[proofplan] The strategy is a two-step reduction. First, we show that any ideal $\mathfrak{a} \leq \mathcal{O}_L$ with $N(\mathfrak{a}) = m$ must divide the principal ideal $\langle m \rangle$; this uses [Norm Lies in the Ideal](/theorems/1592) combined with [Divisibility Equals Containment](/theorems/1587). Second, we invoke [Unique Factorization of Ideals](/theorems/1589) to conclude that $\langle m \rangle$ has only finitely many divisors, since it decomposes into a fixed finite product of primes. Both reductions are purely algebraic: no analytic or geometric input is needed. [/proofplan]

[step:Show that $N(\mathfrak{a}) = m$ forces $m \in \mathfrak{a}$]Suppose $\mathfrak{a} \leq \mathcal{O}_L$ is a non-zero ideal with $N(\mathfrak{a}) = m$. By [Norm Lies in the Ideal](/theorems/1592), the norm satisfies \begin{align*} N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}. \end{align*} Substituting $N(\mathfrak{a}) = m$ yields $m \in \mathfrak{a}$.[/step]

[guided]We begin by extracting an algebraic consequence of the hypothesis $N(\mathfrak{a}) = m$ that we can combine with the ideal-theoretic machinery of $\mathcal{O}_L$. The key input is [Theorem 1592](/theorems/1592), which asserts that for any non-zero ideal $\mathfrak{a} \unlhd \mathcal{O}_L$, \begin{align*} N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}. \end{align*} **Verifying the hypothesis of Theorem 1592.** The theorem requires $\mathfrak{a}$ to be a non-zero ideal of $\mathcal{O}_L$. We are working with $\mathfrak{a}$ of norm $m \in \mathbb{Z}_{>0}$, so $\mathfrak{a} \neq 0$ (the zero ideal has infinite residue, not finite norm $m$). The hypothesis is met. **What the theorem gives us.** The conclusion $m = N(\mathfrak{a}) \in \mathfrak{a}$ is the statement that the positive integer $m$ lies in $\mathfrak{a}$ as an element of $\mathcal{O}_L$. This seemingly innocuous fact is the entire bridge from the numerical data $N(\mathfrak{a}) = m$ to the containment relation $\mathfrak{a} \supseteq \langle m \rangle$, which (in the next step) will become the divisibility $\mathfrak{a} \mid \langle m \rangle$. **Why this works.** The reason $m \in \mathfrak{a}$ is that the additive group $\mathcal{O}_L/\mathfrak{a}$ has order $m$: the additive identity $\bar{1}$ satisfies $m \cdot \bar{1} = \bar{0}$ by [Lagrange's theorem](/theorems/???) applied to the additive group, so $m \cdot 1 = m \in \mathfrak{a}$. (This is the proof sketch of Theorem 1592, included here to emphasise that Lagrange's theorem is the operative tool.)[/guided]

[step:Conclude that $\mathfrak{a}$ divides $\langle m \rangle$]Since $m \in \mathfrak{a}$, every multiple $rm$ with $r \in \mathcal{O}_L$ lies in $\mathfrak{a}$ (because $\mathfrak{a}$ is closed under multiplication by elements of $\mathcal{O}_L$). Thus \begin{align*} \langle m \rangle = m \mathcal{O}_L \subseteq \mathfrak{a}. \end{align*} By [Divisibility Equals Containment](/theorems/1587) part (3), containment $\langle m \rangle \subseteq \mathfrak{a}$ is equivalent to divisibility $\mathfrak{a} \mid \langle m \rangle$. Hence $\mathfrak{a}$ is a divisor of $\langle m \rangle$ in the monoid of non-zero integral ideals of $\mathcal{O}_L$.[/step]

[guided]We now convert the element-level containment $m \in \mathfrak{a}$ into an ideal-level divisibility statement. **Upgrading $m \in \mathfrak{a}$ to $\langle m \rangle \subseteq \mathfrak{a}$.** The principal ideal $\langle m \rangle = m \mathcal{O}_L$ consists of all $\mathcal{O}_L$-multiples of $m$. For any $rm \in \langle m \rangle$ with $r \in \mathcal{O}_L$, since $\mathfrak{a}$ is an ideal it absorbs multiplication by elements of $\mathcal{O}_L$, giving $rm = r \cdot m \in \mathfrak{a}$. Hence every element of $\langle m \rangle$ lies in $\mathfrak{a}$, i.e., $\langle m \rangle \subseteq \mathfrak{a}$. **From containment to divisibility.** In a general ring this step does *not* automatically transfer, but in a Dedekind domain — and $\mathcal{O}_L$ is one by [Ring of Integers is a Dedekind Domain](/theorems/1582) — containment and divisibility of non-zero ideals are the same relation. We verify the hypotheses of [Theorem 1587](/theorems/1587) part (3): it asserts $\mathfrak{a} \mid \mathfrak{b} \iff \mathfrak{b} \subseteq \mathfrak{a}$ for ideals $\mathfrak{a}, \mathfrak{b} \unlhd \mathcal{O}_L$. Our ideals are $\mathfrak{a}$ (the given ideal of norm $m$, non-zero) and $\mathfrak{b} = \langle m \rangle$ (principal and non-zero since $m \in \mathbb{Z}_{>0}$). Both are integral ideals of $\mathcal{O}_L$, so Theorem 1587 part (3) applies. Applying it: $\langle m \rangle \subseteq \mathfrak{a}$ implies $\mathfrak{a} \mid \langle m \rangle$. Thus in the multiplicative monoid of non-zero integral ideals, $\mathfrak{a}$ is a divisor of $\langle m \rangle$. **Conceptually.** The Dedekind-domain identification of "contains" with "divides" is the analogue of the elementary fact for principal ideals in $\mathbb{Z}$: $n \mathbb{Z} \subseteq k \mathbb{Z}$ iff $k \mid n$. In Dedekind domains the monoid of non-zero ideals factors uniquely into primes, and divisibility in that monoid matches reverse inclusion just as for integers.[/guided]

[step:Count divisors of $\langle m \rangle$ via prime factorization]By [Unique Factorization of Ideals](/theorems/1589), $\langle m \rangle$ factors uniquely as a product of prime ideals: \begin{align*} \langle m \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r}, \end{align*} where $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$ are distinct non-zero prime ideals of $\mathcal{O}_L$ and $e_1, \ldots, e_r \in \mathbb{Z}_{\geq 1}$. The divisors of $\langle m \rangle$ in the monoid of non-zero integral ideals are precisely the ideals of the form \begin{align*} \mathfrak{p}_1^{f_1} \cdots \mathfrak{p}_r^{f_r}, \qquad 0 \leq f_i \leq e_i, \end{align*} by unique factorization (each divisor has exponents bounded above by the corresponding $e_i$, and no new primes can appear). Hence the number of divisors of $\langle m \rangle$ is \begin{align*} \prod_{i=1}^r (e_i + 1), \end{align*} which is finite. Since every $\mathfrak{a}$ with $N(\mathfrak{a}) = m$ is a divisor of $\langle m \rangle$ by Steps 1-2, the set of such $\mathfrak{a}$ is a subset of this finite set of divisors, hence finite.[/step]

[guided]The final step combines the divisibility statement of Step 2 with the unique factorization of $\langle m \rangle$ to count the candidates. **Factoring $\langle m \rangle$.** We apply [Unique Factorization of Ideals](/theorems/1589) to $\mathfrak{b} = \langle m \rangle$. Theorem 1589 requires $\mathfrak{b}$ to be a non-zero ideal of $\mathcal{O}_L$. Since $m \geq 1$, $\langle m \rangle$ is non-zero; since it is an ideal of $\mathcal{O}_L$ by construction, both hypotheses are satisfied. Hence there exist distinct prime ideals $\mathfrak{p}_1, \ldots, \mathfrak{p}_r \unlhd \mathcal{O}_L$ and exponents $e_1, \ldots, e_r \in \mathbb{Z}_{\geq 1}$ such that \begin{align*} \langle m \rangle = \prod_{i=1}^r \mathfrak{p}_i^{e_i}. \end{align*} **Describing all divisors of $\langle m \rangle$.** Let $\mathfrak{d} \unlhd \mathcal{O}_L$ be any non-zero integral ideal with $\mathfrak{d} \mid \langle m \rangle$. By [Theorem 1589](/theorems/1589), $\mathfrak{d}$ itself factors uniquely as a product $\prod_{\mathfrak{q}} \mathfrak{q}^{c_\mathfrak{q}}$ over all prime ideals $\mathfrak{q}$ of $\mathcal{O}_L$, with $c_\mathfrak{q} \in \mathbb{Z}_{\geq 0}$ and $c_\mathfrak{q} = 0$ for all but finitely many $\mathfrak{q}$. The relation $\mathfrak{d} \mid \langle m \rangle$ (i.e., $\langle m \rangle = \mathfrak{d} \mathfrak{e}$ for some integral ideal $\mathfrak{e}$) gives \begin{align*} \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_r^{e_r} = \left(\prod_{\mathfrak{q}} \mathfrak{q}^{c_\mathfrak{q}}\right) \cdot \mathfrak{e}. \end{align*} By uniqueness of the prime factorization on both sides, the primes appearing in $\mathfrak{d}$ (those $\mathfrak{q}$ with $c_\mathfrak{q} > 0$) must all appear in $\prod \mathfrak{p}_i^{e_i}$, i.e., $\mathfrak{q} \in \{\mathfrak{p}_1, \ldots, \mathfrak{p}_r\}$; and the exponent $c_{\mathfrak{p}_i}$ of each $\mathfrak{p}_i$ in $\mathfrak{d}$ satisfies $c_{\mathfrak{p}_i} \leq e_i$ (since $\mathfrak{e}$ contributes a non-negative complementary exponent). Thus \begin{align*} \mathfrak{d} = \mathfrak{p}_1^{f_1} \cdots \mathfrak{p}_r^{f_r}, \qquad f_i \in \{0, 1, \ldots, e_i\}. \end{align*} **Counting.** For each $i$ there are $e_i + 1$ choices of the exponent $f_i$, and these choices are independent. The total number of divisors is therefore \begin{align*} \#\{\mathfrak{d} : \mathfrak{d} \mid \langle m \rangle\} = \prod_{i=1}^r (e_i + 1) < \infty. \end{align*} **Applying to the norm problem.** Let \begin{align*} S_m = \{\mathfrak{a} \leq \mathcal{O}_L : \mathfrak{a} \text{ non-zero},\ N(\mathfrak{a}) = m\}. \end{align*} By Step 2, every $\mathfrak{a} \in S_m$ divides $\langle m \rangle$. Hence $S_m$ is contained in the finite set of divisors of $\langle m \rangle$ counted above. A subset of a finite set is finite, so $S_m$ is finite. This completes the proof. **Alternative (sanity check).** One can also see this via the ring structure: an ideal $\mathfrak{a} \supseteq \langle m \rangle$ corresponds bijectively to an ideal $\mathfrak{a}/\langle m \rangle$ of the quotient ring $\mathcal{O}_L/\langle m \rangle$. By [Existence of an Integral Basis](/theorems/1583), $\mathcal{O}_L \cong \mathbb{Z}^n$ as abelian groups, so \begin{align*} \mathcal{O}_L / \langle m \rangle = \mathcal{O}_L / m \mathcal{O}_L \cong (\mathbb{Z}/m\mathbb{Z})^n \end{align*} as abelian groups, a finite set of size $m^n$. Any finite ring has only finitely many ideals (each is a subset of a finite set), so there are only finitely many $\mathfrak{a} \supseteq \langle m \rangle$. This gives an alternative route to the same conclusion and confirms the finiteness bound is at most $m^n$ (in fact typically much smaller, by the $\prod (e_i + 1)$ formula above).[/guided]