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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

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.

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.

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.

For the computation put $s=\sqrt{-6}$. The Kummer-Dedekind theorem says that when $p$ does not divide the power-basis index, a factorization $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ gives $(p)={\prod }_i(p,g_i(s){)}^{e_i}$, with each factor having norm $p^{\mathrm{deg}{g}_i}$. The given full power basis has index one. Since $X^2+6\equiv X^2(\mathrm{mod}2)$ and $X^2+6\equiv (X-1)(X+1)(\mathrm{mod}7)$,Here $N(P)=2$ and $N(Q)=N(Q^{\prime })=7$. A prime ideal dividing an ideal of norm $98=2\cdot 7^2$ lies above $2$ or $7$, and these are all the prime ideals above them. Unique factorization and norm multiplicativity therefore leave exactlyThey are distinct by their prime-ideal valuations, and each has norm $98$.

For an ideal $I$ in a commutative ring $R$, put $I^0=R$ and let $I^n$ for $n\ge 1$ be the product of ideals with $n$ factors equal to $I$. It consists of finite sums of products of $n$ elements of $I$. These ideals form a descending chain $R\supseteq I\supseteq I^2\supseteq \cdots$.

Every positive power has the same radical of an ideal as $I$:One inclusion follows from $I^n\subseteq I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, giving the other inclusion. Different ideals can therefore define the same radical ideal, even when their successive powers are all distinct.