Determine whether the ring \(\mathbb{Z}[3i]\) is a UFD.
Solution
1. The unit group of \(R = \mathbb{Z}[3i]\) is \(R^\times=\{\pm1\}\).
1.1. Elements of \(R = \mathbb{Z}[3i]\) are of the form \(\alpha = a + 3bi\) where \(a, b \in \mathbb{Z}\).Define the multiplicative field norm \(N: R \to \mathbb{Z}_{\ge 0}\) by:\[ N(a + 3bi) = a^2 + 9b^2. \]
1.2. An element \(\alpha \in R\) is a unit if and only if \(N(\alpha) = 1\).The equation \(a^2 + 9b^2 = 1\) in integers has only the solutions \((a, b) = (\pm 1, 0)\).Thus the group of units is \(R^\times = \{ \pm 1 \}\).
2. The elements \(3\) and \(\pm 3i\) are irreducible in \(R\).
2.1. Suppose \(3 = \alpha \beta\) for some \(\alpha, \beta \in R\).Taking norms:\[ 9 = N(3) = N(\alpha) N(\beta). \]If neither \(\alpha\) nor \(\beta\) is a unit, then \(N(\alpha) = 3\) and \(N(\beta) = 3\).
2.2. For any \(\alpha = a + 3bi \in R\), \(N(\alpha) = a^2 + 9b^2\).- If \(b \neq 0\), then \(a^2 + 9b^2 \ge 9 > 3\).- If \(b = 0\), then \(a^2 = 3\), which has no integer solution.Thus there are no elements of norm 3 in \(R\).Therefore \(3\) is irreducible in \(R\).
2.3. Since \(N(\pm 3i) = 9\), the same norm argument shows that \(3i\) and \(-3i\) are also irreducible in \(R\).
3. The element \(9\) has two factorizations into irreducibles whose factors are not associates.
3.1. Consider the element \(9 \in R\), which has two factorizations into irreducibles:\[ 9 = 3 \cdot 3 = (3i) \cdot (-3i). \]
3.2. The irreducible factors \(3\) and \(3i\) are not associates in \(R\), because the only units are \(\pm 1\) and \(3 \cdot (\pm 1) \neq 3i\) (since \(i \notin R\)).Thus \(9\) possesses two distinct factorizations into irreducible elements.
4. \(R\) is not integrally closed, which independently shows that \(R\) is not a UFD.
4.1. The fraction field of \(R\) is \(\mathbb{Q}(i)\), and the integral closure of \(\mathbb{Z}\) in \(\mathbb{Q}(i)\) is the ring of Gaussian integers \(\mathbb{Z}[i]\).The element \(i \in \mathbb{Q}(i)\) is a root of the monic polynomial \(x^2 + 1 \in R[x]\), but \(i \notin R\).Thus \(R = \mathbb{Z}[3i]\) is not integrally closed, and hence cannot be a UFD.
5. Q.E.D.By step 3, \(\mathbb{Z}[3i]\) is not a UFD.