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The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

Suppose ${\alpha }^m=a\in {\mathbb{Q}}^{\times }$ in a number field and $\mathfrak{p}$ lies above $p$. Its normalized integral valuation satisfies $m{v}_{\mathfrak{p}}(\alpha )=e_{\mathfrak{p}/p}{v}_p(a)$. If $m$ and $v_p(a)$ are coprime, then $m\mid e_{\mathfrak{p}/p}$. In a degree-$m$ field the fundamental identity for prime decomposition forces a unique prime, ramification index $m$, and residue degree one. This detects total ramification of a prime even when the displayed defining polynomial is not Eisenstein at $p$.

// Target: algebra.bigb

Ramification is branching in a geometric or algebraic map, detected by multiplicity greater than one. For number fields, ramification of a prime means that some prime ideal occurs with exponent greater than one in the extended prime ideal. For a map of smooth algebraic curves, a local parameter pulls back to a parameter with order greater than one at a ramification point.