Let \(n\) be a squarefree integer greater than 3. Let \(R\) denote the subring \(\mathbb{Z}[\sqrt{-n}] = \{a + b\sqrt{-n} \mid a, b \in \mathbb{Z}\}\) of the field of complex numbers \(\mathbb{C}\).
Show that \(\sqrt{-n}\) and \(1 + \sqrt{-n}\) are irreducible in \(R\).
Prove that \(R\) is not a unique factorization domain (UFD).
Construct an ideal in \(R\) which is not principal.
Hint: In a UFD, an element is irreducible if and only if it is prime.
Solution
Put \(s=\sqrt{-n}\) and define the norm \[ N(a+bs)=(a+bs)(a-bs)=a^2+nb^2. \] It is multiplicative and takes positive integer values on nonzero elements.
1. The element \(s\) is irreducible.
Suppose \[ s=\alpha\beta \] with nonzero \(\alpha,\beta\in R\). Then \[ n=N(s)=N(\alpha)N(\beta). \] If both factors were nonunits, then \[ 1<N(\alpha)<n. \] Write \(\alpha=a+bs\). Since \(N(\alpha)<n\), necessarily \(b=0\), so \(N(\alpha)=a^2\). Thus a square \(a^2>1\) divides the squarefree integer \(n\), impossible. Hence one factor is a unit and \(s\) is irreducible.
2. The element \(1+s\) is irreducible.
Suppose \[ 1+s=\alpha\beta \] with both factors nonzero. If both were nonunits, then \[ 1<N(\alpha)<N(1+s)=n+1. \] Write \(\alpha=a+bs\). If \(b\ne0\), then \(N(\alpha)\ge n\). Because \(N(\alpha)\) divides \(n+1\), the only possibility with \(n\le N(\alpha)<n+1\) is impossible. Hence \(b=0\) and \(\alpha=a\in\mathbb Z\).
But an integer \(a\) divides \(1+s\) in \(R\) only if it divides both coefficients \(1\) and \(1\), so \(a=\pm1\). Thus \(\alpha\) is a unit, a contradiction. Hence \(1+s\) is irreducible.
3. For an integer \(k\), one has\[ s\mid k\quad\Longleftrightarrow\quad n\mid k. \]
Since \[ \frac{k}{s}=-\frac{k}{n}s, \] the quotient lies in \(R\) exactly when \(k/n\in\mathbb Z\).
4. For an integer \(k\), one has\[ 1+s\mid k\quad\Longleftrightarrow\quad n+1\mid k. \]
Since \[ \frac{k}{1+s}=\frac{k(1-s)}{n+1}, \] the quotient belongs to \(R\) exactly when both coefficients \(k/(n+1)\) and \(-k/(n+1)\) are integers, equivalently when \(n+1\mid k\).
5. If \(n\) is even, then the irreducible element \(s\) is not prime.
Because \(n>3\) and is even, \[ n=2\cdot\frac n2 \] with \(0<2<n\) and \(0<n/2<n\). Also \[ n=-s^2, \] so \(s\mid 2(n/2)\). By 3, \(s\) divides neither \(2\) nor \(n/2\). Thus \(s\) is irreducible but not prime.
6. If \(n\) is odd, then the irreducible element \(1+s\) is not prime.
Since \(n>3\) is odd, \[ n+1=2\cdot\frac{n+1}{2} \] with both positive factors strictly smaller than \(n+1\). Moreover \[ n+1=(1+s)(1-s), \] so \(1+s\) divides their product. By 4, it divides neither factor. Hence \(1+s\) is irreducible but not prime.
7. The ring \(R\) is not a UFD.
If \(n\) is even, 5 gives an irreducible element that is not prime. If \(n\) is odd, 6 does so. In a UFD every irreducible element is prime, so in either case \(R\) is not a UFD. This proves part (b).
8. If \(n\) is even, the ideal\[ I=(s,2) \]is not principal.
Suppose \(I=(d)\). Then \(d\mid s\). Since \(s\) is irreducible, either \(d\) is a unit or \(d\) is associate to \(s\).
3. If \(d\) is a unit, then \(I=R\), so\[ 1=rs+2t \]for some \(r,t\in R\). Multiplying by \(n/2\) gives\[ \frac n2=r\frac n2s+nt. \]Both terms on the right are divisible by \(s\) because \(s\mid n\), so \(s\mid n/2\), contradicting
3. Hence \(I\) is not principal.
9. If \(n\) is odd, the ideal\[ J=(1+s,2) \]is not principal.
Suppose \(J=(d)\). Since \(d\mid1+s\) and \(1+s\) is irreducible, either \(d\) is a unit or is associate to \(1+s\).
4. If \(d\) is a unit, then \(J=R\), so\[ 1=r(1+s)+2t \]for some \(r,t\in R\). Multiplying by \((n+1)/2\) gives\[ \frac{n+1}{2}=r\frac{n+1}{2}(1+s)+(n+1)t. \]Both terms on the right are divisible by \(1+s\) because \(1+s\mid n+1\), so \(1+s\mid(n+1)/2\), contradicting
4. Hence \(J\) is not principal.
10. Thus part (c) is solved by \((\sqrt{-n},2)\) when \(n\) is even and by \((1+\sqrt{-n},2)\) when \(n\) is odd.
This is exactly 8 and 9.