[proofplan] The obstruction is a pigeonhole count. By [Dedekind's Criterion](/theorems/1598), $p$ splits completely in $\mathcal{O}_L$ if and only if the reduction $\bar{g}(x) = g(x) \bmod p$ factors as a product of $n$ **distinct** monic linear polynomials in $\mathbb{F}_p[x]$, where $g$ is the minimal polynomial of $\alpha$ and $n = \deg g = [L:\mathbb{Q}]$. A monic linear polynomial in $\mathbb{F}_p[x]$ is determined by its root, so distinct monic linear polynomials correspond to distinct elements of $\mathbb{F}_p$. If $p < n$ then $\mathbb{F}_p$ has only $p < n$ elements — fewer than required — so no such factorization exists. [/proofplan] [step:Translate complete splitting into a factorization requirement via Dedekind's criterion] By hypothesis $p \nmid |\mathcal{O}_L / \mathbb{Z}[\alpha]|$, so [Dedekind's Criterion](/theorems/1598) applies to the prime $p$ and the generator $\alpha$. Let $g \in \mathbb{Z}[x]$ be the minimal polynomial of $\alpha$. Since $\alpha$ generates $L$ over $\mathbb{Q}$ (integrality together with $[\mathbb{Z}[\alpha] : \mathbb{Z}] = [L : \mathbb{Q}] = n$, the latter because $|\mathcal{O}_L / \mathbb{Z}[\alpha]|$ is finite and $\mathcal{O}_L$ has rank $n$ by [Existence of an Integral Basis](/theorems/1585)), we have $\deg g = n$. Factor the reduction in $\mathbb{F}_p[x]$: \begin{align*} \bar{g}(x) = \varphi_1(x)^{e_1} \cdots \varphi_m(x)^{e_m}, \end{align*} where $\varphi_1, \ldots, \varphi_m$ are distinct monic irreducibles. Writing $f_i = \deg \varphi_i$, Dedekind's criterion states \begin{align*} \langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}, \qquad N(\mathfrak{p}_i) = p^{f_i}, \end{align*} with the $\mathfrak{p}_i$ distinct prime ideals of $\mathcal{O}_L$. By definition, $p$ **splits completely** in $\mathcal{O}_L$ when $\langle p \rangle$ factors as a product of $n$ distinct prime ideals each of norm $p$. Comparing with the above, this occurs precisely when \begin{align*} m = n, \qquad e_i = 1 \text{ for all } i, \qquad f_i = 1 \text{ for all } i. \end{align*} Since $f_i = \deg \varphi_i$ and $e_i$ is the multiplicity in the factorization of $\bar{g}$, the three conditions say: $\bar{g}$ factors as a product of $n$ **distinct** monic **linear** polynomials over $\mathbb{F}_p$. [guided] The goal is to obstruct complete splitting of $\langle p \rangle$. The cleanest obstruction will come from Dedekind's criterion, which translates the ideal-theoretic statement into a polynomial-factorization statement over the finite field $\mathbb{F}_p$. Polynomial factorization over finite fields is controlled by the size of the field — this is where the bound $p < n$ will bite. **Verifying Dedekind's hypotheses.** [Theorem 1598](/theorems/1598) requires $\alpha \in \mathcal{O}_L$ with $p \nmid [\mathcal{O}_L : \mathbb{Z}[\alpha]]$. Both are given as theorem hypotheses. **Degree of the minimal polynomial.** The minimal polynomial $g \in \mathbb{Z}[x]$ of $\alpha$ has degree $\deg g = [\mathbb{Q}(\alpha) : \mathbb{Q}]$. We claim $\deg g = n = [L : \mathbb{Q}]$, equivalently that $\alpha$ generates $L$ over $\mathbb{Q}$. Since $|\mathcal{O}_L / \mathbb{Z}[\alpha]|$ is finite (by the hypothesis $p \nmid |\mathcal{O}_L / \mathbb{Z}[\alpha]|$, which presupposes the quantity is a finite integer), the $\mathbb{Z}$-rank of $\mathbb{Z}[\alpha]$ equals the $\mathbb{Z}$-rank of $\mathcal{O}_L$, which is $n$ by [Existence of an Integral Basis](/theorems/1585). Hence $\mathbb{Z}[\alpha]$ has rank $n$ as an abelian group, so $[\mathbb{Q}(\alpha) : \mathbb{Q}] = n$ and $\deg g = n$. **Applying Dedekind.** Factor $\bar{g} = \varphi_1^{e_1} \cdots \varphi_m^{e_m}$ into distinct monic irreducibles in $\mathbb{F}_p[x]$. [Theorem 1598](/theorems/1598) gives \begin{align*} \langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}, \qquad N(\mathfrak{p}_i) = p^{\deg \varphi_i}, \end{align*} with the $\mathfrak{p}_i$ distinct primes. Writing $f_i = \deg \varphi_i$ for notational symmetry with standard splitting invariants, the $f_i$ are the residue degrees and the $e_i$ are the ramification indices. **What does "splits completely" mean?** By definition, a rational prime $p$ splits completely in $L$ if $\langle p \rangle = \mathfrak{P}_1 \cdots \mathfrak{P}_n$ with the $\mathfrak{P}_j$ **distinct** prime ideals each of norm $N(\mathfrak{P}_j) = p$. Equivalently (using the fundamental identity $\sum e_i f_i = n$): there are $n$ distinct primes above $p$, each with $e = f = 1$. **Translating the condition.** Complete splitting of $p$ $\iff$ in Dedekind's factorization we have $m = n$ primes, $e_i = 1$ for each $i$, and $f_i = 1$ for each $i$. Since $f_i = \deg \varphi_i$ and $e_i$ is the multiplicity, this reads: \begin{align*} \bar{g}(x) = \varphi_1(x) \cdots \varphi_n(x), \quad \varphi_i \text{ distinct monic linear polynomials in } \mathbb{F}_p[x]. \end{align*} This is the polynomial-factorization criterion for complete splitting. [/guided] [/step] [step:Count monic linear polynomials over $\mathbb{F}_p$ and apply pigeonhole] A monic linear polynomial in $\mathbb{F}_p[x]$ has the form $x - c$ for some $c \in \mathbb{F}_p$, and the correspondence $c \mapsto x - c$ is a bijection from $\mathbb{F}_p$ to the set of monic linear polynomials in $\mathbb{F}_p[x]$. In particular, there are exactly $|\mathbb{F}_p| = p$ distinct monic linear polynomials in $\mathbb{F}_p[x]$. Suppose for contradiction that $p$ splits completely. By Step 1, $\bar{g}$ is a product of $n$ distinct monic linear polynomials in $\mathbb{F}_p[x]$. Such a collection would consist of $n$ distinct elements of a set of size $p$, which requires $p \geq n$. But the hypothesis $p < n$ gives a strict inequality in the other direction. This contradiction shows $p$ does not split completely. [guided] We now use the pigeonhole argument that sits at the heart of the proof. **Counting monic linear polynomials.** A monic linear polynomial over $\mathbb{F}_p$ is of the form $x - c$, $c \in \mathbb{F}_p$. The map \begin{align*} \mathbb{F}_p &\to \{\text{monic linear polynomials in } \mathbb{F}_p[x]\} \\ c &\mapsto x - c \end{align*} is injective (two different $c$'s yield polynomials with different roots, hence different polynomials) and surjective (every monic linear polynomial has the form $x - c$ for $c$ its unique root in $\mathbb{F}_p$). Hence the number of distinct monic linear polynomials in $\mathbb{F}_p[x]$ equals $|\mathbb{F}_p| = p$. **Pigeonhole.** Suppose for contradiction that $p$ splits completely. Step 1 translates this to: $\bar{g}$ is a product of $n$ **distinct** monic linear polynomials in $\mathbb{F}_p[x]$. Call these polynomials $\varphi_1, \ldots, \varphi_n$. This is a selection of $n$ distinct objects from the set of monic linear polynomials in $\mathbb{F}_p[x]$, which has size $p$. Pigeonhole (in the sense "you cannot select more distinct objects than exist in total") requires $n \leq p$. **Contradiction with the hypothesis.** The theorem hypothesis $p < n$ yields $n > p$, contradicting $n \leq p$. Hence the assumption that $p$ splits completely was false. **Summary of the obstruction.** The obstruction is a counting fact: $\mathbb{F}_p$ has only $p$ elements, so at most $p$ distinct monic linear polynomials exist over $\mathbb{F}_p$. If the minimal polynomial $g$ has degree $n > p$, its reduction $\bar{g}$ cannot possibly factor into $n$ distinct linear factors — the alphabet of allowed factors is too small. The rest (Dedekind's criterion) is just the translation from ideals to polynomials. [/guided] [/step]