For a commutative ring \(R\), let \(U(R) = R^\times\) denote the group of units in \(R\). Recall that in an integral domain \(R\), a non-zero, non-unit element \(r \in R\) is called irreducible if whenever \(r = x y\) with \(x, y \in R\), then either \(x \in U(R)\) or \(y \in U(R)\). A non-zero, non-unit element \(r \in R\) is called prime in \(R\) if \(r \mid a b \implies r \mid a \text{ or } r \mid b\).
Consider the ring \(R = \mathbb{Z}[\sqrt{-5}] = \{a + b \sqrt{-5} \mid a, b \in \mathbb{Z}\} \subset \mathbb{C}\).
Prove that \(R\) is an integral domain.
Show that \(U(R) = \{\pm 1\}\).
Show that \(3\), \(2 + \sqrt{-5}\), and \(2 - \sqrt{-5}\) are irreducible in \(R\).
Show that \(3\) is not prime in \(R\).
Conclude that \(R\) is not a principal ideal domain (PID).
Solution
Goal: Prove properties of the ring \(\mathbb{Z}[\sqrt{-5}]\) using the field norm \(N(a + b\sqrt{-5}) = a^2 + 5b^2\), deduce that \(3\) is irreducible but not prime, and conclude \(R\) is not a PID.
1. Part (a): \(R\) is an integral domain.
1.1. \(R = \mathbb{Z}[\sqrt{-5}]\) is a subring of the field of complex numbers \(\mathbb{C}\):- \(1 = 1 + 0\sqrt{-5} \in R\).- For \(\alpha = a + b\sqrt{-5}\) and \(\beta = c + d\sqrt{-5} \in R\):\[\alpha - \beta = (a - c) + (b - d)\sqrt{-5} \in R,\]\[\alpha \beta = (a c - 5 b d) + (a d + b c)\sqrt{-5} \in R.\]
1.2. Since \(\mathbb{C}\) is a field, \(\mathbb{C}\) has no non-zero zero-divisors.
1.3. Any subring of a field is an integral domain. Thus \(R\) is an integral domain.
2. Part (b): \(U(R) = \{\pm 1\}\).
2.1. Define the algebraic norm \(N: R \to \mathbb{Z}_{\ge 0}\) by\[N(a + b\sqrt{-5}) = |a + b\sqrt{-5}|^2 = a^2 + 5b^2.\]
2.2. The norm is multiplicative: \(N(\alpha \beta) = N(\alpha) N(\beta)\) for all \(\alpha, \beta \in R\).
2.3. If \(u \in U(R)\), then there exists \(v \in R\) such that \(u v = 1\).
2.4. Taking norms gives \(N(u) N(v) = N(1) = 1\) in \(\mathbb{Z}_{\ge 0}\), which forces \(N(u) = 1\).
2.5. Write \(u = a + b\sqrt{-5}\) with \(a, b \in \mathbb{Z}\). Then \(a^2 + 5b^2 = 1\).
2.6. Since \(a, b \in \mathbb{Z}\), if \(b \ne 0\), then \(a^2 + 5b^2 \ge 5 > 1\). Thus \(b = 0\), which leaves \(a^2 = 1 \implies a = \pm 1\).
2.7. Conversely, \((\pm 1)(\pm 1) = 1\), so \(\pm 1 \in U(R)\).
2.8. Therefore \(U(R) = \{\pm 1\}\).
3. Part (c): \(3\), \(2 + \sqrt{-5}\), and \(2 - \sqrt{-5}\) are irreducible in \(R\).
3.1. Compute the norms:\[N(3) = 3^2 + 5(0)^2 = 9,\]\[N(2 + \sqrt{-5}) = 2^2 + 5(1)^2 = 4 + 5 = 9,\]\[N(2 - \sqrt{-5}) = 2^2 + 5(-1)^2 = 4 + 5 = 9.\]
3.2. Suppose \(\alpha \in \{3, 2 + \sqrt{-5}, 2 - \sqrt{-5}\}\) factors as \(\alpha = x y\) for some \(x, y \in R\).
3.3. Taking norms gives \(N(x) N(y) = N(\alpha) = 9\).
3.4. In \(\mathbb{Z}_{\ge 0}\), the divisors of 9 are 1, 3, and 9.
3.5. If \(N(x) = 3\), write \(x = a + b\sqrt{-5}\) with \(a, b \in \mathbb{Z}\). Then \(a^2 + 5b^2 = 3\).- If \(|b| \ge 1\), then \(a^2 + 5b^2 \ge 5 > 3\).- If \(b = 0\), then \(a^2 = 3\), which has no integer solutions since 3 is not a square in \(\mathbb{Z}\).- Thus no element in \(R\) has norm 3.
3.6. Therefore, the only possibilities for \((N(x), N(y))\) are \((1, 9)\) or \((9, 1)\).
3.7. By Part (b), an element of norm 1 is a unit (\(\pm 1\)). Thus either \(x \in U(R)\) or \(y \in U(R)\).
3.8. Since \(N(\alpha) = 9 > 1\), \(\alpha\) is not a unit. Hence \(3\), \(2 + \sqrt{-5}\), and \(2 - \sqrt{-5}\) are irreducible in \(R\).
4. Part (d): \(3\) is not prime in \(R\).
4.1. Consider the factorization:\[(2 + \sqrt{-5})(2 - \sqrt{-5}) = 2^2 - (-5) = 4 + 5 = 9 = 3 \cdot 3.\]
4.2. Thus \(3 \mid (2 + \sqrt{-5})(2 - \sqrt{-5})\) in \(R\).
4.3. Suppose for contradiction that \(3 \mid (2 + \sqrt{-5})\).- Then \(2 + \sqrt{-5} = 3(a + b\sqrt{-5}) = 3a + 3b\sqrt{-5}\) for some \(a, b \in \mathbb{Z}\).- Equating real parts gives \(3a = 2\), so \(a = 2/3 \notin \mathbb{Z}\), a contradiction.
4.4. Similarly, suppose \(3 \mid (2 - \sqrt{-5})\).- Then \(2 - \sqrt{-5} = 3(c + d\sqrt{-5})\) implies \(3c = 2\), again impossible in \(\mathbb{Z}\).
4.5. Thus \(3 \nmid (2 + \sqrt{-5})\) and \(3 \nmid (2 - \sqrt{-5})\).
4.6. Therefore \(3\) is not a prime element in \(R\).
5. Part (e): \(R\) is not a PID.
5.1. In any principal ideal domain \(D\), every irreducible element is prime:- Let \(p \in D\) be irreducible. If \(p \mid a b\) and \(p \nmid a\), then \(\gcd(p, a) = 1\).- In a PID, Bézout’s identity gives \(x p + y a = 1\) for some \(x, y \in D\).- Multiplying by \(b\) gives \(x p b + y a b = b\).- Since \(p \mid (x p b)\) and \(p \mid (y a b)\), \(p \mid b\).
5.2. In \(R = \mathbb{Z}[\sqrt{-5}]\), the element \(3\) is irreducible (by Part (c)) but is not prime (by Part (d)).
5.3. By contraposition, \(R\) cannot be a principal ideal domain.
6. Conclusion:
$R = \mathbb{Z}[\sqrt{-5}]$ is an integral domain with units $\{\pm 1\}$, has $3$ as an irreducible non-prime element, and is not a PID.