Show that a commutative Noetherian domain \(R\) has the following property: given any nonzero, nonunit \(x \in R\) one has \(x = z_1 z_2 \cdots z_m\) for some irreducible elements \(z_i \in R\).
Prove that any UFD is integrally closed in its field of fractions.
Give, with proof, an example of a domain \(R\) which is Noetherian, but not a UFD.
Solution
1. Every nonzero nonunit in a Noetherian domain is a finite product of irreducibles.
Assume, for contradiction, that some nonzero nonunit \(x_0\in R\) is not a finite product of irreducibles. Then \(x_0\) is not irreducible, so \[ x_0=x_1y_1 \] with \(x_1,y_1\) nonunits. If both \(x_1\) and \(y_1\) were finite products of irreducibles, then so would be \(x_0\). Hence at least one factor is again a nonzero nonunit that is not a finite product of irreducibles; rename it \(x_1\).
Since \(x_0=x_1y_1\), \[ (x_0)\subseteq(x_1). \] This inclusion is strict. Indeed, if \((x_0)=(x_1)\), then \(x_1=x_0r=x_1y_1r\) for some \(r\in R\). Because \(R\) is a domain and \(x_1\ne0\), \[ 1=y_1r, \] contradicting that \(y_1\) is a nonunit.
Repeating the construction gives an infinite strictly ascending chain \[ (x_0)\subsetneq(x_1)\subsetneq(x_2)\subsetneq\cdots. \] But a Noetherian ring satisfies the ascending chain condition on ideals. This contradiction proves that every nonzero nonunit factors into finitely many irreducibles.
2. Every UFD is integrally closed in its field of fractions.
Let \(R\) be a UFD with fraction field \(K\), and suppose \[ \alpha=\frac ab\in K \] is integral over \(R\), where \(a,b\in R\), \(b\ne0\), and \(a,b\) have no common irreducible factor. Since \(\alpha\) is integral, there are \(c_0,\ldots,c_{n-1}\in R\) such that \[ \alpha^n+c_{n-1}\alpha^{n-1}+\cdots+c_1\alpha+c_0=0. \] Multiplying by \(b^n\) gives \[ a^n+c_{n-1}a^{n-1}b+\cdots+c_1ab^{n-1}+c_0b^n=0. \] Hence \[ b\mid a^n. \] If \(b\) were not a unit, some irreducible \(q\) would divide \(b\). Then \(q\mid a^n\). In a UFD every irreducible is prime, so \(q\mid a\), contradicting the choice of \(a,b\) with no common irreducible factor. Therefore \(b\) is a unit, and \[ \alpha\in R. \] Thus \(R\) is integrally closed in \(K\).
3. The domain\[ R=\mathbb Z[\sqrt{-5}] \]is Noetherian.
As a \(\mathbb Z\)-module, \[ R=\mathbb Z\oplus\mathbb Z\sqrt{-5}, \] so \(R\) is finitely generated over the Noetherian ring \(\mathbb Z\). Hence \(R\) is a Noetherian \(\mathbb Z\)-module. Every ideal of \(R\) is, in particular, a \(\mathbb Z\)-submodule of \(R\). Therefore every ascending chain of ideals of \(R\) stabilizes, so \(R\) is a Noetherian ring. It is a domain because it is a subring of \(\mathbb C\).
4. The ring \(\mathbb Z[\sqrt{-5}]\) is not a UFD.
For \[ z=a+b\sqrt{-5}\in R, \] define \[ N(z)=z\overline z=a^2+5b^2\in\mathbb Z_{\ge0}. \] The norm is multiplicative: \[ N(zw)=N(z)N(w). \] The only units have norm \(1\). There are no elements of norm \(2\) or \(3\), since \[ a^2+5b^2=2 \quad\text{or}\quad 3 \] has no integer solutions.
Now \[ 6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}). \] The elements \(2\) and \(3\) are irreducible: a nontrivial factorization of \(2\) would force a factor of norm \(2\), and a nontrivial factorization of \(3\) would force a factor of norm \(3\). Likewise \[ N(1\pm\sqrt{-5})=6, \] so a nontrivial factorization of either element would force factors of norms \(2\) and \(3\); these do not exist. Thus all four displayed factors are irreducible.
The two factorizations are inequivalent up to units and ordering. Indeed associates have the same norm, whereas \[ N(2)=4, \qquad N(3)=9, \qquad N(1\pm\sqrt{-5})=6. \] Hence neither \(2\) nor \(3\) is associate to either \(1+\sqrt{-5}\) or \(1-\sqrt{-5}\). Therefore \(R\) does not have unique factorization.
5. Consequently \(\mathbb Z[\sqrt{-5}]\) is a Noetherian domain that is not a UFD.
Combine 3 and 4.