Let's solve the problem step by step: --- ## **Step 1: Find the ring of integers \(\mathcal{O}_K\) of \(K = \mathbb{Q}(\sqrt{3})\)** For a quadratic field \(K = \mathbb{Q}(\sqrt{d})\), the ring of integers is: - \(\mathbb{Z}[\sqrt{d}]\) if \(d \equiv 2,3 \pmod{4}\) - \(\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right]\) if \(d \equiv 1 \pmod{4}\) Here, \(d = 3\), and \(3 \equiv 3 \pmod{4}\), so: \[ \mathcal{O}_K = \mathbb{Z}[\sqrt{3}] \] **Explanation:** We determined the ring of integers to be \(\mathbb{Z}[\sqrt{3}]\) using the standard classification for quadratic fields. --- ## **Step 2: Use Theorem 10.1 to factor ideals \(\langle p \rangle\) in \(\mathcal{O}_K\)** Theorem 10.1 (likely the Dedekind-Kummer theorem) tells us how a rational prime \(p\) factors in \(\mathcal{O}_K\): For \(\mathbb{Q}(\sqrt{d})\): - If the minimal polynomial \(x^2 - d\) mod \(p\) is irreducible, \(p\) is inert. - If \(x^2 - d\) splits into distinct linear factors mod \(p\), \(p\) splits. - If \(p\) divides the discriminant (\(d_K\)), \(p\) ramifies. For \(d = 3\), the discriminant is \(d_K = 4 \cdot 3 = 12\). Now, examine each prime: ### For \(p = 2\): - \(2 \mid 12\) (divides discriminant) ⇒ **ramifies**. - \(x^2 - 3 \equiv x^2 + 1 \pmod{2}\) has only one root mod 2. - So, \(2\) ramifies: \(\langle 2 \rangle = \mathfrak{p}^2\), where \(\mathfrak{p} = \langle 2, \sqrt{3} \rangle\). ### For \(p = 3\): - \(3 \mid 12\) ⇒ **ramifies**. - \(x^2 - 3 \equiv x^2 \pmod{3}\) has one root (0). - So, \(3\) ramifies: \(\langle 3 \rangle = \mathfrak{q}^2\), where \(\mathfrak{q} = \langle 3, 1 + \sqrt{3} \rangle\). ### For \(p = 5\): - \(5 \nmid 12\), so check \(x^2 - 3 \pmod{5}\). - \(x^2 \equiv 3 \pmod{5}\): \(2^2 = 4\), \(3^2 = 9 \equiv 4\), so no solution. - Not a quadratic residue ⇒ **inert**. - So, \(\langle 5 \rangle\) is prime in \(\mathcal{O}_K\). **Explanation:** We used splitting/ramification/inertia criteria for each prime in the ring of integers. --- ## **Step 3: Factor the composite ideals \(\langle 10 \rangle\) and \(\langle 30 \rangle\)** Recall: - \(\langle 10 \rangle = \langle 2 \cdot 5 \rangle = \langle 2 \rangle \langle 5 \rangle\) - \(\langle 30 \rangle = \langle 2 \cdot 3 \cdot 5 \rangle = \langle 2 \rangle \langle 3 \rangle \langle 5 \rangle\) From above: - \(\langle 2 \rangle = \mathfrak{p}^2\) - \(\langle 3 \rangle = \mathfrak{q}^2\) - \(\langle 5 \rangle\) is prime So, - \(\langle 10 \rangle = \mathfrak{p}^2 \langle 5 \rangle\) - \(\langle 30 \rangle = \mathfrak{p}^2 \mathfrak{q}^2 \langle 5 \rangle\) **Explanation:** We used the multiplicativity of ideal factorization. --- ## **Final Answer:** \[ \begin{align*} \langle 2 \rangle &= \mathfrak{p}^2, \quad \text{where } \mathfrak{p} = \langle 2, \sqrt{3} \rangle\\ \langle 3 \rangle &= \mathfrak{q}^2, \quad \text{where } \mathfrak{q} = \langle 3, 1+\sqrt{3} \rangle\\ \langle 5 \rangle &\text{ is prime in } \mathcal{O}_K\\ \langle 10 \rangle &= \mathfrak{p}^2 \langle 5 \rangle\\ \langle 30 \rangle &= \mathfrak{p}^2 \mathfrak{q}^2 \langle 5 \rangle \end{align*} \] Each ideal is factored into prime ideals in \(\mathbb{Z}[\sqrt{3}]\) as above.

Let's approach the problem systematically in two steps: --- ## **Step 1: Determine the structure of the ring of integers \(\mathcal{O}_K\) and analyze the behavior of the primes 2, 3, and 5** **Explanation:** Since \(K = \mathbb{Q}(\sqrt{3})\), the discriminant \(d_K\) is given by: - \(d_K = 4 \times 3 = 12\). For quadratic fields, the ring of integers \(\mathcal{O}_K\) is: - \(\mathbb{Z}[\sqrt{3}]\) because \(3 \equiv 3 \pmod{4}\). Next, analyze how each prime \(p\) factors in \(\mathcal{O}_K\): - **Prime 2:** - 2 divides the discriminant (\(12\)), so 2 ramifies. - The minimal polynomial \(x^2 - 3\) modulo 2 is \(x^2 + 1\), which has no roots in \(\mathbb{F}_2\), so 2 ramifies into a prime ideal to the power 2: \(\langle 2 \rangle = \mathfrak{p}^2\). - **Prime 3:** - 3 divides discriminant, so 3 ramifies. - \(x^2 - 3 \equiv x^2 \pmod{3}\), which has root 0, so \(\langle 3 \rangle = \mathfrak{q}^2\). - **Prime 5:** - 5 does not divide discriminant. - \(x^2 \equiv 3 \pmod{5}\); check quadratic residues: - \(2^2=4\), \(3^2=9 \equiv 4\), so 3 is not a quadratic residue mod 5. - Thus, 5 remains inert: \(\langle 5 \rangle\) is prime in \(\mathcal{O}_K\). --- ## **Step 2: Factorize the ideals \((2)\), \((3)\), \((5)\), \((10)\), and \((30)\) in \(\mathcal{O}_K\)** **Explanation:** - \(\langle 2 \rangle = \mathfrak{p}^2\), where \(\mathfrak{p} = \langle 2, \sqrt{3} \rangle\). - \(\langle 3 \rangle = \mathfrak{q}^2\), where \(\mathfrak{q} = \langle 3, 1+\sqrt{3} \rangle\). - \(\langle 5 \rangle\) remains prime. Now, express the composite ideals: - \(\langle 10 \rangle = \langle 2 \rangle \langle 5 \rangle = \mathfrak{p}^2 \langle 5 \rangle\). - \(\langle 30 \rangle = \langle 2 \rangle \langle 3 \rangle \langle 5 \rangle = \mathfrak{p}^2 \mathfrak{q}^2 \langle 5 \rangle\). --- ## **Final Answer:** \[ \boxed{ \begin{aligned} &\langle 2 \rangle = \mathfrak{p}^2, \quad \text{where } \mathfrak{p} = \langle 2, \sqrt{3} \rangle, \\ &\langle 3 \rangle = \mathfrak{q}^2, \quad \text{where } \mathfrak{q} = \langle 3, 1+\sqrt{3} \rangle, \\ &\langle 5 \rangle \text{ is prime in } \mathcal{O}_K, \\ &\langle 10 \rangle = \mathfrak{p}^2 \langle 5 \rangle, \\ &\langle 30 \rangle = \mathfrak{p}^2 \mathfrak{q}^2 \langle 5 \rangle. \end{aligned} } \] This completes the factorization of the principal ideals in \(\mathcal{O}_K\).