[proofplan] The proof has two logically distinct halves: establishing that each $\mathfrak{p}_i$ is a prime ideal of $\mathcal{O}_L$, and establishing the product formula $\langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$. For primality, we show the ring isomorphism $\mathcal{O}_L/\mathfrak{p}_i \cong \mathbb{F}_p[x]/\langle \varphi_i \rangle$; since $\varphi_i$ is irreducible, the right-hand side is a field, making $\mathfrak{p}_i$ maximal. The isomorphism passes through $\mathbb{Z}[\alpha]/\mathfrak{p}_i \cap \mathbb{Z}[\alpha]$ via the bridge identification $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \cong \mathcal{O}_L/p\mathcal{O}_L$, which is where the coprimality hypothesis $p \nmid |\mathcal{O}_L/\mathbb{Z}[\alpha]|$ is consumed — it makes multiplication by $p$ on the index-quotient invertible. For the product formula, we first show the containment $\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \langle p \rangle$ by expanding products of generators, then upgrade to equality by comparing norms: both sides have norm $p^n$, using [Element Norm Equals Ideal Norm](/theorems/1596) for $N(\langle p \rangle) = p^n$ and counting residue degrees for the left-hand side. Well-definedness of $\mathfrak{p}_i$ in the lift $\tilde{\varphi}_i$ follows because any two lifts differ by an element of $p\mathbb{Z}[x]$, so $\tilde{\varphi}_i(\alpha) - \tilde{\varphi}_i'(\alpha) \in p\mathbb{Z}[\alpha] \subseteq \langle p \rangle \subseteq \mathfrak{p}_i$. [/proofplan] [step:Establish the bridge $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \cong \mathcal{O}_L/p\mathcal{O}_L$ using the coprimality hypothesis] We prove: [claim:The natural map $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \to \mathcal{O}_L/p\mathcal{O}_L$ is a ring isomorphism] Let $\pi: \mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \to \mathcal{O}_L/p\mathcal{O}_L$ be the ring homomorphism induced by the inclusion $\mathbb{Z}[\alpha] \hookrightarrow \mathcal{O}_L$ (well-defined since $p\mathbb{Z}[\alpha] \subseteq p\mathcal{O}_L$). Under the hypothesis $p \nmid |\mathcal{O}_L/\mathbb{Z}[\alpha]|$, $\pi$ is an isomorphism. [/claim] [proof] Let $Q := \mathcal{O}_L/\mathbb{Z}[\alpha]$, a finite abelian group of order coprime to $p$ by hypothesis. Consider the multiplication-by-$p$ map \begin{align*} \mu_p: Q &\to Q \\ x + \mathbb{Z}[\alpha] &\mapsto px + \mathbb{Z}[\alpha]. \end{align*} **$\mu_p$ is injective.** Suppose $\mu_p(x + \mathbb{Z}[\alpha]) = 0$, i.e., $px \in \mathbb{Z}[\alpha]$. Viewing $x + \mathbb{Z}[\alpha] \in Q$, we have $p \cdot (x + \mathbb{Z}[\alpha]) = 0$ in $Q$, so the order of $x + \mathbb{Z}[\alpha]$ divides $p$. Combined with the order dividing $|Q|$ (Lagrange's theorem), the order divides $\gcd(p, |Q|) = 1$ (since $p \nmid |Q|$). Hence $x + \mathbb{Z}[\alpha] = 0$, i.e., $x \in \mathbb{Z}[\alpha]$. **$\mu_p$ is surjective.** A map of finite sets is surjective iff it is injective, so $\mu_p$ is surjective. **Injectivity of $\mu_p$ gives $\mathbb{Z}[\alpha] \cap p\mathcal{O}_L = p\mathbb{Z}[\alpha]$.** The inclusion $p\mathbb{Z}[\alpha] \subseteq \mathbb{Z}[\alpha] \cap p\mathcal{O}_L$ is immediate. For the reverse, let $y \in \mathbb{Z}[\alpha] \cap p\mathcal{O}_L$, so $y = pz$ for some $z \in \mathcal{O}_L$. In $Q$, $\mu_p(z + \mathbb{Z}[\alpha]) = y + \mathbb{Z}[\alpha] = 0$ (since $y \in \mathbb{Z}[\alpha]$). By injectivity, $z \in \mathbb{Z}[\alpha]$, so $y = pz \in p\mathbb{Z}[\alpha]$. **Surjectivity of $\mu_p$ gives $\mathbb{Z}[\alpha] + p\mathcal{O}_L = \mathcal{O}_L$.** Let $\beta \in \mathcal{O}_L$ and consider $\beta + \mathbb{Z}[\alpha] \in Q$. By surjectivity of $\mu_p$, there exists $z + \mathbb{Z}[\alpha] \in Q$ with $\mu_p(z + \mathbb{Z}[\alpha]) = \beta + \mathbb{Z}[\alpha]$, i.e., $pz \equiv \beta \pmod{\mathbb{Z}[\alpha]}$. So $\beta - pz \in \mathbb{Z}[\alpha]$, i.e., $\beta = (\beta - pz) + pz \in \mathbb{Z}[\alpha] + p\mathcal{O}_L$. **Isomorphism of quotients.** The natural homomorphism $\pi$ factors through \begin{align*} \mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \xrightarrow{\pi} \mathcal{O}_L/p\mathcal{O}_L. \end{align*} - **Injectivity of $\pi$.** If $y \in \mathbb{Z}[\alpha]$ maps to $0$ in $\mathcal{O}_L/p\mathcal{O}_L$, then $y \in p\mathcal{O}_L$, so $y \in \mathbb{Z}[\alpha] \cap p\mathcal{O}_L = p\mathbb{Z}[\alpha]$ by the identity above. Hence $y = 0$ in $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]$. - **Surjectivity of $\pi$.** Let $\beta + p\mathcal{O}_L \in \mathcal{O}_L/p\mathcal{O}_L$. Write $\beta = y + pz$ with $y \in \mathbb{Z}[\alpha]$ and $z \in \mathcal{O}_L$ (from the decomposition above). Then $\beta \equiv y \pmod{p\mathcal{O}_L}$, and $y + p\mathbb{Z}[\alpha] \in \mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]$ maps to $y + p\mathcal{O}_L = \beta + p\mathcal{O}_L$. Therefore $\pi$ is a ring isomorphism. [/proof] [guided] **The role of the hypothesis $p \nmid |\mathcal{O}_L/\mathbb{Z}[\alpha]|$.** The quotient $Q := \mathcal{O}_L/\mathbb{Z}[\alpha]$ is a finite abelian group (finite by [Nonzero Ideals Have Bounded Quotients](/theorems/1583) and the fact that $\mathbb{Z}[\alpha]$ has finite index in $\mathcal{O}_L$ — both are free $\mathbb{Z}$-modules of rank $n$). Its order is what the hypothesis is about. The multiplication-by-$p$ map $\mu_p: Q \to Q$ is well-defined ($Q$ is a $\mathbb{Z}$-module). When is it invertible? A group homomorphism $\mu_p: Q \to Q$ on a finite abelian group is a bijection iff its kernel is trivial, iff no element of $Q$ has order dividing $p$, iff $\gcd(p, |Q|) = 1$. This last condition is exactly $p \nmid |Q|$, the hypothesis. **The two consequences of invertibility.** 1. **Injectivity of $\mu_p$ translates to**: "$pz \in \mathbb{Z}[\alpha]$ forces $z \in \mathbb{Z}[\alpha]$." Equivalently, \begin{align*} \mathbb{Z}[\alpha] \cap p\mathcal{O}_L = p\mathbb{Z}[\alpha]. \end{align*} Why this identity matters: we want to compare the two reductions $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]$ and $\mathcal{O}_L/p\mathcal{O}_L$, and the left-to-right map fails to be injective exactly when some element of $\mathbb{Z}[\alpha]$ becomes divisible by $p$ in $\mathcal{O}_L$ without already being divisible by $p$ in $\mathbb{Z}[\alpha]$. The identity rules this out. 2. **Surjectivity of $\mu_p$ translates to**: every class in $Q$ is divisible by $p$, so \begin{align*} \mathbb{Z}[\alpha] + p\mathcal{O}_L = \mathcal{O}_L. \end{align*} Equivalently, every $\beta \in \mathcal{O}_L$ differs from some $y \in \mathbb{Z}[\alpha]$ by an element of $p\mathcal{O}_L$. This is surjectivity of the map $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \to \mathcal{O}_L/p\mathcal{O}_L$. Together these say: the natural reduction map is both injective and surjective, hence an isomorphism. **The snake-lemma perspective.** The short exact sequence of $\mathbb{Z}$-modules \begin{align*} 0 \to \mathbb{Z}[\alpha] \to \mathcal{O}_L \to Q \to 0 \end{align*} tensored (derived-functor style) with the map "multiplication by $p$" produces a six-term exact sequence; the hypothesis $p \nmid |Q|$ makes multiplication by $p$ on $Q$ an isomorphism, and the snake lemma then identifies the relevant kernels and cokernels to yield the isomorphism $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \cong \mathcal{O}_L/p\mathcal{O}_L$. **Counterexample when the hypothesis fails.** If $p \mid |Q|$, $\mathbb{Z}[\alpha]$ is not $p$-maximal, and the criterion fails. For instance, in $L = \mathbb{Q}(\sqrt{5})$ with $\alpha = \sqrt{5}$, the ring $\mathbb{Z}[\sqrt{5}] \subsetneq \mathcal{O}_L = \mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ has index $2$. Applying Dedekind's criterion to $g(x) = x^2 - 5$ with $p = 2$ would give $\bar{g}(x) = x^2 - 1 = (x-1)(x+1)$, suggesting $2$ splits. But in fact $2$ is inert in $\mathbb{Q}(\sqrt{5})$ (since $5 \equiv 5 \pmod 8$, see [Behaviour of 2 in Quadratic Fields](/theorems/1599)). The hypothesis is violated, and the criterion gives the wrong answer. [/guided] [/step] [step:Identify $\mathcal{O}_L/\mathfrak{p}_i$ with $\mathbb{F}_p[x]/\langle \varphi_i \rangle$] Using the claim of Step 1, we now compute $\mathcal{O}_L/\mathfrak{p}_i$ explicitly. Recall $\mathfrak{p}_i = \langle p, \tilde{\varphi}_i(\alpha) \rangle \unlhd \mathcal{O}_L$. Consider the composition of natural reductions \begin{align*} \mathbb{Z}[x] \twoheadrightarrow \mathbb{Z}[x]/\langle g(x) \rangle \xrightarrow{\sim} \mathbb{Z}[\alpha] \twoheadrightarrow \mathcal{O}_L, \end{align*} where the middle isomorphism sends $x \mapsto \alpha$ (valid since $g$ is the minimal polynomial of $\alpha$). Then: \begin{align*} \frac{\mathcal{O}_L}{\mathfrak{p}_i} = \frac{\mathcal{O}_L}{\langle p, \tilde{\varphi}_i(\alpha) \rangle} \cong \frac{\mathcal{O}_L / p\mathcal{O}_L}{\langle \tilde{\varphi}_i(\alpha) \rangle \pmod p} \end{align*} (where the last quotient is taken modulo the image of $\tilde{\varphi}_i(\alpha)$). Applying the Step 1 claim $\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha] \cong \mathcal{O}_L/p\mathcal{O}_L$, \begin{align*} \frac{\mathcal{O}_L}{\mathfrak{p}_i} \cong \frac{\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]}{\langle \tilde{\varphi}_i(\alpha) \rangle \pmod p} \cong \frac{\mathbb{Z}[\alpha]}{\langle p, \tilde{\varphi}_i(\alpha) \rangle}. \end{align*} Using $\mathbb{Z}[\alpha] \cong \mathbb{Z}[x]/\langle g(x) \rangle$, \begin{align*} \frac{\mathbb{Z}[\alpha]}{\langle p, \tilde{\varphi}_i(\alpha) \rangle} \cong \frac{\mathbb{Z}[x]}{\langle p, \tilde{\varphi}_i(x), g(x) \rangle} \cong \frac{\mathbb{F}_p[x]}{\langle \varphi_i(x), \bar{g}(x) \rangle}. \end{align*} Now $\bar{g}(x) = \prod_j \varphi_j(x)^{e_j}$ in $\mathbb{F}_p[x]$ by hypothesis. Since the $\varphi_j$ are pairwise coprime distinct monic irreducibles, $\langle \varphi_i \rangle + \langle \varphi_j^{e_j} \rangle = \mathbb{F}_p[x]$ for $j \neq i$ (because $\gcd(\varphi_i, \varphi_j^{e_j}) = 1$). Hence in $\mathbb{F}_p[x]/\langle \varphi_i \rangle$, each factor $\varphi_j^{e_j}$ with $j \neq i$ is a unit. Therefore \begin{align*} \langle \varphi_i, \bar{g} \rangle = \langle \varphi_i, \varphi_i^{e_i} \prod_{j \neq i} \varphi_j^{e_j} \rangle = \langle \varphi_i \rangle \end{align*} in $\mathbb{F}_p[x]$ (since $\varphi_i^{e_i}$ is already a multiple of $\varphi_i$). Substituting, \begin{align*} \frac{\mathcal{O}_L}{\mathfrak{p}_i} \cong \frac{\mathbb{F}_p[x]}{\langle \varphi_i \rangle}. \end{align*} [guided] **The goal.** We want to identify $\mathcal{O}_L/\mathfrak{p}_i$ with the finite field $\mathbb{F}_p[x]/\langle \varphi_i \rangle$. Then primality of $\mathfrak{p}_i$ is immediate. **The chain of reductions.** Starting from $\mathcal{O}_L$, we quotient in stages: first by $p\mathcal{O}_L$ to reduce coefficients mod $p$, then by the residual class of $\tilde{\varphi}_i(\alpha)$. The first quotient is where we apply the Step 1 bridge. **Working through the chain.** Step A. The definition: $\mathfrak{p}_i = \langle p, \tilde{\varphi}_i(\alpha) \rangle$, so \begin{align*} \frac{\mathcal{O}_L}{\mathfrak{p}_i} = \frac{\mathcal{O}_L}{\langle p, \tilde{\varphi}_i(\alpha) \rangle}. \end{align*} Step B. We can compute the quotient by $\langle p, \tilde{\varphi}_i(\alpha) \rangle$ by first quotienting by $\langle p \rangle = p\mathcal{O}_L$ and then by the image of $\tilde{\varphi}_i(\alpha)$ (the "second isomorphism theorem" for ideals): \begin{align*} \frac{\mathcal{O}_L}{\langle p, \tilde{\varphi}_i(\alpha) \rangle} \cong \frac{\mathcal{O}_L/p\mathcal{O}_L}{\overline{\langle \tilde{\varphi}_i(\alpha) \rangle}}. \end{align*} Step C. By the claim of Step 1, $\mathcal{O}_L/p\mathcal{O}_L \cong \mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]$ as rings. The isomorphism sends $\tilde{\varphi}_i(\alpha) + p\mathcal{O}_L \mapsto \tilde{\varphi}_i(\alpha) + p\mathbb{Z}[\alpha]$ (preimages match since $\tilde{\varphi}_i(\alpha) \in \mathbb{Z}[\alpha]$). So \begin{align*} \frac{\mathcal{O}_L/p\mathcal{O}_L}{\overline{\langle \tilde{\varphi}_i(\alpha) \rangle}} \cong \frac{\mathbb{Z}[\alpha]/p\mathbb{Z}[\alpha]}{\overline{\langle \tilde{\varphi}_i(\alpha) \rangle}} \cong \frac{\mathbb{Z}[\alpha]}{\langle p, \tilde{\varphi}_i(\alpha) \rangle}. \end{align*} Step D. The first isomorphism theorem gives $\mathbb{Z}[\alpha] \cong \mathbb{Z}[x]/\langle g \rangle$ via $x \mapsto \alpha$ (valid since $g$ is the minimal polynomial of $\alpha$ and so generates the kernel of $\mathbb{Z}[x] \to \mathbb{Z}[\alpha]$, $x \mapsto \alpha$). So \begin{align*} \frac{\mathbb{Z}[\alpha]}{\langle p, \tilde{\varphi}_i(\alpha) \rangle} \cong \frac{\mathbb{Z}[x]/\langle g(x) \rangle}{\langle p, \tilde{\varphi}_i(x) \rangle \pmod{g(x)}} \cong \frac{\mathbb{Z}[x]}{\langle p, \tilde{\varphi}_i(x), g(x) \rangle}. \end{align*} Step E. Quotienting $\mathbb{Z}[x]$ by $p$ replaces $\mathbb{Z}[x]$ with $\mathbb{F}_p[x]$: \begin{align*} \frac{\mathbb{Z}[x]}{\langle p, \tilde{\varphi}_i(x), g(x) \rangle} \cong \frac{\mathbb{F}_p[x]}{\langle \varphi_i(x), \bar{g}(x) \rangle}, \end{align*} where $\varphi_i(x) = \tilde{\varphi}_i(x) \bmod p$ and $\bar{g}(x) = g(x) \bmod p$. Step F. Simplifying $\langle \varphi_i, \bar{g} \rangle$ in $\mathbb{F}_p[x]$. We have $\bar{g} = \varphi_i^{e_i} \prod_{j \neq i} \varphi_j^{e_j}$. The ideal $\langle \varphi_i, \bar{g} \rangle$ is generated by $\varphi_i$ and this product. In $\mathbb{F}_p[x]/\langle \varphi_i \rangle$, the image of $\varphi_i^{e_i}$ is zero (since $\varphi_i \equiv 0$). So $\bar{g} \equiv 0 \cdot \prod_{j \neq i} \varphi_j^{e_j} = 0 \pmod{\varphi_i}$, i.e., $\bar{g} \in \langle \varphi_i \rangle$. This gives $\langle \varphi_i, \bar{g} \rangle \subseteq \langle \varphi_i \rangle$. The reverse inclusion $\langle \varphi_i \rangle \subseteq \langle \varphi_i, \bar{g} \rangle$ holds since $\varphi_i$ is already one of the listed generators. So \begin{align*} \langle \varphi_i, \bar{g} \rangle = \langle \varphi_i \rangle. \end{align*} **Final identification.** Stringing all the isomorphisms together: \begin{align*} \frac{\mathcal{O}_L}{\mathfrak{p}_i} \cong \frac{\mathbb{F}_p[x]}{\langle \varphi_i \rangle}. \end{align*} This is the key computational identity. [/guided] [/step] [step:Conclude that $\mathfrak{p}_i$ is prime and compute $N(\mathfrak{p}_i)$] Since $\varphi_i \in \mathbb{F}_p[x]$ is irreducible and $\mathbb{F}_p[x]$ is a PID, the ideal $\langle \varphi_i \rangle$ is maximal. Therefore $\mathbb{F}_p[x]/\langle \varphi_i \rangle$ is a field. By the isomorphism of Step 2, $\mathcal{O}_L/\mathfrak{p}_i$ is a field, so $\mathfrak{p}_i$ is a maximal ideal of $\mathcal{O}_L$, hence prime. For the norm, $\mathbb{F}_p[x]/\langle \varphi_i \rangle$ is a finite field of cardinality $p^{\deg \varphi_i}$ (since it is an $\mathbb{F}_p$-vector space with basis $\{1, x, \ldots, x^{\deg \varphi_i - 1}\}$). Hence \begin{align*} N(\mathfrak{p}_i) = |\mathcal{O}_L/\mathfrak{p}_i| = |\mathbb{F}_p[x]/\langle \varphi_i \rangle| = p^{\deg \varphi_i}, \end{align*} establishing $f_i = \deg \varphi_i$ and $N(\mathfrak{p}_i) = p^{\deg \varphi_i}$. [guided] **Primality from maximality.** An ideal $\mathfrak{p} \unlhd R$ is **maximal** if $R/\mathfrak{p}$ is a field, **prime** if $R/\mathfrak{p}$ is an integral domain. Since fields are integral domains, maximal $\implies$ prime. Step 2 gives $\mathcal{O}_L/\mathfrak{p}_i \cong \mathbb{F}_p[x]/\langle \varphi_i \rangle$. Now $\mathbb{F}_p[x]$ is a principal ideal domain, and in a PID an ideal $\langle f \rangle$ is maximal iff $f$ is irreducible (or equivalently, $f$ is prime). Since $\varphi_i$ is irreducible by hypothesis, $\langle \varphi_i \rangle$ is maximal in $\mathbb{F}_p[x]$, so $\mathbb{F}_p[x]/\langle \varphi_i \rangle$ is a field. Therefore $\mathcal{O}_L/\mathfrak{p}_i$ is a field, and $\mathfrak{p}_i$ is maximal — hence prime — in $\mathcal{O}_L$. **Counting the cardinality.** $\mathbb{F}_p[x]/\langle \varphi_i \rangle$ is an $\mathbb{F}_p$-vector space. A basis is given by the classes $\{1, x, x^2, \ldots, x^{\deg \varphi_i - 1}\}$ (by polynomial division against the monic $\varphi_i$: every polynomial class has a unique representative of degree $< \deg \varphi_i$). So \begin{align*} \dim_{\mathbb{F}_p} \left( \mathbb{F}_p[x]/\langle \varphi_i \rangle \right) = \deg \varphi_i, \end{align*} and the cardinality is $|\mathbb{F}_p|^{\deg \varphi_i} = p^{\deg \varphi_i}$. **Ideal norm.** By definition, $N(\mathfrak{p}_i) = |\mathcal{O}_L/\mathfrak{p}_i|$. Combining with the isomorphism, \begin{align*} N(\mathfrak{p}_i) = p^{\deg \varphi_i}, \end{align*} so the residue degree $f_i = \deg \varphi_i$. [/guided] [/step] [step:Well-definedness of $\mathfrak{p}_i$ under change of lift] Let $\tilde{\varphi}_i, \tilde{\varphi}_i' \in \mathbb{Z}[x]$ be two lifts of $\varphi_i$, i.e., $\tilde{\varphi}_i \equiv \tilde{\varphi}_i' \equiv \varphi_i \pmod p$. Then $\tilde{\varphi}_i - \tilde{\varphi}_i' \in p\mathbb{Z}[x]$, so \begin{align*} \tilde{\varphi}_i(\alpha) - \tilde{\varphi}_i'(\alpha) \in p\mathbb{Z}[\alpha] \subseteq p\mathcal{O}_L = \langle p \rangle. \end{align*} Therefore $\tilde{\varphi}_i(\alpha) \equiv \tilde{\varphi}_i'(\alpha) \pmod{\langle p \rangle}$, so \begin{align*} \langle p, \tilde{\varphi}_i(\alpha) \rangle = \langle p, \tilde{\varphi}_i'(\alpha) \rangle, \end{align*} i.e., $\mathfrak{p}_i$ does not depend on the choice of lift. [guided] **The concern.** The ideal $\mathfrak{p}_i = \langle p, \tilde{\varphi}_i(\alpha) \rangle$ apparently depends on the choice of lift $\tilde{\varphi}_i \in \mathbb{Z}[x]$ of $\varphi_i \in \mathbb{F}_p[x]$ — there are many lifts, differing by elements of $p\mathbb{Z}[x]$. **The resolution.** Two lifts $\tilde{\varphi}_i, \tilde{\varphi}_i'$ have a difference in $p\mathbb{Z}[x]$: \begin{align*} \tilde{\varphi}_i - \tilde{\varphi}_i' = p \cdot h(x) \quad \text{for some } h \in \mathbb{Z}[x]. \end{align*} Evaluating at $\alpha$, \begin{align*} \tilde{\varphi}_i(\alpha) - \tilde{\varphi}_i'(\alpha) = p \cdot h(\alpha) \in p\mathcal{O}_L = \langle p \rangle. \end{align*} **Ideals generated by two elements modulo $\langle p \rangle$.** Two elements $\beta, \beta' \in \mathcal{O}_L$ with $\beta - \beta' \in \langle p \rangle$ generate the same ideal together with $p$: \begin{align*} \langle p, \beta \rangle = \langle p, \beta + p h(\alpha) \rangle = \langle p, \beta' \rangle, \end{align*} since adding a multiple of $p$ (which is already a generator) does not change the ideal. Therefore $\mathfrak{p}_i$ is independent of the choice of lift. [/guided] [/step] [step:Show the containment $\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \langle p \rangle$] We claim [claim:Each $\mathfrak{p}_i^{e_i}$ is contained in $\langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$] $\mathfrak{p}_i^{e_i} \subseteq \langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$. [/claim] [proof] $\mathfrak{p}_i = \langle p, \tilde{\varphi}_i(\alpha) \rangle$. An element of $\mathfrak{p}_i^{e_i}$ is a $\mathbb{Z}$-linear combination of products of the form $u_1 u_2 \cdots u_{e_i}$ with each $u_k \in \{p, \tilde{\varphi}_i(\alpha)\} \cdot \mathcal{O}_L$. Expanding, \begin{align*} u_1 u_2 \cdots u_{e_i} = \prod_k (p \cdot r_k + \tilde{\varphi}_i(\alpha) \cdot s_k) \quad \text{with } r_k, s_k \in \mathcal{O}_L. \end{align*} This is a product of $e_i$ factors, each a combination of $p$ and $\tilde{\varphi}_i(\alpha)$. Expand fully by distributivity: each term in the expansion is a product of chosen summands — either $p \cdot r_k$ or $\tilde{\varphi}_i(\alpha) \cdot s_k$ — from each factor. Any term in which at least one summand is $p \cdot r_k$ is a multiple of $p$, hence lies in $\langle p \rangle \subseteq \langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$. The remaining term, in which every summand is $\tilde{\varphi}_i(\alpha) \cdot s_k$, equals \begin{align*} \tilde{\varphi}_i(\alpha)^{e_i} \cdot \prod_k s_k \in \langle \tilde{\varphi}_i(\alpha)^{e_i} \rangle \subseteq \langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle. \end{align*} Hence each product $u_1 \cdots u_{e_i} \in \langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$. $\mathbb{Z}$-linear combinations of such products remain in this ideal. [/proof] Now we compute the containment for the full product: \begin{align*} \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \prod_{i=1}^m \langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle \subseteq \left\langle p, \prod_{i=1}^m \tilde{\varphi}_i(\alpha)^{e_i} \right\rangle. \end{align*} The second inclusion holds because: a product $\mathfrak{b}_1 \cdot \mathfrak{b}_2$ of ideals with $\mathfrak{b}_i = \langle p, x_i \rangle$ is generated by $\{p^2, p x_1, p x_2, x_1 x_2\}$, which in turn is contained in $\langle p, x_1 x_2 \rangle$ (each generator is either a multiple of $p$ or equals $x_1 x_2$). By induction on $m$, this extends to arbitrary products. Finally, \begin{align*} \prod_{i=1}^m \tilde{\varphi}_i(x)^{e_i} \equiv \bar{g}(x) \pmod p, \end{align*} so $\prod_{i=1}^m \tilde{\varphi}_i(\alpha)^{e_i} \equiv g(\alpha) = 0 \pmod{p}$, meaning $\prod_{i=1}^m \tilde{\varphi}_i(\alpha)^{e_i} \in p\mathcal{O}_L + 0 = p\mathcal{O}_L = \langle p \rangle$. Hence \begin{align*} \left\langle p, \prod_{i=1}^m \tilde{\varphi}_i(\alpha)^{e_i} \right\rangle = \langle p \rangle. \end{align*} Combining, \begin{align*} \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \langle p \rangle. \end{align*} [guided] **Structure of the argument.** We prove containment in two stages: first absorb the powers into the ideals $\langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$ (the claim above), then aggregate across all $i$ and use the fact that the product $\prod_i \tilde{\varphi}_i(\alpha)^{e_i}$ congruence-equals $g(\alpha) = 0$ modulo $p$. **Why the first claim holds.** An ideal $\mathfrak{a} = \langle x_1, \ldots, x_k \rangle$ has the property that $\mathfrak{a}^e$ is generated by products of $e$ factors drawn from the generators (times ring elements). Expanding: \begin{align*} (\langle p, \tilde{\varphi}_i(\alpha) \rangle)^{e_i} \subseteq \langle p^{e_i}, p^{e_i - 1}\tilde{\varphi}_i(\alpha), \ldots, p\, \tilde{\varphi}_i(\alpha)^{e_i - 1}, \tilde{\varphi}_i(\alpha)^{e_i} \rangle. \end{align*} Every generator on the right is divisible by $p$ — except the last, which is $\tilde{\varphi}_i(\alpha)^{e_i}$. So all generators lie in $\langle p, \tilde{\varphi}_i(\alpha)^{e_i} \rangle$. **Aggregating products.** For ideals $\mathfrak{b}_i = \langle p, y_i \rangle$ (where $y_i := \tilde{\varphi}_i(\alpha)^{e_i}$), the product $\mathfrak{b}_1 \cdots \mathfrak{b}_m$ is generated by products $\prod_i z_i$ with $z_i \in \{p, y_i\}$. Any such product with at least one $z_i = p$ is a multiple of $p$. The single product with all $z_i = y_i$ is $y_1 \cdots y_m = \prod_i \tilde{\varphi}_i(\alpha)^{e_i}$. So \begin{align*} \mathfrak{b}_1 \cdots \mathfrak{b}_m \subseteq \left\langle p, \prod_i \tilde{\varphi}_i(\alpha)^{e_i} \right\rangle. \end{align*} **Using $g(\alpha) = 0$.** The key polynomial identity is \begin{align*} \prod_i \tilde{\varphi}_i(x)^{e_i} \equiv \bar{g}(x) \pmod p \quad \text{in } \mathbb{Z}[x], \end{align*} because the product of lifts reduces mod $p$ to the product of the $\varphi_i^{e_i}$, which is $\bar{g}$ by the factorisation hypothesis. Writing this as $\prod_i \tilde{\varphi}_i(x)^{e_i} = g(x) + p \cdot r(x)$ for some $r \in \mathbb{Z}[x]$ (since lifts of $g$ and of the product $\prod \varphi_i^{e_i}$ are both congruent to $\bar{g}$ mod $p$, their difference is in $p\mathbb{Z}[x]$). Evaluating at $\alpha$: \begin{align*} \prod_i \tilde{\varphi}_i(\alpha)^{e_i} = g(\alpha) + p\, r(\alpha) = 0 + p\, r(\alpha) = p\, r(\alpha) \in \langle p \rangle, \end{align*} using $g(\alpha) = 0$ (since $g$ is the minimal polynomial of $\alpha$). **Assembling the containment.** \begin{align*} \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \left\langle p, \prod_i \tilde{\varphi}_i(\alpha)^{e_i} \right\rangle = \left\langle p, p\, r(\alpha) \right\rangle = \langle p \rangle. \end{align*} [/guided] [/step] [step:Upgrade containment to equality by comparing norms] By the containment of Step 5 and [Divisibility Equals Containment](/theorems/1587), $\langle p \rangle \mid \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$... wait, the direction: containment $\mathfrak{a} \subseteq \mathfrak{b}$ means $\mathfrak{b} \mid \mathfrak{a}$. So from $\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \langle p \rangle$, we get $\langle p \rangle \mid \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$. By the [Multiplicativity of the Ideal Norm](/theorems/1593), \begin{align*} N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = \prod_{i=1}^m N(\mathfrak{p}_i)^{e_i} = \prod_{i=1}^m p^{e_i \deg \varphi_i} = p^{\sum_i e_i \deg \varphi_i}. \end{align*} By the factorisation $\bar{g}(x) = \prod_i \varphi_i(x)^{e_i}$, \begin{align*} \deg \bar{g} = \sum_{i=1}^m e_i \deg \varphi_i. \end{align*} The polynomial $\bar{g}$ is obtained from $g$ by reducing coefficients mod $p$, which preserves degree when the leading coefficient is not divisible by $p$. Since $g$ is monic (being the minimal polynomial of the algebraic integer $\alpha$), its leading coefficient is $1$, not divisible by $p$. So $\deg \bar{g} = \deg g$. Also, $\deg g = n = [L:\mathbb{Q}]$. Indeed, $g$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$; the hypothesis $p \nmid |\mathcal{O}_L/\mathbb{Z}[\alpha]|$ in particular implies $\mathcal{O}_L/\mathbb{Z}[\alpha]$ is finite, so $\mathbb{Z}[\alpha]$ is a free $\mathbb{Z}$-module of rank $n$, meaning $[L:\mathbb{Q}(\alpha)] = 1$ and so $[\mathbb{Q}(\alpha):\mathbb{Q}] = \deg g = n$. Combining, \begin{align*} N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = p^n. \end{align*} On the other hand, by [Element Norm Equals Ideal Norm](/theorems/1596), \begin{align*} N(\langle p \rangle) = |N_{L/\mathbb{Q}}(p)| = p^n, \end{align*} where we used $N_{L/\mathbb{Q}}(p) = p^n$ (multiplication by $p$ on $L$ has determinant $p^n$, as computed in Step 5 of proof 1597). Thus $\langle p \rangle \mid \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$ with $N(\langle p \rangle) = N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = p^n$. Now, if $\mathfrak{a} \mid \mathfrak{b}$ (both nonzero) with $N(\mathfrak{a}) = N(\mathfrak{b})$, then $\mathfrak{a} = \mathfrak{b}$: write $\mathfrak{b} = \mathfrak{a} \mathfrak{c}$; taking norms gives $N(\mathfrak{b}) = N(\mathfrak{a}) N(\mathfrak{c})$, so $N(\mathfrak{c}) = 1$, hence $\mathfrak{c} = \mathcal{O}_L$, and $\mathfrak{b} = \mathfrak{a}$. Applying this, \begin{align*} \langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}. \end{align*} [guided] **Strategy.** We have containment $\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m} \subseteq \langle p \rangle$ from Step 5. Equivalently, $\langle p \rangle$ divides $\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$ (by [Divisibility Equals Containment](/theorems/1587), the smaller ideal is the product/multiple). Since divisibility plus equal norms implies equality, we compute both norms and show they agree. **Computing $N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m})$.** By [Multiplicativity of the Ideal Norm](/theorems/1593), \begin{align*} N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = \prod_i N(\mathfrak{p}_i)^{e_i}. \end{align*} Step 3 gives $N(\mathfrak{p}_i) = p^{\deg \varphi_i}$. So \begin{align*} N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = \prod_i p^{e_i \deg \varphi_i} = p^{\sum_i e_i \deg \varphi_i}. \end{align*} **Identifying the exponent.** The factorisation $\bar{g} = \prod_i \varphi_i^{e_i}$ in $\mathbb{F}_p[x]$ gives \begin{align*} \deg \bar{g} = \sum_i e_i \deg \varphi_i. \end{align*} **$\deg \bar{g} = \deg g$.** When reducing a polynomial mod $p$, the degree drops only if the leading coefficient is divisible by $p$. Since $g$ is monic (leading coefficient $1$), reduction mod $p$ preserves degree: $\deg \bar{g} = \deg g$. **$\deg g = n$.** The minimal polynomial $g$ of $\alpha$ has degree $[\mathbb{Q}(\alpha):\mathbb{Q}]$. We claim $\mathbb{Q}(\alpha) = L$, i.e., $[\mathbb{Q}(\alpha):\mathbb{Q}] = n$. The hypothesis $p \nmid |\mathcal{O}_L/\mathbb{Z}[\alpha]|$ implies the index is **finite**, so $\mathbb{Z}[\alpha]$ has rank $n$ as a $\mathbb{Z}$-module (same rank as $\mathcal{O}_L$). This rank equals $[\mathbb{Q}(\alpha):\mathbb{Q}]$ (a generating set of $\mathbb{Z}[\alpha]$ as a free $\mathbb{Z}$-module tensored with $\mathbb{Q}$ gives a $\mathbb{Q}$-basis of $\mathbb{Q}(\alpha)$). Hence $[\mathbb{Q}(\alpha):\mathbb{Q}] = n$, so $\mathbb{Q}(\alpha) = L$, and $\deg g = n$. Conclusion: \begin{align*} N(\mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}) = p^n. \end{align*} **Computing $N(\langle p \rangle)$.** By [Element Norm Equals Ideal Norm](/theorems/1596), $N(\langle p \rangle) = |N_{L/\mathbb{Q}}(p)|$. Since $p \in \mathbb{Q}$, the multiplication-by-$p$ map on $L$ is scalar: $m_p = p \cdot \operatorname{id}_L$. Its determinant is $p^{\dim_\mathbb{Q} L} = p^n$ (positive). Hence $|N_{L/\mathbb{Q}}(p)| = p^n$. **Divisibility + equal norms = equality.** Let $\mathfrak{a} \mid \mathfrak{b}$ (both nonzero) mean $\mathfrak{b} = \mathfrak{a}\mathfrak{c}$ for some ideal $\mathfrak{c}$. Taking norms: $N(\mathfrak{b}) = N(\mathfrak{a})N(\mathfrak{c})$. If also $N(\mathfrak{a}) = N(\mathfrak{b})$, then $N(\mathfrak{c}) = 1$, i.e., $|\mathcal{O}_L/\mathfrak{c}| = 1$, so $\mathfrak{c} = \mathcal{O}_L$, and $\mathfrak{b} = \mathfrak{a}$. Applying to $\mathfrak{a} = \langle p \rangle$ and $\mathfrak{b} = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}$: \begin{align*} \langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}. \end{align*} **Summary of what each piece contributed.** - The containment $\mathfrak{p}_1^{e_1} \cdots \subseteq \langle p \rangle$ (Step 5) came from expanding ideals and using $g(\alpha) = 0$. - The norm of the product (this step) came from [Multiplicativity of the Ideal Norm](/theorems/1593) plus Step 3's $N(\mathfrak{p}_i) = p^{\deg \varphi_i}$. - The norm of $\langle p \rangle$ came from [Element Norm Equals Ideal Norm](/theorems/1596). - The bridge between divisibility and equality came from multiplicativity of the norm plus the basic fact $|\mathcal{O}_L/\mathcal{O}_L| = 1$. [/guided] [/step]