[proofplan] The argument splits into two halves. First, the intersection $\mathfrak{p} \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$, hence principal — say $\mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$. We show $p$ is a rational prime using the primality of $\mathfrak{p}$ together with the ring structure of $\mathbb{Z}$: if $p = ab$ with $a, b \in \mathbb{Z}$, then $ab \in \mathfrak{p}$ forces one of $a, b \in \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$, so $p$ divides one of its factors. Uniqueness of $p$ follows from the uniqueness of the generator of the principal ideal $\mathfrak{p} \cap \mathbb{Z}$, up to sign. Second, for the norm statement, we use the fact that $\langle p \rangle \subseteq \mathfrak{p}$ translates (via divisibility equals containment) to $\mathfrak{p} \mid \langle p \rangle$, so $\langle p \rangle = \mathfrak{p}\mathfrak{a}$ for some ideal $\mathfrak{a}$. Taking norms and using [Multiplicativity of the Ideal Norm](/theorems/1593) together with [Element Norm Equals Ideal Norm](/theorems/1596) gives $p^n = N(\mathfrak{p}) N(\mathfrak{a})$, whence $N(\mathfrak{p}) = p^f$ with $1 \leq f \leq n$. [/proofplan]

[step:Exhibit the rational prime via the intersection $\mathfrak{p} \cap \mathbb{Z}$]Consider the inclusion of rings \begin{align*} \iota: \mathbb{Z} \hookrightarrow \mathcal{O}_L. \end{align*} The preimage $\iota^{-1}(\mathfrak{p}) = \mathfrak{p} \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$ (ring-homomorphism preimages of ideals are ideals). Since $\mathbb{Z}$ is a principal ideal domain, there exists a unique non-negative integer $p$ with \begin{align*} \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}. \end{align*}[/step]

[guided]**Why $\mathfrak{p} \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$.** A standard fact: if $\varphi: R \to S$ is a ring homomorphism and $J \unlhd S$, then $\varphi^{-1}(J) \unlhd R$. Applied to the inclusion $\iota: \mathbb{Z} \hookrightarrow \mathcal{O}_L$ and $J = \mathfrak{p}$, we get $\iota^{-1}(\mathfrak{p}) = \mathfrak{p} \cap \mathbb{Z} \unlhd \mathbb{Z}$. We can also check directly: if $a, b \in \mathfrak{p} \cap \mathbb{Z}$, then $a, b \in \mathfrak{p}$ and $\mathfrak{p}$ is closed under addition, so $a + b \in \mathfrak{p}$; also $a + b \in \mathbb{Z}$, so $a + b \in \mathfrak{p} \cap \mathbb{Z}$. For $z \in \mathbb{Z}$ and $a \in \mathfrak{p} \cap \mathbb{Z}$, we have $za \in \mathfrak{p}$ (since $\mathfrak{p}$ is an ideal in $\mathcal{O}_L$ and $z \in \mathbb{Z} \subseteq \mathcal{O}_L$) and $za \in \mathbb{Z}$. **Principal generator.** $\mathbb{Z}$ is a PID: every ideal is of the form $m\mathbb{Z}$ for a unique $m \in \mathbb{Z}_{\geq 0}$ (the non-negative generator). Let $p \in \mathbb{Z}_{\geq 0}$ be this generator of $\mathfrak{p} \cap \mathbb{Z}$: \begin{align*} \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}. \end{align*}[/guided]

[step:Show that $p > 0$ by using that $\mathcal{O}_L/\mathfrak{p}$ is a finite field]We claim $p \neq 0$. By [Nonzero Ideals Have Bounded Quotients](/theorems/1583), $\mathcal{O}_L/\mathfrak{p}$ is a finite ring (since $\mathfrak{p}$ is a nonzero prime ideal — primality rules out $\mathfrak{p} = \mathcal{O}_L$, and the standing assumption in the theorem excludes $\mathfrak{p} = \{0\}$ by virtue of the word "prime ideal" taking $\mathfrak{p}$ to be a *nonzero* prime ideal in the Dedekind-domain convention used on this page). Since $\mathfrak{p}$ is prime and the quotient $\mathcal{O}_L/\mathfrak{p}$ is a finite integral domain, it is in fact a finite field of positive characteristic $p'$ (every finite integral domain is a field). The characteristic $p'$ is the smallest positive integer $m$ with $m \cdot 1 = 0$ in $\mathcal{O}_L/\mathfrak{p}$, equivalently the smallest positive integer $m$ with $m \in \mathfrak{p}$. Hence $p' \in \mathfrak{p} \cap \mathbb{Z}$, so $p'$ is a positive element of $p\mathbb{Z}$. Therefore $p\mathbb{Z}$ contains a positive integer, so $p\mathbb{Z} \neq \{0\}$, and $p > 0$.[/step]

[guided]**Why $p \neq 0$.** If $p = 0$, then $\mathfrak{p} \cap \mathbb{Z} = \{0\}$, meaning no nonzero integer lies in $\mathfrak{p}$. But this contradicts a more basic fact: $\mathfrak{p}$ is a nonzero prime ideal of $\mathcal{O}_L$, and $\mathcal{O}_L/\mathfrak{p}$ is a finite field (see below), so $\mathfrak{p}$ must contain the prime characteristic of that field, which is a positive integer. **$\mathcal{O}_L/\mathfrak{p}$ is a finite integral domain.** - **Finite**: By [Nonzero Ideals Have Bounded Quotients](/theorems/1583), $|\mathcal{O}_L/\mathfrak{p}|$ is finite since $\mathfrak{p}$ is nonzero. - **Integral domain**: By definition of a prime ideal, $\mathfrak{p}$ is prime iff $\mathcal{O}_L/\mathfrak{p}$ is an integral domain. - **Field**: Every finite integral domain is a field — given $a \neq 0$ in $\mathcal{O}_L/\mathfrak{p}$, the map $x \mapsto ax$ is injective (by cancellation in an integral domain) on a finite set, hence surjective, so $a$ has an inverse. **Positive characteristic.** A finite field has some positive characteristic $p'$: the smallest positive $m$ with $m \cdot 1 \equiv 0 \pmod{\mathfrak{p}}$. By definition, $p' \in \mathfrak{p} \cap \mathbb{Z}$, and $p'$ is positive, so $\mathfrak{p} \cap \mathbb{Z}$ contains a positive integer. Consequently, $p\mathbb{Z} = \mathfrak{p} \cap \mathbb{Z}$ contains a positive integer, hence $p \neq 0$, and by the convention $p \geq 0$ we conclude $p > 0$.[/guided]

[step:Show that $p$ is prime in $\mathbb{Z}$]We verify $p$ is a prime integer by showing (i) $p > 1$ and (ii) if $p = ab$ with $a, b \in \mathbb{Z}_{> 0}$, then $a = 1$ or $b = 1$. (i) $p \neq 1$: otherwise $\mathfrak{p} \cap \mathbb{Z} = \mathbb{Z}$, so $1 \in \mathfrak{p}$, forcing $\mathfrak{p} = \mathcal{O}_L$. This contradicts the primality of $\mathfrak{p}$ (primes are proper). (ii) Suppose $p = ab$ with $a, b \in \mathbb{Z}$, $a, b > 0$. Then $p \in \mathfrak{p} \cap \mathbb{Z}$ gives $ab = p \in \mathfrak{p}$. By the primality of $\mathfrak{p}$ applied to the pair $a, b \in \mathcal{O}_L$, either $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$. Since $a, b \in \mathbb{Z}$, we get $a \in \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$ or $b \in \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$, i.e., $p \mid a$ or $p \mid b$. But also $a \mid p$ and $b \mid p$ from $p = ab$. Combined with $p \mid a$ (resp. $p \mid b$), and positivity, we obtain $a = p, b = 1$ or $a = 1, b = p$. Therefore $p$ is a prime integer.[/step]

[guided]**Ruling out $p = 1$.** If $p = 1$, the ideal $p\mathbb{Z} = \mathbb{Z}$ equals all of $\mathbb{Z}$, and $1 \in \mathfrak{p}$. But $\mathfrak{p}$ is an ideal of $\mathcal{O}_L$ containing $1$, so $\mathfrak{p} = \mathcal{O}_L$. Primality of $\mathfrak{p}$ requires $\mathfrak{p} \neq \mathcal{O}_L$ (primes are proper by definition), contradiction. **Proving $p$ is prime using primality of $\mathfrak{p}$.** We use the Euclidean characterisation: $p$ is prime iff for $a, b \in \mathbb{Z}$ with $p \mid ab$ we have $p \mid a$ or $p \mid b$. Equivalently (for $p > 1$): any factorisation $p = ab$ with $a, b > 0$ forces $\{a, b\} = \{1, p\}$. Suppose $p = ab$ with $a, b > 0$. Then $p \in \mathfrak{p}$, i.e., $ab \in \mathfrak{p}$. Primality of $\mathfrak{p}$ says: \begin{align*} ab \in \mathfrak{p} \implies a \in \mathfrak{p} \text{ or } b \in \mathfrak{p}. \end{align*} WLOG $a \in \mathfrak{p}$. Since $a \in \mathbb{Z}$, this means $a \in \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$, so $p \mid a$. Combined with $a \mid p$ (from $p = ab$) and positivity, $a = p$ and $b = 1$. Thus $p$ is a rational prime.[/guided]

[step:Prove uniqueness of the prime $p$]Suppose $p, p'$ are rational primes with $p \in \mathfrak{p}$ and $p' \in \mathfrak{p}$. Then $p\mathbb{Z} \subseteq \mathfrak{p} \cap \mathbb{Z}$ and $p'\mathbb{Z} \subseteq \mathfrak{p} \cap \mathbb{Z}$. Since $\mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$ by Step 1 (where $p$ is *the* non-negative generator), we must have $p'\mathbb{Z} \subseteq p\mathbb{Z}$, i.e., $p \mid p'$. But $p, p'$ are both rational primes, so $p' = p$ (a prime divides another prime iff they are equal). Hence $p$ is the unique rational prime lying in $\mathfrak{p}$. The equivalence $\mathfrak{p} \mid \langle p \rangle \iff \langle p \rangle \subseteq \mathfrak{p}$ follows: $\langle p \rangle = p\mathbb{Z} \cdot \mathcal{O}_L = p\mathcal{O}_L \subseteq \mathfrak{p}$ uses $p \in \mathfrak{p}$, while conversely $\langle p \rangle \subseteq \mathfrak{p}$ gives $p \in \mathfrak{p}$.[/step]

[guided]**Uniqueness.** Suppose two rational primes $p, p'$ both lie in $\mathfrak{p}$. Then $p\mathbb{Z} \subseteq \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$ and $p'\mathbb{Z} \subseteq \mathfrak{p} \cap \mathbb{Z} = p\mathbb{Z}$. The inclusion $p'\mathbb{Z} \subseteq p\mathbb{Z}$ means $p'$ is a multiple of $p$; since both are positive primes and $p \mid p'$, we must have $p = p'$. **Containment $\iff$ divisibility.** The two statements "$\mathfrak{p} \mid \langle p \rangle$" and "$\langle p \rangle \subseteq \mathfrak{p}$" are equivalent by [Divisibility Equals Containment](/theorems/1587). We now double-check them both: - $p \in \mathfrak{p}$ (by construction). - Hence $p \cdot \mathcal{O}_L \subseteq \mathfrak{p}$ (ideal property: $\mathfrak{p}$ is closed under multiplication by $\mathcal{O}_L$), i.e., $\langle p \rangle \subseteq \mathfrak{p}$. Conversely, $\langle p \rangle \subseteq \mathfrak{p}$ immediately gives $p \in \mathfrak{p}$.[/guided]

[step:Compute $N(\mathfrak{p}) = p^f$ via the norm of $\langle p \rangle$]We have $\langle p \rangle \subseteq \mathfrak{p}$, which by [Divisibility Equals Containment](/theorems/1587) is equivalent to $\mathfrak{p} \mid \langle p \rangle$. By unique factorisation in the ideal group — the [Ideal Group Is Free Abelian](/theorems/1590) result combined with the [Unique Factorization of Ideals](/theorems/1589) — we can write \begin{align*} \langle p \rangle = \mathfrak{p} \cdot \mathfrak{a} \end{align*} for some ideal $\mathfrak{a} \unlhd \mathcal{O}_L$. Taking norms, by the [Multiplicativity of the Ideal Norm](/theorems/1593), \begin{align*} N(\langle p \rangle) = N(\mathfrak{p}) \cdot N(\mathfrak{a}). \end{align*} By [Element Norm Equals Ideal Norm](/theorems/1596), $N(\langle p \rangle) = |N_{L/\mathbb{Q}}(p)|$. Since $p \in \mathbb{Z} \subseteq L$, we have $N_{L/\mathbb{Q}}(p) = p^n$ (the norm of a rational integer $p$ is $p^{[L:\mathbb{Q}]}$, as multiplication by $p$ on $L$ has the same characteristic polynomial as multiplication by $p$ on $\mathbb{Q}$, namely $(x - p)^n$, after base change). Therefore \begin{align*} N(\langle p \rangle) = p^n. \end{align*} Substituting, \begin{align*} p^n = N(\mathfrak{p}) \cdot N(\mathfrak{a}). \end{align*} Both $N(\mathfrak{p})$ and $N(\mathfrak{a})$ are positive integers, and their product equals $p^n$. By the fundamental theorem of arithmetic, $N(\mathfrak{p})$ and $N(\mathfrak{a})$ are each a power of $p$: $N(\mathfrak{p}) = p^f$ and $N(\mathfrak{a}) = p^{n-f}$ for some integer $f \geq 0$, $n - f \geq 0$. Finally, $f \geq 1$: if $f = 0$, then $N(\mathfrak{p}) = 1$, so $|\mathcal{O}_L/\mathfrak{p}| = 1$, forcing $\mathfrak{p} = \mathcal{O}_L$, which contradicts primality. Hence $1 \leq f \leq n$, and \begin{align*} N(\mathfrak{p}) = p^f. \end{align*}[/step]

[guided]**From containment to factorisation.** We know $\langle p \rangle \subseteq \mathfrak{p}$ from Steps 1-4. By [Divisibility Equals Containment](/theorems/1587), this is the same as $\mathfrak{p} \mid \langle p \rangle$. The [Unique Factorization of Ideals](/theorems/1589) says every nonzero ideal of $\mathcal{O}_L$ factors uniquely as a product of prime ideals. In particular, $\langle p \rangle$ has such a factorisation, and the primes appearing are exactly those dividing $\langle p \rangle$. Since $\mathfrak{p}$ is one of them, we can write \begin{align*} \langle p \rangle = \mathfrak{p} \cdot \mathfrak{a} \end{align*} for some ideal $\mathfrak{a}$ (the product of the remaining prime factors with their multiplicities). **Taking norms.** The [Multiplicativity of the Ideal Norm](/theorems/1593) theorem states $N(\mathfrak{b}_1 \mathfrak{b}_2) = N(\mathfrak{b}_1) N(\mathfrak{b}_2)$ for nonzero ideals. Applying this, \begin{align*} N(\langle p \rangle) = N(\mathfrak{p})\, N(\mathfrak{a}). \end{align*} **Computing $N(\langle p \rangle)$.** The [Element Norm Equals Ideal Norm](/theorems/1596) theorem says $N(\langle \beta \rangle) = |N_{L/\mathbb{Q}}(\beta)|$ for $\beta \in \mathcal{O}_L$. Take $\beta = p$: we need $N_{L/\mathbb{Q}}(p)$. The field norm $N_{L/\mathbb{Q}}(p)$ is by definition $\det(m_p)$, where $m_p: L \to L$, $x \mapsto px$. Since $p \in \mathbb{Q}$, $m_p = p \cdot \operatorname{id}_L$, so $\det(m_p) = p^{\dim_\mathbb{Q} L} = p^n$. Hence $N_{L/\mathbb{Q}}(p) = p^n$ (positive since $p > 0$), and $|N_{L/\mathbb{Q}}(p)| = p^n$. Plugging in: \begin{align*} p^n = N(\mathfrak{p})\, N(\mathfrak{a}). \end{align*} **Extracting $N(\mathfrak{p}) = p^f$.** Both $N(\mathfrak{p})$ and $N(\mathfrak{a})$ are positive integers multiplying to $p^n$. By unique factorisation in $\mathbb{Z}$, the only divisors of $p^n$ are $1, p, p^2, \ldots, p^n$. So $N(\mathfrak{p}) = p^f$ for some integer $0 \leq f \leq n$. **Ruling out $f = 0$.** If $f = 0$, then $N(\mathfrak{p}) = 1$, i.e., $|\mathcal{O}_L/\mathfrak{p}| = 1$, so $\mathcal{O}_L/\mathfrak{p} = \{0\}$, i.e., $\mathfrak{p} = \mathcal{O}_L$. This contradicts primality of $\mathfrak{p}$ (primes are proper ideals). Hence $f \geq 1$. **Conclusion.** $N(\mathfrak{p}) = p^f$ with $1 \leq f \leq n$, as claimed. **Terminology.** The integer $f$ is called the **residue degree** (or **inertial degree**) of $\mathfrak{p}$ over $p$. Geometrically, $\mathcal{O}_L/\mathfrak{p}$ is a finite field extension of $\mathbb{F}_p$ of degree $f$.[/guided]