Every nontrivial number field has a ramified finite prime

Fix an integral basis ${\alpha }_1,\dots ,{\alpha }_n$ of ${\mathcal{O}}_K$, which exists by [F2]. By [F1] and [F3] the integer $\mathrm{\mid }{d}_K\mathrm{\mid }$ is greater than $1$, so [F12] gives a rational prime $p$ dividing $d_K$.

Put $A:={\mathcal{O}}_K\mathrm{/}p{\mathcal{O}}_K$. Since ${\mathcal{O}}_K$ is free with $\mathbb{Z}$-basis ${\alpha }_1,\dots ,{\alpha }_n$ by [F2], the classes ${\bar{\alpha }}_1,\dots ,{\bar{\alpha }}_n$ form an ${\mathbb{F}}_p$-basis of $A$; in particular ${\dim }_{{\mathbb{F}}_p}A=n$.

Factorisation: by [F5], write $p{\mathcal{O}}_K={\mathfrak{P}}_1^{e_1}\cdots {\mathfrak{P}}_r^{e_r}$ with distinct nonzero primes ${\mathfrak{P}}_i$ and $e_i\ge 1$. Distinct maximal ideals satisfy ${\mathfrak{P}}_i+{\mathfrak{P}}_j={\mathcal{O}}_K$; choosing $u+v=1$ with $u\in {\mathfrak{P}}_i$, $v\in {\mathfrak{P}}_j$ and expanding $(u+v{)}^{e_i+e_j-1}$ exhibits every term as an element of ${\mathfrak{P}}_i^{e_i}+{\mathfrak{P}}_j^{e_j}$, so $1$ lies in that sum and the powers are pairwise comaximal. Applying [F6] to the ideals ${\mathfrak{P}}_i^{e_i}$ gives an isomorphism $A\cong {\prod }_{i=1}^r{A}_i$ with $A_i:={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i^{e_i}$.

Reducedness of the factors: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$, so the maximal ideals of $A_i$ are the images of maximal ideals of ${\mathcal{O}}_K$ containing ${\mathfrak{P}}_i^{e_i}$; a maximal ideal containing ${\mathfrak{P}}_i^{e_i}$ contains the prime ${\mathfrak{P}}_i$, hence equals it, and ${\mathfrak{m}}_i:={\mathfrak{P}}_i\mathrm{/}{\mathfrak{P}}_i^{e_i}$ is the unique maximal ideal of $A_i$, with $A_i\mathrm{/}{\mathfrak{m}}_i={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i$ a finite field by [F5]. If $e_i=1$ then $A_i={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i$ is a field and reduced. If $e_i\ge 2$ then ${\mathfrak{P}}_i^{e_i}⊊{\mathfrak{P}}_i$: otherwise ${\mathfrak{P}}_i^{e_i}={\mathfrak{P}}_i$, and multiplying by the inverse ideal ${\mathfrak{P}}_i^{-1}$, which exists by [F11], gives ${\mathfrak{P}}_i^{e_i-1}={\mathcal{O}}_K\subseteq {\mathfrak{P}}_i$, a contradiction; so some $x\in {\mathfrak{P}}_i\setminus {\mathfrak{P}}_i^{e_i}$ has nonzero image in $A_i$ with $x^{e_i}\in {\mathfrak{P}}_i^{e_i}$, a nonzero nilpotent. Therefore $A_i$ is reduced exactly when $e_i=1$, and since a finite product of nonzero rings is reduced exactly when each factor is, $A$ is reduced exactly when all $e_i=1$; by [F5] this is exactly the case that $p$ is unramified.

Trace form and discriminant: for $x\in {\mathcal{O}}_K$, multiplication by $x$ on ${\mathcal{O}}_K$ has matrix with integer entries in the basis ${\alpha }_i$ and trace ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(x)$ by [F4]. Reducing modulo $p$ shows that multiplication by $\bar{x}$ on $A$ has ${\mathbb{F}}_p$-trace ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(x)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p$. Hence $T(\bar{x},\bar{y}):={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(xy)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p$ defines an ${\mathbb{F}}_p$-bilinear form on $A$ whose matrix in the basis ${\bar{\alpha }}_i$ is $({\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}({\alpha }_i{\alpha }_j)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)$, with determinant $d_K<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p$ by [F3]. By [F8] this form is degenerate exactly when that determinant vanishes, that is, exactly when $p\mid d_K$.

Trace form versus reducedness over the perfect field ${\mathbb{F}}_p$: (a) if every $e_i=1$, then $A\cong {\prod }_i{F}_i$ with $F_i={\mathcal{O}}_K\mathrm{/}{\mathfrak{P}}_i$ a finite field; each $F_i\mathrm{/}{\mathbb{F}}_p$ is finite, hence separable by [F9], so each factor trace pairing is nondegenerate by [F7]. Multiplication by an element of the product acts blockwise on the direct sum ${⨁}_i{F}_i$, so the trace form of $A$ is the orthogonal direct sum of the factor pairings; a vector orthogonal to everything has every component orthogonal to its own factor, hence is zero, and by [F8] the form is nondegenerate. (b) if some $e_i>1$, choose $0\mathrm{\ne }\bar{x}$ in the nilpotent maximal ideal of the factor $A_i$ as in step 1.4; for every $\bar{y}\in A$ the product $\bar{x}\bar{y}$ is nilpotent, so multiplication by it is a nilpotent endomorphism and has trace $0$ by [F10]. Thus $\bar{x}\mathrm{\ne }0$ lies in the radical of $T$ and $T$ is degenerate. Consequently $T$ is nondegenerate exactly when $A$ is reduced.

Combining steps 2.1, 1.4 and 2.2, for the rational prime $p$ the following are equivalent: $p\mid d_K$; the trace form $T$ on $A={\mathcal{O}}_K\mathrm{/}p{\mathcal{O}}_K$ is degenerate; $A$ is not reduced; some ramification index exceeds $1$; and $p$ ramifies in $K$. This verifies the published ramification-discriminant criterion [F13] for this field and prime in full.

By step 1.1 the prime $p$ divides $d_K$, so step 3.1, equivalently the criterion [F13], shows that $p$ ramifies in $K$. Therefore every number field of degree $n>1$ has a rational prime that ramifies in it.