The ring $\mathcal{O}_{\mathbb{Q}(\sqrt{-5})} = \mathbb{Z}\left[\frac{1+\sqrt{-5}}{2}\right]$ is not a UFD since $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ gives two distinct element factorizations.

At the ideal level, let $\mathfrak{p}_2 = (2, 1+\sqrt{-5})$ and $\mathfrak{p}_3 = (3, 1+\sqrt{-5})$. Then: $(2) = \mathfrak{p}_2^2, \quad (3) = \mathfrak{p}_3\bar{\mathfrak{p}}_3, \quad (1+\sqrt{-5}) = \mathfrak{p}_2\mathfrak{p}_3$

Thus $(6) = \mathfrak{p}_2^2\mathfrak{p}_3\bar{\mathfrak{p}}_3$ uniquely, reconciling the two element factorizations.

In $\mathbb{Z}[\zeta_5]$ where $\zeta_5 = e^{2\pi i/5}$, consider factoring $(11)$. The polynomial $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$ factors modulo 11 as $\Phi_5(x) \equiv (x-2)(x-6)(x-7)(x-8) \pmod{11}$

This reveals $(11)$ splits completely: $(11) = \mathfrak{p}_2\mathfrak{p}_6\mathfrak{p}_7\mathfrak{p}_8$ where $\mathfrak{p}_a = (11, \zeta_5 - a)$ for each root $a \bmod 11$.

To find generators for an ideal in $\mathcal{O}_K$, use the Hermite normal form algorithm. For instance, in $\mathbb{Z}[\sqrt{-5}]$, the ideal $I = (6, 2+2\sqrt{-5})$ can be written in matrix form and reduced:

Working modulo the $\mathbb{Z}$-basis $\{1, \sqrt{-5}\}$, we find $I = (2, \sqrt{-5}+1)$ with norm $N(I) = 2$. This is the prime ideal $\mathfrak{p}_2$ above 2.

The Hasse-Minkowski theorem exemplifies how local information determines global solvability. A quadratic form $Q$ over $\mathbb{Q}$ represents zero non-trivially if and only if it represents zero over $\mathbb{R}$ and all $\mathbb{Q}_p$.

For the form $Q(x,y,z) = x^2 + y^2 - 5z^2$, checking local conditions:

• Over $\mathbb{R}$: clearly has solutions
• Over $\mathbb{Q}_2, \mathbb{Q}_5$: verify using Hensel's lemma
• Over $\mathbb{Q}_p$ for $p \neq 2, 5$: use quadratic reciprocity

All local conditions satisfied implies existence of rational solutions, which can be found as $(x,y,z) = (1,2,1)$.

In $\mathbb{Z}[i]$, compute $(2+3i) \cap (5)$ as ideals:

Since $2+3i$ is irreducible (norm 13 is prime), $(2+3i)$ is prime. The ideal $(5) = (2+i)(2-i)$ factors. Using $\gcd(2+3i, 5) = 1$ in $\mathbb{Z}[i]$, we have $(2+3i) \cap (5) = (2+3i)(5) = (10+15i)$

This intersection has norm $N(10+15i) = 325 = 13 \cdot 25 = N(2+3i) \cdot N(5)$.