Factor the principal ideals in an algebraic integer ring which are generated by ramified rational primes

Let K = \mathbb{Q}(\sqrt{-5}) and let \mathcal{O} be the ring of integers in K. Factor (p) in \mathcal{O}, where p is a ramified rational prime.

By Theorem 9.6, if p does not divide the discriminant d = -20 of K (using Theorem 6.11), then p is not ramified. Now 20 = 2^2 \cdot 5, so that if p is ramified in K, it is either 2 or 5.

We claim that (2) = (2,1+\sqrt{-5})^2. Indeed, the (\supseteq) direction is clear, and we have 2 = -(1+\sqrt{-5})^2 + 2 \cdot (1+\sqrt{-2}) - 2 \cdot 2. We claim also that P = (2,1+\sqrt{-5}) is maximal. To this end, let a+b \sqrt{-5} \in \mathcal{O}, and say a - b \equiv c mod 2 where c \in \{0,1\}. Evidently, a+b\sqrt{-5} \equiv c mod P; if c \equiv 0 mod 2 then a+b\sqrt{-5} \equiv 0 mod P, and if c \equiv 1 mod 2 then a+b\sqrt{-5} \equiv 1 mod P. Now suppose 1 \in P; then 1 = 2(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5}). Comparing coefficients mod 2, we have 0 \equiv h+k \equiv 1 mod 2, a contradiction. So \mathcal{O}/(2,1+\sqrt{-5}) = \{ \overline{0}, \overline{1}\}, and thus \mathcal{O}/(2,1+\sqrt{-5}) \cong \mathbb{Z}/(2) is a field. Hence P is maximal, and (2) = (2,1+\sqrt{-5})^2 is the prime factorization of (2).

Certianly (5) = (\sqrt{-5})^2. We claim that Q = (\sqrt{-5}) is prime. To see this, note that 5 \in (\sqrt{-5}). If a+b\sqrt{-5} \in \mathcal{O}, then a+b\sqrt{-5} \equiv a \equiv a_0 mod Q, where a_0 \in \{0,1,2,3,4\}. Suppose t \in Q \cap \mathbb{Z}. Then t = (a+b\sqrt{-5})\sqrt{-5} for some a,b \in \mathbb{Z}. Comparing coefficients, we have t \equiv 0 mod 5. In particular, r \not\equiv s mod Q, where r,s \in \{0,1,2,3,4\} are distinct. Thus \mathcal{O}/(\sqrt{-5}) \cong \mathbb{Z}/(5) is a field, so that (\sqrt{-5}) is maximal. Thus (5) = (\sqrt{-5})^2 is the prime factorization of (5).

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