Give an example of each of the following:
A Euclidean domain.
A Unique Factorization Domain that is not a PID.
An integral domain that is not a Unique Factorization Domain.
Solution
Examples are, respectively, \(\mathbb Z\), \(\mathbb Z[x]\), and \(\mathbb Z[\sqrt{-5}]\).
1. The ring \(\mathbb Z\) is Euclidean.
Use the Euclidean function \(d(b)=|b|\) for \(b\ne0\). For integers \(a,b\) with \(b\ne0\), integer division gives \(a=qb+r\) with \(0\leq r<|b|\). Thus \(r=0\) or \(d(r)<d(b)\), as required.
2. The ring \(\mathbb Z[x]\) is a unique factorization domainbut not a principal ideal domain.
Unique prime factorization makes \(\mathbb Z\) a unique factorization domain. Gauss’s lemma implies that a polynomial ring in one variable over a unique factorization domain is again a unique factorization domain [@DF04]. Hence \(\mathbb Z[x]\) has that property.
Its ideal \(J=(2,x)\) is not principal. Every element of \(J\) has even constant coefficient, so \(1\notin J\) and \(J\) is proper. If \(J=(f)\), then \(2=fg\) for some \(g\in\mathbb Z[x]\). Degrees of nonzero polynomials add, so \(f\) is a constant integer dividing \(2\): it is \(1,-1,2\), or \(-2\). The first two possibilities would give the whole ring. The last two give \((2)\), which does not contain \(x\) because its coefficient of \(x\) is odd. No generator is possible.
3. The integral domain \(R=\mathbb Z[\sqrt{-5}]\) is not aunique factorization domain.
The ring embeds in \(\mathbb C\), so it is a domain. For \(z=a+b\sqrt{-5}\) define \[ N(z)=z\overline z=a^2+5b^2. \] This is a nonnegative integer, is positive when \(z\ne0\), and satisfies \(N(zw)=N(z)N(w)\) by multiplication and conjugation. A unit must have norm \(1\); conversely norm \(1\) gives inverse \(\overline z\in R\). Solving \(a^2+5b^2=1\) shows that the only units are \(\pm1\). There is no element of norm \(2\) or \(3\): if \(b\ne0\) the norm is at least \(5\), and if \(b=0\) it is a square integer.
Now \[ 6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}). \] The element \(2\), of norm \(4\), is irreducible: a factorization into two nonunits would require both factors to have norm \(2\). Similarly \(3\), of norm \(9\), would require factors of norm \(3\). Each element \(1\pm\sqrt{-5}\) has norm \(6\), so a factorization into nonunits would require norms \(2\) and \(3\). All these possibilities have been excluded.
Associates have the same norm because units have norm \(1\). The norms \(4\) and \(9\) of the factors on the left differ from the norm \(6\) of either factor on the right. Thus no rearrangement and multiplication by units can identify the two irreducible factorizations. This disproves unique factorization in \(R\).