[proofplan] Part (1) is a determinant–adjugate argument, the same mechanism as in the "finitely generated implies integral" direction. Since $\mathcal{O}_L$ is Noetherian (from [Ring of Integers is a Dedekind Domain](/theorems/1582)), $\mathfrak{a}$ has finitely many generators $\alpha_1, \ldots, \alpha_m$. The stability condition $x\mathfrak{a} \subseteq \mathfrak{a}$ writes multiplication-by-$x$ on $\mathfrak{a}$ as an $\mathcal{O}_L$-linear map, so $(xI - A)\vec\alpha = 0$ for a matrix $A \in M_m(\mathcal{O}_L)$; the adjugate trick then gives $\det(xI - A) = 0$ in $L$, showing $x$ is integral over $\mathcal{O}_L$ and therefore in $\mathcal{O}_L$ by integral closure. Part (2) reduces to the case of a maximal (equivalently prime) ideal, then uses the key lemma that every nonzero integral ideal contains a product of primes — a Noetherian induction argument — to manufacture an explicit $\beta \in \mathfrak{p}_2 \cdots \mathfrak{p}_r \setminus \langle \alpha \rangle$ for which $\beta/\alpha \in L \setminus \mathcal{O}_L$ sends $\mathfrak{p}$ into $\mathcal{O}_L$. [/proofplan]

[step:Reduce (1) to a monic polynomial identity for $x$ over $\mathcal{O}_L$ using finite generation]Suppose $x \in L$ satisfies $x\mathfrak{a} \subseteq \mathfrak{a}$. By [Ring of Integers is a Dedekind Domain](/theorems/1582), $\mathcal{O}_L$ is Noetherian, so $\mathfrak{a}$ is finitely generated as an ideal: write $\mathfrak{a} = \langle \alpha_1, \ldots, \alpha_m \rangle$ with $\alpha_1, \ldots, \alpha_m \in \mathfrak{a}$. The hypothesis $x\mathfrak{a} \subseteq \mathfrak{a}$ means $x \alpha_i \in \mathfrak{a}$ for each $i$, so we may expand each $x\alpha_i$ as an $\mathcal{O}_L$-linear combination of the generators: \begin{align*} x \alpha_i &= \sum_{j=1}^m a_{ij} \alpha_j, \qquad a_{ij} \in \mathcal{O}_L \quad (1 \leq i \leq m). \end{align*} Let $A = (a_{ij}) \in M_m(\mathcal{O}_L)$ and $\vec\alpha = (\alpha_1, \ldots, \alpha_m)^\top \in L^m$. The relations can be written in matrix form as \begin{align*} x \vec\alpha &= A \vec\alpha, \qquad \text{equivalently} \qquad (xI_m - A)\vec\alpha = 0, \end{align*} where $xI_m - A \in M_m(L)$ and the computation is performed inside the $L$-vector space $L^m$.[/step]

[guided]We are given $x \in L$ with $x\mathfrak{a} \subseteq \mathfrak{a}$, and we want to prove $x \in \mathcal{O}_L$. The strategy is the standard "Cayley–Hamilton / determinant–adjugate" argument: writing multiplication-by-$x$ as a linear operator on a finite generating set yields a monic polynomial relation, after which integral closure of $\mathcal{O}_L$ in $L$ finishes the proof. **Using Noetherianity to get finitely many generators.** By [Ring of Integers is a Dedekind Domain](/theorems/1582), $\mathcal{O}_L$ is Noetherian; equivalently, every ideal is finitely generated. Therefore we may fix a finite set of generators $\alpha_1, \ldots, \alpha_m$ of $\mathfrak{a}$, so that \begin{align*} \mathfrak{a} = \sum_{i=1}^m \mathcal{O}_L \alpha_i. \end{align*} **Encoding stability as a matrix equation.** The stability condition says $x \alpha_i \in \mathfrak{a}$ for each $i \in \{1, \ldots, m\}$ — because $\alpha_i \in \mathfrak{a}$ and $x\mathfrak{a} \subseteq \mathfrak{a}$. Since the $\alpha_j$ generate $\mathfrak{a}$, each $x\alpha_i$ admits an expansion \begin{align*} x \alpha_i &= \sum_{j=1}^m a_{ij} \alpha_j \quad \text{for some } a_{ij} \in \mathcal{O}_L. \end{align*} (The coefficients $a_{ij}$ need not be unique, but we simply choose any valid choice.) Collecting these $m$ scalar relations into a single matrix relation, let $A = (a_{ij})_{1 \leq i,j \leq m} \in M_m(\mathcal{O}_L)$ and $\vec\alpha = (\alpha_1, \ldots, \alpha_m)^\top$. Then \begin{align*} x \vec\alpha &= A \vec\alpha, \qquad \text{i.e.,} \qquad (xI_m - A)\vec\alpha = 0. \end{align*} **Where are we working?** The matrix $xI_m - A$ has entries in $L$ (because $x \in L$ and the $a_{ij} \in \mathcal{O}_L \subseteq L$). The vector $\vec\alpha$ has entries in $\mathcal{O}_L \subseteq L$. The product and equation are performed in $L^m$. Nothing so far requires $x \in \mathcal{O}_L$; the statement holds at the level of $L$-linear algebra.[/guided]

[step:Apply the adjugate to produce a monic integer polynomial annihilating $x$]Multiply the equation $(xI_m - A)\vec\alpha = 0$ on the left by the adjugate matrix $\operatorname{adj}(xI_m - A) \in M_m(L)$. Using the identity $\operatorname{adj}(M) \cdot M = \det(M) \cdot I_m$ valid for any square matrix $M$ over a commutative ring (here $L$), \begin{align*} \det(xI_m - A) \cdot \vec\alpha &= 0 \in L^m. \end{align*} Componentwise, $\det(xI_m - A) \cdot \alpha_j = 0$ for each $j$. Since $\mathfrak{a}$ is nonzero, at least one $\alpha_j \neq 0$ — in fact, a nonzero ideal cannot have all generators zero (otherwise $\mathfrak{a} = \{0\}$). Working in the field $L$, from $\det(xI_m - A)\alpha_j = 0$ with $\alpha_j \neq 0$ we deduce \begin{align*} \det(xI_m - A) &= 0 \quad \text{in } L. \end{align*} The polynomial $f(T) := \det(T I_m - A) \in \mathcal{O}_L[T]$ is monic of degree $m$ (its leading term is $T^m$, since the diagonal entries of $TI_m - A$ are $T - a_{ii}$ and off-diagonals are the constants $-a_{ij}$). Evaluating at $T = x$: $f(x) = \det(xI_m - A) = 0$. So $x$ satisfies a monic polynomial with coefficients in $\mathcal{O}_L$, i.e., $x$ is integral over $\mathcal{O}_L$. By [Ring of Integers is a Dedekind Domain](/theorems/1582), $\mathcal{O}_L$ is integrally closed in its fraction field $L$. Since $x \in L$ is integral over $\mathcal{O}_L$, we conclude $x \in \mathcal{O}_L$. This proves part (1).[/step]

[guided]We now extract a polynomial identity from the matrix equation, then invoke integral closure. **The adjugate identity.** For any square matrix $M \in M_m(R)$ over a commutative ring $R$, the adjugate (classical adjoint) $\operatorname{adj}(M) \in M_m(R)$ satisfies \begin{align*} \operatorname{adj}(M) \cdot M = M \cdot \operatorname{adj}(M) = \det(M) \cdot I_m. \end{align*} We apply this with $R = L$ and $M = xI_m - A \in M_m(L)$. **Multiplying the matrix equation.** From Step 1, $(xI_m - A)\vec\alpha = 0$. Multiplying on the left by $\operatorname{adj}(xI_m - A)$: \begin{align*} 0 &= \operatorname{adj}(xI_m - A) \cdot (xI_m - A) \cdot \vec\alpha \\ &= \det(xI_m - A) \cdot I_m \cdot \vec\alpha \\ &= \det(xI_m - A) \cdot \vec\alpha. \end{align*} Componentwise: \begin{align*} \det(xI_m - A) \cdot \alpha_j = 0 \quad \text{for } j = 1, \ldots, m. \end{align*} **From "kills every generator" to "equals zero".** Since $\mathfrak{a} \neq \{0\}$, not every $\alpha_j$ vanishes; otherwise $\mathfrak{a} = \sum \mathcal{O}_L \alpha_j = 0$. Pick $\alpha_j \neq 0$. Then $\det(xI_m - A) \cdot \alpha_j = 0$ in $L$; since $L$ is a field and $\alpha_j \neq 0$, multiplication by $\alpha_j$ is injective, forcing $\det(xI_m - A) = 0$ in $L$. **Reading this as integrality.** Define the polynomial \begin{align*} f(T) &:= \det(T I_m - A) \in \mathcal{O}_L[T]. \end{align*} Why $\mathcal{O}_L[T]$? Because the entries of $TI_m - A$ lie in $\mathcal{O}_L[T]$: the diagonal entries are $T - a_{ii}$ and the off-diagonals are $-a_{ij}$ with $a_{ij} \in \mathcal{O}_L$. The determinant is a polynomial expression in the entries, hence lies in $\mathcal{O}_L[T]$. The polynomial $f$ is monic of degree $m$: expanding $\det(TI_m - A)$ by the Leibniz formula, the identity permutation contributes the product $\prod_{i}(T - a_{ii}) = T^m + \text{(lower-degree terms)}$; every non-identity permutation $\sigma$ contributes a term $\prod_i m_{i,\sigma(i)}$ where at least two indices $i$ have $\sigma(i) \neq i$, meaning at least two factors are off-diagonal entries (degree-$0$ in $T$) — giving a term of degree $\leq m - 2 < m$ in $T$. So the coefficient of $T^m$ in $f$ is $1$. Evaluating at $T = x$: $f(x) = \det(xI_m - A) = 0$. So $x \in L$ satisfies a monic polynomial in $\mathcal{O}_L[T]$ of degree $m$. By definition, $x$ is **integral over $\mathcal{O}_L$**. **Closing with integral closure.** By [Ring of Integers is a Dedekind Domain](/theorems/1582), $\mathcal{O}_L$ is integrally closed in its fraction field $\operatorname{Frac}(\mathcal{O}_L) = L$. This means: any element of $L$ integral over $\mathcal{O}_L$ already lies in $\mathcal{O}_L$. Applied to $x$: $x \in \mathcal{O}_L$. This proves part (1): $x\mathfrak{a} \subseteq \mathfrak{a} \implies x \in \mathcal{O}_L$.[/guided]

[step:Reduce (2) to the case $\mathfrak{a} = \mathfrak{p}$ a maximal ideal]For part (2), we must find $y \in L \setminus \mathcal{O}_L$ with $y\mathfrak{a} \subseteq \mathcal{O}_L$, given that $\mathfrak{a}$ is a proper ideal. Every proper ideal of the ring $\mathcal{O}_L$ is contained in some maximal ideal. (This is a standard consequence of Zorn's Lemma applied to the partially ordered set of proper ideals containing $\mathfrak{a}$; alternatively, since $\mathcal{O}_L$ is Noetherian, a direct ascending-chain argument suffices.) Let $\mathfrak{p} \unlhd \mathcal{O}_L$ be a maximal ideal with $\mathfrak{a} \subseteq \mathfrak{p}$; in particular, $\mathfrak{p}$ is also prime (a maximal ideal in a commutative ring is prime). Suppose we have found $y \in L \setminus \mathcal{O}_L$ with $y\mathfrak{p} \subseteq \mathcal{O}_L$. Then for any $\gamma \in \mathfrak{a}$, since $\gamma \in \mathfrak{p}$, $y\gamma \in \mathcal{O}_L$; i.e., $y\mathfrak{a} \subseteq \mathcal{O}_L$. The same $y$ witnesses the claim for $\mathfrak{a}$. Thus it suffices to prove the statement when $\mathfrak{a} = \mathfrak{p}$ is a maximal (equivalently, nonzero prime) ideal.[/step]

[guided]We are given a proper ideal $\mathfrak{a} \unlhd \mathcal{O}_L$, i.e., $\mathfrak{a} \neq \mathcal{O}_L$. We want to find $y \in L \setminus \mathcal{O}_L$ with $y\mathfrak{a} \subseteq \mathcal{O}_L$. **Why reduce to maximal ideals?** The structural tools available to us — Noetherianity, finiteness of quotients, and prime-ideal manipulations — are most powerful for prime ideals. A maximal ideal is prime (standard ring theory: $R/\mathfrak{m}$ is a field, hence an integral domain, so $\mathfrak{m}$ is prime). So we reduce the problem from "proper ideal" to "maximal ideal" — a stronger hypothesis that makes the argument tractable. **Existence of a maximal ideal containing $\mathfrak{a}$.** This is a standard fact about Noetherian rings (or, more generally, any ring, using Zorn's Lemma). Since $\mathcal{O}_L$ is Noetherian (by [Ring of Integers is a Dedekind Domain](/theorems/1582)), we can argue directly: form the set of proper ideals containing $\mathfrak{a}$, partially ordered by inclusion. Every ascending chain stabilizes, so there exists a maximal element $\mathfrak{p}$. This $\mathfrak{p}$ is proper, contains $\mathfrak{a}$, and is maximal among proper ideals — i.e., $\mathfrak{p}$ is a maximal ideal. **Why a witness for $\mathfrak{p}$ gives one for $\mathfrak{a}$.** Suppose $y \in L \setminus \mathcal{O}_L$ satisfies $y\mathfrak{p} \subseteq \mathcal{O}_L$. Since $\mathfrak{a} \subseteq \mathfrak{p}$, \begin{align*} y\mathfrak{a} \subseteq y\mathfrak{p} \subseteq \mathcal{O}_L. \end{align*} So the same $y$ (which is not in $\mathcal{O}_L$) witnesses the containment for $\mathfrak{a}$. Hence we may assume $\mathfrak{a} = \mathfrak{p}$ is maximal, and WLOG nonzero: if $\mathfrak{a}$ is nonzero and contained in $\mathfrak{p}$, and $\mathfrak{a} \neq \{0\}$, then $\mathfrak{p} \neq \{0\}$ either. (If $\mathfrak{a} = \{0\}$, the theorem's hypothesis "proper ideal" is still satisfied, but we may replace $\mathfrak{a}$ with any fixed nonzero maximal ideal $\mathfrak{p}$ of $\mathcal{O}_L$ — which exists because $\mathcal{O}_L$ is not a field, since $\mathcal{O}_L \cong \mathbb{Z}^n$ as abelian groups and $\mathbb{Z}^n$ is not a field for $n \geq 1$.) In what follows, we assume $\mathfrak{a} = \mathfrak{p}$ is a nonzero maximal (hence prime) ideal of $\mathcal{O}_L$.[/guided]

[step:Prove the auxiliary lemma: every nonzero integral ideal contains a product of prime ideals]We isolate a key intermediate fact about the Noetherian ring $\mathcal{O}_L$: [claim:Prime Product Dominance] For every nonzero ideal $\mathfrak{b} \unlhd \mathcal{O}_L$, there exist (not necessarily distinct) nonzero prime ideals $\mathfrak{q}_1, \ldots, \mathfrak{q}_r$ such that $\mathfrak{q}_1 \cdots \mathfrak{q}_r \subseteq \mathfrak{b}$. [/claim] [proof] Let $\mathcal{S}$ be the set of nonzero ideals $\mathfrak{b} \unlhd \mathcal{O}_L$ that do **not** contain any product of nonzero prime ideals. We want $\mathcal{S} = \varnothing$; suppose for contradiction $\mathcal{S} \neq \varnothing$. Since $\mathcal{O}_L$ is Noetherian (by [Ring of Integers is a Dedekind Domain](/theorems/1582)), every nonempty collection of ideals has a maximal element. Let $\mathfrak{m}$ be a maximal element of $\mathcal{S}$. We show $\mathfrak{m}$ cannot be prime: if $\mathfrak{m}$ were prime and nonzero, then $\mathfrak{m} \subseteq \mathfrak{m}$ is already a (length-$1$) product of nonzero primes contained in $\mathfrak{m}$, contradicting $\mathfrak{m} \in \mathcal{S}$. So $\mathfrak{m}$ is not prime: there exist $a, b \in \mathcal{O}_L$ with $a, b \notin \mathfrak{m}$ but $ab \in \mathfrak{m}$. Consider the two ideals \begin{align*} \mathfrak{m} + \langle a \rangle \quad \text{and} \quad \mathfrak{m} + \langle b \rangle. \end{align*} Both strictly contain $\mathfrak{m}$ (because $a \notin \mathfrak{m}$ and $b \notin \mathfrak{m}$, yet both $a, b$ lie in the respective sums). Both are nonzero (they contain the nonzero ideal $\mathfrak{m}$ — which is nonzero because $\mathcal{S}$ only contains nonzero ideals). By maximality of $\mathfrak{m}$ in $\mathcal{S}$, neither $\mathfrak{m} + \langle a \rangle$ nor $\mathfrak{m} + \langle b \rangle$ belongs to $\mathcal{S}$. Hence each contains a product of nonzero primes: \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r &\subseteq \mathfrak{m} + \langle a \rangle, \\ \mathfrak{q}_1 \cdots \mathfrak{q}_s &\subseteq \mathfrak{m} + \langle b \rangle. \end{align*} We now multiply these inclusions: \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r \cdot \mathfrak{q}_1 \cdots \mathfrak{q}_s \subseteq (\mathfrak{m} + \langle a \rangle)(\mathfrak{m} + \langle b \rangle). \end{align*} Expanding the right-hand side using bilinearity of ideal multiplication, \begin{align*} (\mathfrak{m} + \langle a \rangle)(\mathfrak{m} + \langle b \rangle) &= \mathfrak{m}^2 + \mathfrak{m}\langle b \rangle + \langle a \rangle \mathfrak{m} + \langle a \rangle \langle b \rangle \\ &= \mathfrak{m}^2 + \mathfrak{m}\langle b \rangle + \langle a \rangle \mathfrak{m} + \langle ab \rangle. \end{align*} Each term is contained in $\mathfrak{m}$: $\mathfrak{m}^2 \subseteq \mathfrak{m}$ (since $\mathfrak{m}$ is an ideal), $\mathfrak{m}\langle b \rangle \subseteq \mathfrak{m}$ (since $\mathfrak{m}$ is an ideal and $\langle b \rangle \subseteq \mathcal{O}_L$), similarly $\langle a \rangle \mathfrak{m} \subseteq \mathfrak{m}$, and $\langle ab \rangle \subseteq \mathfrak{m}$ because $ab \in \mathfrak{m}$. Thus \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r \mathfrak{q}_1 \cdots \mathfrak{q}_s &\subseteq \mathfrak{m}. \end{align*} This is a product of $r + s$ nonzero primes contained in $\mathfrak{m}$, contradicting $\mathfrak{m} \in \mathcal{S}$. The contradiction forces $\mathcal{S} = \varnothing$: every nonzero ideal contains a product of nonzero prime ideals. [/proof][/step]

[guided]The proof of (2) rests on being able to "sandwich" a given ideal between a product of prime ideals and something more manageable. We establish this now. **Statement.** Every nonzero ideal $\mathfrak{b} \unlhd \mathcal{O}_L$ contains a product of nonzero prime ideals. Concretely: there exist $\mathfrak{q}_1, \ldots, \mathfrak{q}_r$ prime (nonzero) with $\mathfrak{q}_1 \cdots \mathfrak{q}_r \subseteq \mathfrak{b}$. **Strategy — Noetherian induction.** Suppose for contradiction the claim fails. Let \begin{align*} \mathcal{S} &= \{\mathfrak{b} \unlhd \mathcal{O}_L : \mathfrak{b} \neq 0 \text{ and no product of nonzero primes is contained in } \mathfrak{b}\}. \end{align*} If $\mathcal{S} \neq \varnothing$, then since $\mathcal{O}_L$ is Noetherian, $\mathcal{S}$ has a maximal element $\mathfrak{m}$. We derive a contradiction by showing $\mathfrak{m}$ would have to contain a product of primes after all. **Step A: $\mathfrak{m}$ is not prime.** A length-one product of a nonzero prime ideal is just the prime itself. If $\mathfrak{m}$ were nonzero and prime, then $\mathfrak{m} \subseteq \mathfrak{m}$ would exhibit a product of primes inside $\mathfrak{m}$, contradicting $\mathfrak{m} \in \mathcal{S}$. Hence $\mathfrak{m}$ is not prime. **Step B: Exhibit $a, b \notin \mathfrak{m}$ with $ab \in \mathfrak{m}$.** By definition of "prime ideal", an ideal $\mathfrak{p}$ is prime iff for all $a, b$, $ab \in \mathfrak{p} \implies a \in \mathfrak{p}$ or $b \in \mathfrak{p}$. Negating: $\mathfrak{m}$ not prime means there exist $a, b \in \mathcal{O}_L$ with $a, b \notin \mathfrak{m}$ but $ab \in \mathfrak{m}$. **Step C: Enlarge $\mathfrak{m}$.** Form $\mathfrak{m} + \langle a \rangle$ and $\mathfrak{m} + \langle b \rangle$. Each is a strictly larger ideal than $\mathfrak{m}$ (because $a \notin \mathfrak{m}$, so $\mathfrak{m} + \langle a \rangle$ contains $a$ which $\mathfrak{m}$ does not). So each is strictly larger than $\mathfrak{m}$, and hence by maximality of $\mathfrak{m}$ in $\mathcal{S}$, neither lies in $\mathcal{S}$. Since they are nonzero ideals, the only way to escape $\mathcal{S}$ is to contain a product of nonzero primes. Say \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r &\subseteq \mathfrak{m} + \langle a \rangle, \\ \mathfrak{q}_1 \cdots \mathfrak{q}_s &\subseteq \mathfrak{m} + \langle b \rangle. \end{align*} **Step D: Multiply.** Ideal multiplication is monotonic: if $I_1 \subseteq J_1$ and $I_2 \subseteq J_2$, then $I_1 I_2 \subseteq J_1 J_2$. Hence \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r \cdot \mathfrak{q}_1 \cdots \mathfrak{q}_s &\subseteq (\mathfrak{m} + \langle a \rangle)(\mathfrak{m} + \langle b \rangle). \end{align*} Expanding by bilinearity: \begin{align*} (\mathfrak{m} + \langle a \rangle)(\mathfrak{m} + \langle b \rangle) &= \mathfrak{m} \cdot \mathfrak{m} + \mathfrak{m} \cdot \langle b \rangle + \langle a \rangle \cdot \mathfrak{m} + \langle a \rangle \cdot \langle b \rangle. \end{align*} We argue each summand is inside $\mathfrak{m}$: - $\mathfrak{m} \cdot \mathfrak{m} \subseteq \mathfrak{m}$: since $\mathfrak{m}$ is an ideal closed under multiplication by $\mathcal{O}_L$, and $\mathfrak{m} \subseteq \mathcal{O}_L$. - $\mathfrak{m} \cdot \langle b \rangle \subseteq \mathfrak{m}$: same reason — the generator $b$ multiplies $\mathfrak{m}$ into itself since $\mathfrak{m}$ is an ideal. - $\langle a \rangle \cdot \mathfrak{m} \subseteq \mathfrak{m}$: same. - $\langle a \rangle \cdot \langle b \rangle = \langle ab \rangle \subseteq \mathfrak{m}$: because $ab \in \mathfrak{m}$ (from Step B), hence the entire principal ideal $\langle ab \rangle \subseteq \mathfrak{m}$. Summing these inclusions: $(\mathfrak{m} + \langle a \rangle)(\mathfrak{m} + \langle b \rangle) \subseteq \mathfrak{m}$. Composing: \begin{align*} \mathfrak{p}_1 \cdots \mathfrak{p}_r \mathfrak{q}_1 \cdots \mathfrak{q}_s \subseteq \mathfrak{m}. \end{align*} **Step E: Conclude.** This exhibits a product of $r + s$ nonzero primes inside $\mathfrak{m}$, contradicting $\mathfrak{m} \in \mathcal{S}$. Hence $\mathcal{S} = \varnothing$, and the claim holds.[/guided]

[step:Use the auxiliary lemma to build a witness $y = \beta/\alpha \in L \setminus \mathcal{O}_L$ with $y\mathfrak{p} \subseteq \mathcal{O}_L$]Let $\alpha \in \mathfrak{p}$ be nonzero (possible since, as reduced in Step 3, we may assume $\mathfrak{p}$ is a nonzero maximal prime). By the Prime Product Dominance claim from Step 4, there exist nonzero primes $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ with \begin{align*} \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle. \end{align*} Since $\langle \alpha \rangle \neq \{0\}$ (as $\alpha \neq 0$), we may choose $r$ minimal among all such representations. **Step A — some $\mathfrak{p}_i$ equals $\mathfrak{p}$.** From $\mathfrak{p}_1 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle \subseteq \mathfrak{p}$ and primality of $\mathfrak{p}$, inducting on $r$: if $r = 1$, then $\mathfrak{p}_1 \subseteq \mathfrak{p}$; if $r \geq 2$, writing $\mathfrak{p}_1 (\mathfrak{p}_2 \cdots \mathfrak{p}_r) \subseteq \mathfrak{p}$ and applying the primality of $\mathfrak{p}$ (i.e., the contrapositive: if neither factor is in $\mathfrak{p}$, neither is the product — see [Prime Ideals Divide Products](/theorems/???) applied at the level of ideals), at least one of $\mathfrak{p}_1$ or $\mathfrak{p}_2 \cdots \mathfrak{p}_r$ lies in $\mathfrak{p}$. Iterating on the second factor, some $\mathfrak{p}_i \subseteq \mathfrak{p}$. Since $\mathfrak{p}_i$ is prime and nonzero, it is maximal (by [Ring of Integers is a Dedekind Domain](/theorems/1582), nonzero primes are maximal). As $\mathfrak{p}_i \subseteq \mathfrak{p}$ with $\mathfrak{p}_i$ maximal, equality holds: $\mathfrak{p}_i = \mathfrak{p}$. Relabeling, WLOG $\mathfrak{p}_1 = \mathfrak{p}$. **Step B — build $\beta$.** By minimality of $r$, the shorter product $\mathfrak{p}_2 \cdots \mathfrak{p}_r$ is **not** contained in $\langle \alpha \rangle$ (otherwise we could drop $\mathfrak{p}_1$ and contradict minimality of $r$). If $r = 1$, this reads "the empty product" $(1) = \mathcal{O}_L$ is not contained in $\langle \alpha \rangle$ — equivalent to $\langle \alpha \rangle \neq \mathcal{O}_L$, which holds because $\alpha \in \mathfrak{p} \subsetneq \mathcal{O}_L$ ($\mathfrak{p}$ is proper). Hence we may pick $\beta \in (\mathfrak{p}_2 \cdots \mathfrak{p}_r) \setminus \langle \alpha \rangle$ (using $\mathcal{O}_L$ in place of the empty product when $r = 1$). **Step C — compute $\beta \mathfrak{p}$.** We have \begin{align*} \beta \mathfrak{p} &= \beta \mathfrak{p}_1 \subseteq \mathfrak{p}_1 (\mathfrak{p}_2 \cdots \mathfrak{p}_r) \subseteq \langle \alpha \rangle. \end{align*} The first inclusion uses $\beta \in \mathfrak{p}_2 \cdots \mathfrak{p}_r$, so $\beta \mathfrak{p}_1 \subseteq \mathfrak{p}_1 \cdot (\mathfrak{p}_2 \cdots \mathfrak{p}_r)$. The second is our hypothesis. Dividing by $\alpha$ (working in $L$, where $\alpha \neq 0$ is invertible), \begin{align*} \frac{\beta}{\alpha} \mathfrak{p} &\subseteq \frac{1}{\alpha} \langle \alpha \rangle = \mathcal{O}_L. \end{align*} **Step D — the candidate $y = \beta/\alpha$ is not in $\mathcal{O}_L$.** Suppose for contradiction $\beta/\alpha \in \mathcal{O}_L$. Then $\beta = \alpha \cdot (\beta/\alpha) \in \alpha \mathcal{O}_L = \langle \alpha \rangle$, contradicting $\beta \notin \langle \alpha \rangle$ (from Step B). So $y := \beta/\alpha \in L \setminus \mathcal{O}_L$ and $y\mathfrak{p} \subseteq \mathcal{O}_L$, establishing \begin{align*} \{y \in L : y\mathfrak{p} \subseteq \mathcal{O}_L\} \supsetneq \mathcal{O}_L. \end{align*} Combined with the reduction in Step 3, we conclude: for every proper ideal $\mathfrak{a} \unlhd \mathcal{O}_L$, $\{y \in L : y\mathfrak{a} \subseteq \mathcal{O}_L\}$ strictly contains $\mathcal{O}_L$. This proves (2).[/step]

[guided]The lemma from Step 4 says every nonzero ideal contains a product of primes. Applying this to the principal ideal $\langle \alpha \rangle$ (for some $0 \neq \alpha \in \mathfrak{p}$) produces primes $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ with $\mathfrak{p}_1 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle \subseteq \mathfrak{p}$. The magic lies in choosing $r$ minimal and extracting from it an element $\beta$ one step away from $\langle \alpha \rangle$ but still close enough that $\beta/\alpha$ behaves well on $\mathfrak{p}$. **Picking $\alpha$.** Choose any $\alpha \in \mathfrak{p} \setminus \{0\}$ (possible since $\mathfrak{p}$ is nonzero). **Applying Prime Product Dominance.** The principal ideal $\langle \alpha \rangle = \alpha \mathcal{O}_L$ is a nonzero ideal in $\mathcal{O}_L$. The claim from Step 4 applies: there exist nonzero primes $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ with \begin{align*} \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle. \end{align*} **Choose the minimum-length such representation.** Among all tuples $(\mathfrak{p}_1, \ldots, \mathfrak{p}_r)$ of nonzero primes with product inside $\langle \alpha \rangle$, pick one minimizing $r$. The minimum exists (every nonempty subset of $\mathbb{N}$ has a least element). **Why some $\mathfrak{p}_i = \mathfrak{p}$.** We use $\langle \alpha \rangle \subseteq \mathfrak{p}$ and compose: $\mathfrak{p}_1 \cdots \mathfrak{p}_r \subseteq \mathfrak{p}$. Since $\mathfrak{p}$ is prime, the primality condition on ideals (contrapositive form) says: if an ideal product $I \cdot J \subseteq \mathfrak{p}$, then $I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$. Iterating this for the $r$-fold product: there exists $i$ with $\mathfrak{p}_i \subseteq \mathfrak{p}$. Now $\mathfrak{p}_i$ is nonzero prime in $\mathcal{O}_L$, so by [Ring of Integers is a Dedekind Domain](/theorems/1582) it is maximal. An inclusion of a maximal ideal into a proper ideal $\mathfrak{p}_i \subseteq \mathfrak{p} \subsetneq \mathcal{O}_L$ forces equality: $\mathfrak{p}_i = \mathfrak{p}$. Relabel so that $i = 1$, i.e., $\mathfrak{p}_1 = \mathfrak{p}$. **Why the shorter product escapes $\langle \alpha \rangle$.** Consider $\mathfrak{p}_2 \cdots \mathfrak{p}_r$ (interpreted as $\mathcal{O}_L$ if $r = 1$). Suppose for contradiction $\mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle$. Then this is a length-$(r-1)$ representation contradicting minimality of $r$ — unless $r = 1$, in which case $\mathfrak{p}_2 \cdots \mathfrak{p}_r = \mathcal{O}_L \subseteq \langle \alpha \rangle$ would mean $\alpha$ is a unit, i.e., $\langle \alpha \rangle = \mathcal{O}_L$, contradicting $\alpha \in \mathfrak{p} \subsetneq \mathcal{O}_L$. Either way, we get a contradiction; hence $\mathfrak{p}_2 \cdots \mathfrak{p}_r \not\subseteq \langle \alpha \rangle$. Pick $\beta \in (\mathfrak{p}_2 \cdots \mathfrak{p}_r) \setminus \langle \alpha \rangle$. **The element $\beta/\alpha \in L$ and the containment $(\beta/\alpha) \mathfrak{p} \subseteq \mathcal{O}_L$.** We compute: \begin{align*} \beta \cdot \mathfrak{p} &= \beta \cdot \mathfrak{p}_1 \subseteq (\mathfrak{p}_2 \cdots \mathfrak{p}_r) \cdot \mathfrak{p}_1 = \mathfrak{p}_1 \cdot \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq \langle \alpha \rangle. \end{align*} (The first inclusion: $\beta \in \mathfrak{p}_2 \cdots \mathfrak{p}_r$, so $\beta\mathfrak{p}_1 \subseteq (\mathfrak{p}_2 \cdots \mathfrak{p}_r)\mathfrak{p}_1$. Commutativity of ideal multiplication then gives the equality. The final inclusion is the displayed equation for our minimum-$r$ representation.) Now divide by $\alpha$ in the field $L$. For any element $\gamma \in \beta\mathfrak{p}$, we have $\gamma \in \langle \alpha \rangle$, so $\gamma = \alpha \cdot \delta$ for some $\delta \in \mathcal{O}_L$. Thus $\gamma/\alpha = \delta \in \mathcal{O}_L$. This shows \begin{align*} \frac{\beta}{\alpha} \mathfrak{p} = \left\{\frac{\gamma}{\alpha} : \gamma \in \beta\mathfrak{p}\right\} \subseteq \mathcal{O}_L. \end{align*} **Why $\beta/\alpha \notin \mathcal{O}_L$.** If $\beta/\alpha$ were in $\mathcal{O}_L$, then $\beta = \alpha \cdot (\beta/\alpha) \in \alpha \mathcal{O}_L = \langle \alpha \rangle$ — contradicting $\beta \notin \langle \alpha \rangle$ from our construction. **Putting it together.** The element $y := \beta/\alpha$ satisfies $y \in L$, $y \notin \mathcal{O}_L$, and $y\mathfrak{p} \subseteq \mathcal{O}_L$. Hence \begin{align*} \{y \in L : y\mathfrak{p} \subseteq \mathcal{O}_L\} \supsetneq \mathcal{O}_L, \end{align*} establishing (2) for $\mathfrak{a} = \mathfrak{p}$ a nonzero maximal ideal. By Step 3, (2) follows for every proper ideal $\mathfrak{a}$. **Closing remark on the structural role of the Prime Product Dominance claim.** This lemma is the engine behind Dedekind-domain ideal theory. It says that although $\mathcal{O}_L$ may not be a unique factorization domain (so principal ideals need not be prime), at the level of ideals there is always a "prime-dominance" relation: every nonzero ideal sits above some product of primes. Unique factorization of ideals (a later theorem) upgrades this to a precise equality.[/guided]