ is an ideal: addition and multiplication by constants preserve the required degree bound. Multiplying a representing polynomial by $t$ gives $A_n\subseteq A_{n+1}$. The ascending chain condition therefore supplies $N$ such that $A_n=A_N$ for every $n\ge N$.

For each $0\le n\le N$, choose a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$ for $A_n$, omitting zero generators, and choose $f_{nj}\in K$ of polynomial degree $n$ with leading coefficient $a_{nj}$. If $A_n=0$, choose no such polynomials. We claim that the finitely many $f_{nj}$ generate $K$.

Take a nonzero $f\in K$ of polynomial degree $d$. Put $n=\min (d,N)$. Its leading coefficient belongs to $A_d=A_n$, so write it as ${\sum }_j{b}_j{a}_{nj}$. The polynomialbelongs to $K$ and is either zero or has polynomial degree strictly below $d$. Induction on polynomial degree now expresses $f$ as a combination of the chosen $f_{nj}$; when $d=0$, the displayed difference is zero. Hence $K$ is finitely generated. Since $K$ was arbitrary, $R[t]$ is a Noetherian ring.

For $x\in X$, let $e_x$ be the indicator function of $\{x\}$. In the ring of functions $R={\mathbb{R}}^X$, multiplication is pointwise, soSuppose first that $X$ is finite, and let $J$ be an ideal of $R$. Define $S=\{x\in X:f(x)\ne 0\text{ for some }f\in J\}$. If $x\in S$, choose such an $f$; multiplying it by $e_x/f(x)$ shows that $e_x\in J$. Every $f\in J$ satisfieswhere the real scalars are constant functions. Thus $J=(e_x:x\in S)$ has a finite generating set of an ideal, so $R$ is a Noetherian ring. If $X$ is empty, $R$ is the zero ring, and the conclusion still holds.

If $X$ is infinite, consider the idealIt is an ideal because a sum has support contained in the union of the two supports, and multiplication by any function cannot enlarge support. If $f_1,\dots ,f_r$ generated $F$, their supports would have a finite union $T$. Every element of $(f_1,\dots ,f_r)$ would vanish outside $T$, whereas $e_x\in F$ for any $x\in X\setminus T$. This contradiction proves

Suppose first that $\mathfrak{a}$ is $\mathfrak{m}$-primary. Since $R$ is a Noetherian ring, choose a finite generating set of an ideal $\mathfrak{m}=(x_1,\dots ,x_d)$. The equality of radicals of ideals $\sqrt{\mathfrak{a}}=\mathfrak{m}$ supplies positive integers $e_i$ with $x_i^{e_i}\in \mathfrak{a}$. SetEvery product of $n$ generators has some $x_i$ appearing at least $e_i$ times. Such products generate the product of ideals ${\mathfrak{m}}^n$, so ${\mathfrak{m}}^n\subseteq \mathfrak{a}$. Also $\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}=\mathfrak{m}$. If $\mathfrak{m}=0$, take $n=1$ directly.

Conversely, suppose ${\mathfrak{m}}^n\subseteq \mathfrak{a}\subseteq \mathfrak{m}$. Taking radicals of ideals givesHence $\sqrt{\mathfrak{a}}=\mathfrak{m}$. The only maximal ideal containing $\mathfrak{a}$ is $\mathfrak{m}$: any such maximal ideal contains $\sqrt{\mathfrak{a}}$ and therefore equals $\mathfrak{m}$. By the correspondence theorem for ideals, the quotient ring $R/\mathfrak{a}$ is a local ring with maximal ideal $\mathfrak{m}/\mathfrak{a}$. Every element outside this maximal ideal is a unit.

If $ab\in \mathfrak{a}$ and $a\notin \mathfrak{a}$, then $b$ cannot be outside $\mathfrak{m}$, since multiplying by its inverse in the quotient ring would give $a\in \mathfrak{a}$. Thus $b\in \mathfrak{m}$ and $b^n\in \mathfrak{a}$, proving the required primary ideal condition. We have proved the maximal-primary ideal criterion

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

Yes: $R[[x]]$ is Noetherian. We prove Noetherianity of a formal power series ring directly, using only the equivalence in part (a). Let $A=R[[x]]$, a formal power series ring, and fix an ideal $I\subseteq A$. For $n\ge 0$, let $C_n$ be the set of coefficients of $x^n$ in elements of $I\cap x^{n}A$. It is an ideal of $R$, because coefficient extraction is additive and respects multiplication by constants. Multiplication by $x$ givesSince $R$ is Noetherian, there is an $N$ with $C_n=C_N$ for all $n\ge N$. Each $C_n$, for $0\le n\le N$, has a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$. Choose $f_{nj}\in I\cap x^{n}A$ whose coefficient of $x^n$ is $a_{nj}$. If $C_n=0$, use no generators at that index.

We claim that these finitely many $f_{nj}$ generate $I$ over $A$. Given $f\in I$, start with $g_0=f$ and recursively construct $g_n\in I\cap x^{n}A$. For $n<N$, express the coefficient of $x^n$ in $g_n$ as ${\sum }_j{c}_{nj}{a}_{nj}$, with $c_{nj}\in R$, and subtract ${\sum }_j{c}_{nj}{f}_{nj}$ to obtain $g_{n+1}\in I\cap x^{n+1}A$. For $n\ge N$, use $C_n=C_N$ and subtractto cancel that coefficient. Every step involves a finite sum and leaves a residual still in $I$.

For each $j$, define the formal power seriesThe residual after the $n$th cancellation lies in $x^{n+1}A$, so comparison of each coefficient gives the exact identityThe right side is a finite $A$-linear combination of the chosen generators. No assertion that arbitrary ideals are closed under limits is being used: the equality is verified coefficient by coefficient, and the multipliers $h_j$ belong to $A$. Thus every ideal of $A$ is finitely generated, proving

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.

For a commutative ring $k$ and a set $X$, the set $k^X$ of all functions $X\to k$ is a commutative ring under pointwise addition and multiplication. It is the direct product of rings with one copy of $k$ for each $x\in X$.

If $k$ is a field and $X$ is finite, every ideal is generated by some of the singleton indicator functions $e_x$. Indeed, whenever $f$ in the ideal has $f(x)\ne 0$, multiplication by $e_x/f(x)$ puts $e_x$ in the ideal. Every function in that ideal is a finite sum of its coordinate multiples of these $e_x$.

If $X$ is infinite and $k$ is a nonzero commutative ring, the functions with finite support form an ideal without a finite generating set of an ideal. A finite collection of such functions vanishes outside one finite subset, as does the ideal it generates, whereas a singleton indicator function outside that subset belongs to the ideal of all finitely supported functions. Thus, for a field $k$, $k^X$ is a Noetherian ring exactly when $X$ is finite.