[proofplan] The ideal norm $N(\mathfrak{a}) := |\mathcal{O}_L / \mathfrak{a}|$ is a positive integer (by [Nonzero Ideals Have Bounded Quotients](/theorems/1583)). The key observation is Lagrange's theorem applied to the additive structure: the order of the element $1 + \mathfrak{a}$ in the finite additive group $\mathcal{O}_L/\mathfrak{a}$ divides the group order $N(\mathfrak{a})$. Equivalently, $N(\mathfrak{a}) \cdot (1 + \mathfrak{a}) = 0 + \mathfrak{a}$, which unpacks to $N(\mathfrak{a}) \in \mathfrak{a}$. Combined with $N(\mathfrak{a}) \in \mathbb{Z}$ by definition, we obtain $N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}$. [/proofplan] [step:Interpret $\mathcal{O}_L/\mathfrak{a}$ as a finite additive abelian group of order $N(\mathfrak{a})$] Let $\mathfrak{a} \unlhd \mathcal{O}_L$ be a nonzero ideal. By [Nonzero Ideals Have Bounded Quotients](/theorems/1583), the quotient ring $\mathcal{O}_L/\mathfrak{a}$ is finite. Viewing $\mathcal{O}_L/\mathfrak{a}$ only as an abelian group under addition (forgetting multiplication momentarily), its order is \begin{align*} N(\mathfrak{a}) = |\mathcal{O}_L/\mathfrak{a}| \in \mathbb{Z}_{\geq 1}. \end{align*} [guided] Two preliminary observations. **Additive group structure.** The quotient $\mathcal{O}_L/\mathfrak{a}$ is naturally a commutative ring (quotient of the commutative ring $\mathcal{O}_L$ by the ideal $\mathfrak{a}$). Forgetting multiplication yields its additive abelian group structure: elements $x + \mathfrak{a}$, addition $(x + \mathfrak{a}) + (y + \mathfrak{a}) = (x+y) + \mathfrak{a}$, identity $0 + \mathfrak{a} = \mathfrak{a}$, inverses $-(x + \mathfrak{a}) = (-x) + \mathfrak{a}$. **Finiteness.** The key input: [Nonzero Ideals Have Bounded Quotients](/theorems/1583) tells us that for any nonzero ideal $\mathfrak{a} \unlhd \mathcal{O}_L$, the quotient $\mathcal{O}_L/\mathfrak{a}$ is a finite set. Its cardinality, by definition of the ideal norm, is $N(\mathfrak{a})$: \begin{align*} N(\mathfrak{a}) := |\mathcal{O}_L/\mathfrak{a}|. \end{align*} Since $\mathfrak{a} \neq \mathcal{O}_L$ (a nonzero proper ideal — if $\mathfrak{a} = \mathcal{O}_L$ the statement is vacuous, with $N(\mathfrak{a}) = 1 \in \mathcal{O}_L = \mathfrak{a}$), in fact $N(\mathfrak{a}) \geq 2$ (if $\mathfrak{a}$ is nonzero and proper). For any nonzero $\mathfrak{a}$ (including the $\mathfrak{a} = \mathcal{O}_L$ edge case), $N(\mathfrak{a})$ is a positive integer. **Why the additive structure.** The multiplicative structure of $\mathcal{O}_L/\mathfrak{a}$ is relevant for other theorems (multiplicativity of the norm, for instance), but for this particular result the additive group theory suffices — and indeed the crucial input is Lagrange's theorem on the order of elements in a finite abelian group. [/guided] [/step] [step:Apply Lagrange's theorem to the element $1 + \mathfrak{a}$] Consider the element \begin{align*} \bar{1} := 1 + \mathfrak{a} \in \mathcal{O}_L/\mathfrak{a}. \end{align*} By Lagrange's theorem for finite abelian groups (or its corollary: the order of every element divides the order of the group), the order of $\bar{1}$ in the additive group $\mathcal{O}_L/\mathfrak{a}$ divides $N(\mathfrak{a})$. In particular, taking the $N(\mathfrak{a})$-fold sum of $\bar{1}$ with itself in the additive group yields the identity: \begin{align*} \underbrace{\bar{1} + \bar{1} + \cdots + \bar{1}}_{N(\mathfrak{a}) \text{ terms}} = \bar{0} \in \mathcal{O}_L/\mathfrak{a}, \end{align*} where $\bar{0} = 0 + \mathfrak{a} = \mathfrak{a}$ is the additive identity. [guided] **Lagrange's theorem in the additive form.** If $G$ is a finite abelian group with $|G| = m$, then for every $g \in G$, the order of $g$ divides $m$. Equivalently, $m \cdot g = 0_G$ (writing the operation additively). **Hypothesis verification.** - $G = \mathcal{O}_L/\mathfrak{a}$: finite abelian group under addition (Step 1). In particular, $G$ is finite, so Lagrange applies. - $|G| = N(\mathfrak{a})$: by definition. - $g = \bar{1} = 1 + \mathfrak{a} \in G$: well-defined element. **Conclusion of Lagrange.** $N(\mathfrak{a}) \cdot \bar{1} = 0_G$ in $G$. In additive notation, \begin{align*} \underbrace{\bar{1} + \bar{1} + \cdots + \bar{1}}_{N(\mathfrak{a}) \text{ terms}} = \bar{0}. \end{align*} **An alternative view.** The cyclic subgroup $\langle \bar{1} \rangle \leq \mathcal{O}_L/\mathfrak{a}$ generated by $\bar{1}$ is a subgroup of the finite abelian group $\mathcal{O}_L/\mathfrak{a}$. By Lagrange, $|\langle \bar{1} \rangle|$ divides $|\mathcal{O}_L/\mathfrak{a}| = N(\mathfrak{a})$. The order of $\bar{1}$ is $|\langle \bar{1} \rangle|$, so the order divides $N(\mathfrak{a})$, confirming $N(\mathfrak{a}) \cdot \bar{1} = 0$. (This is the standard trick in number-theoretic applications of Lagrange: group-theoretic order dividing group order, applied to the specific element $1$, gives an identity of the form "$m = 0$ in the quotient", which unpacks to "$m$ lies in the ideal".) [/guided] [/step] [step:Unpack the identity $N(\mathfrak{a}) \cdot \bar{1} = \bar{0}$ into a statement in $\mathcal{O}_L$] The identity in $\mathcal{O}_L/\mathfrak{a}$, \begin{align*} N(\mathfrak{a}) \cdot \bar{1} = \bar{0}, \end{align*} is equivalent to the statement that $N(\mathfrak{a}) \cdot 1 = N(\mathfrak{a}) \in \mathcal{O}_L$ lies in the kernel of the quotient map $\mathcal{O}_L \twoheadrightarrow \mathcal{O}_L/\mathfrak{a}$, which is $\mathfrak{a}$. That is, \begin{align*} N(\mathfrak{a}) \in \mathfrak{a}. \end{align*} Note we have used $N(\mathfrak{a}) \cdot \bar{1} = \overline{N(\mathfrak{a}) \cdot 1} = \overline{N(\mathfrak{a})}$, which combines the additive-group action of $\mathbb{Z}$ on $\mathcal{O}_L/\mathfrak{a}$ (repeated addition) with the ring embedding $\mathbb{Z} \hookrightarrow \mathcal{O}_L$ sending $n$ to $n \cdot 1_{\mathcal{O}_L} = n$. [guided] **Decoding the $\mathbb{Z}$-action.** For any abelian group $G$ written additively and any integer $m \in \mathbb{Z}$, the element $m \cdot g$ is well-defined as $g + g + \cdots + g$ ($m$ times, interpreting negative $m$ as adding inverses). In particular, when $G = \mathcal{O}_L/\mathfrak{a}$ and $g = \bar{1} = 1 + \mathfrak{a}$: \begin{align*} m \cdot \bar{1} = \underbrace{\bar{1} + \cdots + \bar{1}}_{m \text{ times}} = m \cdot 1 + \mathfrak{a} = m + \mathfrak{a} = \bar{m}, \end{align*} using that the embedding $\mathbb{Z} \hookrightarrow \mathcal{O}_L$, $m \mapsto m \cdot 1_{\mathcal{O}_L}$, sends $m$ to the ring element $m$ itself. **The specific case.** With $m = N(\mathfrak{a})$, \begin{align*} N(\mathfrak{a}) \cdot \bar{1} = \overline{N(\mathfrak{a})} = N(\mathfrak{a}) + \mathfrak{a}. \end{align*} The equation $N(\mathfrak{a}) \cdot \bar{1} = \bar{0}$ from Step 2 reads \begin{align*} N(\mathfrak{a}) + \mathfrak{a} = 0 + \mathfrak{a}, \end{align*} i.e., the cosets of $N(\mathfrak{a})$ and $0$ are equal in $\mathcal{O}_L/\mathfrak{a}$. This means $N(\mathfrak{a}) - 0 = N(\mathfrak{a}) \in \mathfrak{a}$. Alternatively: the kernel of the quotient map $\pi: \mathcal{O}_L \to \mathcal{O}_L/\mathfrak{a}$ is $\mathfrak{a}$. The equation $\pi(N(\mathfrak{a})) = \bar{0}$ is equivalent to $N(\mathfrak{a}) \in \ker \pi = \mathfrak{a}$. **Conclusion so far.** $N(\mathfrak{a}) \in \mathfrak{a}$. [/guided] [/step] [step:Combine $N(\mathfrak{a}) \in \mathfrak{a}$ and $N(\mathfrak{a}) \in \mathbb{Z}$ to conclude] By Step 3, $N(\mathfrak{a}) \in \mathfrak{a}$. By definition of the ideal norm, $N(\mathfrak{a}) = |\mathcal{O}_L/\mathfrak{a}| \in \mathbb{Z}$ (in fact, a positive integer). Intersecting the two memberships: \begin{align*} N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}. \end{align*} This is the conclusion of the theorem. [guided] **Two memberships to intersect.** - **$N(\mathfrak{a}) \in \mathfrak{a}$** (Step 3, via Lagrange). - **$N(\mathfrak{a}) \in \mathbb{Z}$**: directly from the definition $N(\mathfrak{a}) := |\mathcal{O}_L/\mathfrak{a}|$, which is the cardinality of a finite set, hence a nonnegative integer (positive if $\mathfrak{a} \neq (0)$). **Intersection.** By elementary set theory, $x \in A$ and $x \in B$ implies $x \in A \cap B$. With $x = N(\mathfrak{a})$, $A = \mathfrak{a}$, $B = \mathbb{Z}$: \begin{align*} N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}. \end{align*} **Here $\mathbb{Z} \subseteq \mathcal{O}_L$ is understood.** The intersection takes place inside $\mathcal{O}_L$, using the inclusion $\mathbb{Z} \subseteq \mathcal{O}_L$ (every rational integer is an algebraic integer). So $\mathfrak{a} \cap \mathbb{Z}$ is the set of rational integers $n$ with $n \in \mathfrak{a}$, or equivalently, $n\mathcal{O}_L \subseteq \mathfrak{a}$. **Why this matters.** The conclusion $N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}$ provides an explicit rational integer that lies in $\mathfrak{a}$. This has two important consequences used in subsequent theory: 1. **The ideal $\mathfrak{a}$ contains $N(\mathfrak{a}) \cdot \mathcal{O}_L$**, giving the inclusion $\langle N(\mathfrak{a}) \rangle \subseteq \mathfrak{a}$. Equivalently, $\mathfrak{a}$ divides $\langle N(\mathfrak{a}) \rangle$ — the rational integer $N(\mathfrak{a})$ is a "multiple" of $\mathfrak{a}$ in the divisibility order on ideals (via [Divisibility Equals Containment](/theorems/1587)). 2. **Every prime ideal lies above a rational prime.** If $\mathfrak{p}$ is a nonzero prime ideal, Step 3 gives $N(\mathfrak{p}) \in \mathfrak{p} \cap \mathbb{Z}$. Hence $\mathfrak{p} \cap \mathbb{Z}$ is a nonzero ideal of $\mathbb{Z}$; being the preimage of the prime ideal $\mathfrak{p}$ under the inclusion $\mathbb{Z} \hookrightarrow \mathcal{O}_L$, it is itself a prime ideal of $\mathbb{Z}$. The nonzero prime ideals of $\mathbb{Z}$ are $\langle p \rangle$ for $p$ a rational prime. Hence $\mathfrak{p} \cap \mathbb{Z} = \langle p \rangle$ for some rational prime $p$, i.e., "$\mathfrak{p}$ lies above $p$". This is the starting point of the theory of prime decomposition in number fields. 3. **Finiteness of ideals of bounded norm.** The set of integral ideals with norm bounded by some $M$ is finite. The proof uses that each such $\mathfrak{a}$ contains $\mathcal{O}_L$ times an integer $\leq M$, so $\mathfrak{a} \supseteq \langle N(\mathfrak{a}) \rangle \supseteq \langle M! \rangle$, reducing the count to the finitely many divisors of $\langle M! \rangle$. [/guided] [/step]