## 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 $(5)$.