Ramification is detected by the number-field discriminant

Fix an integral basis ${\alpha }_1,\dots ,{\alpha }_n$ as in [F1], and put $A:={\mathcal{O}}_K\mathrm{/}p{\mathcal{O}}_K$. The classes ${\bar{\alpha }}_1,\dots ,{\bar{\alpha }}_n$ form an ${\mathbb{F}}_p$-basis of $A$, since ${\mathcal{O}}_K$ is free over $\mathbb{Z}$ with that basis. Thus ${\dim }_{{\mathbb{F}}_p}A=n\ge 1$, so $p{\mathcal{O}}_K$ is proper; it is nonzero because $p\mathrm{\ne }0$ in the characteristic zero field $K$. By [F5], write $p{\mathcal{O}}_K={\prod }_{i=1}^r{\mathfrak{P}}_i^{e_i}$ with distinct nonzero prime ideals and positive exponents. The list is nonempty because $p{\mathcal{O}}_K$ is proper. Define ${\mathrm{Tr}}_A(a):=\mathrm{tr}(m_a)$ for the ${\mathbb{F}}_p$-linear multiplication map on $A$, and $T_A(x,y):={\mathrm{Tr}}_A(xy)$; matrix-trace linearity makes $T_A$ a bilinear form.

Each ${\mathfrak{P}}_i$ is maximal. Indeed, [F6] makes ${\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i$ finite, and [F6] also makes it an integral domain. For any nonzero class, multiplication by it is injective on this finite set, hence surjective, so the class has an inverse and the quotient is a field; [F6] then gives maximality. Therefore distinct ${\mathfrak{P}}_i,{\mathfrak{P}}_j$ are comaximal. Choose $u\in {\mathfrak{P}}_i$ and $v\in {\mathfrak{P}}_j$ with $u+v=1$. In the expansion of $(u+v{)}^{e_i+e_j-1}$, a term $u^k{v}^{e_i+e_j-1-k}$ lies in ${\mathfrak{P}}_i^{e_i}$ if $k\ge e_i$, and otherwise its exponent of $v$ is at least $e_j$, so it lies in ${\mathfrak{P}}_j^{e_j}$. The sum is $1$; hence the powers ${\mathfrak{P}}_i^{e_i}$ and ${\mathfrak{P}}_j^{e_j}$ are comaximal.

Put $A_i:={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i^{e_i}$. Applying [F7] to the pairwise comaximal powers gives $A\cong {\prod }_{i=1}^r{A}_i\mathrm{.}$ If $e_i=1$, then $A_i={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i$ is a finite field. Since ${\mathfrak{P}}_i$ lies above $(p)$, the map $\mathbb{Z}\mathrm{/}p\mathbb{Z}\to A_i$ induced by inclusion of $\mathbb{Z}$ is injective: its kernel is $({\mathfrak{P}}_i\cap \mathbb{Z})\mathrm{/}(p)=0$. Thus $A_i\mathrm{/}{\mathbb{F}}_p$ is a finite field extension, of degree $f_i=f({\mathfrak{P}}_i\mathrm{/}p)$ by definition.

We first prove that the trace pairing of every finite extension $E\mathrm{/}{\mathbb{F}}_p$ of finite fields is nondegenerate, without passing to an algebraic closure. Let $f=[E:{\mathbb{F}}_p]$ and $\sigma (x)=x^p$. By [F8], ${\sigma }^0,\dots ,{\sigma }^{f-1}$ are the distinct automorphisms of $E$ over ${\mathbb{F}}_p$. Fix an ${\mathbb{F}}_p$-basis $b_1,\dots ,b_f$ and form the square matrix $C=({\sigma }^j(b_k){)}_{0\le j<f,1\le k\le f}$. If a row combination with coefficients $c_j\in E$ vanishes, then ${\sum }_j{c}_j{\sigma }^j(b_k)=0$ for each basis vector. Each ${\sigma }^j$ is ${\mathbb{F}}_p$-linear, so ${\sum }_j{c}_j{\sigma }^j(x)=0$ for every $x\in E$, and in particular for every $x\in E^{\times }$. The restrictions ${\sigma }^j{\mathrm{\mid }}_{E^{\times }}$ are distinct group homomorphisms $E^{\times }\to E^{\times }$; [F9] makes them $E$-linearly independent, so every $c_j=0$. Hence $C$ is invertible.

If every $e_i=1$, then [F7] identifies $A$ with the product of the finite fields $A_i$ from step 1.3. Multiplication by $(x_i{y}_i{)}_i$ on this product is block diagonal, so its trace is ${\sum }_i{\mathrm{Tr}}_{A_i\mathrm{/}{\mathbb{F}}_p}(x_i{y}_i)$. The form on $A$ is therefore the orthogonal direct sum of the finite field trace pairings from step 1.4. Each is nondegenerate, so the product pairing is nondegenerate: for a nonzero vector, choose a nonzero component and pair it with a witness in that factor, taking zero in all other factors.

If some $e_i>1$, then ${\mathfrak{P}}_i^{e_i}⊊{\mathfrak{P}}_i$. The containment is immediate; equality would contradict uniqueness in [F5], because the factorizations of ${\mathfrak{P}}_i$ and ${\mathfrak{P}}_i^{e_i}$ have exponents $1$ and $e_i$. Choose $a\in {\mathfrak{P}}_i\setminus {\mathfrak{P}}_i^{e_i}$. Its class in $A_i$ is nonzero and nilpotent, since its $e_i$-th power is zero. Through the product isomorphism in step 1.3, it gives a nonzero nilpotent $\eta \in A$. For any $y\in A$, $u=\eta y$ is nilpotent, so multiplication $T=m_u$ on the finite-dimensional ${\mathbb{F}}_p$-space $A$ is nilpotent.

Steps 2.1 and 2.2 show that the trace pairing on $A$ is nondegenerate exactly when every $e_i=1$. By [F5], this is exactly when $p$ is unramified; equivalently, the pairing is degenerate exactly when $p$ ramifies.

For $z\in {\mathcal{O}}_K$, multiplication by $z$ on ${\mathcal{O}}_K$ has an integer matrix in the integral basis ${\alpha }_1,\dots ,{\alpha }_n$. Viewed over $\mathbb{Q}$, the same matrix represents multiplication by $z$ on $K$, so [F3] makes its matrix trace ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(z)$. Reducing the integer matrix modulo $p$ gives multiplication by $\bar{z}$ on $A$; hence ${\mathrm{tr}}_{{\mathbb{F}}_p}(m_{\bar{z}})={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(z)(\mathrm{m}\mathrm{o}\mathrm{d}p)\mathrm{.}$ Consequently the matrix of $T_A$ in the basis ${\bar{\alpha }}_1,\dots ,{\bar{\alpha }}_n$ is the reduction modulo $p$ of $G=({\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}({\alpha }_i{\alpha }_j){)}_{i,j}$. Since each ${\alpha }_i{\alpha }_j\in {\mathcal{O}}_K$ by [F1], [F13] makes every entry of $G$ an integer, and [F4] gives $\det G=d_K$. By [F12], the trace pairing $T_A$ is degenerate exactly when its matrix is not invertible, equivalently when $\det G<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p=0$; this is equivalent to $p\mid d_K$. Combining with step 3.1 proves that $p$ ramifies exactly when $p\mid d_K$.