is it this one

Example 18.10. Not every integral domain is a unique factorization domain. The subring Z[v/3 i] = {a + bv/3 i} of the complex numbers is an integral domain by v(z) = |z|? = a? + 3b?. It is clear that v(z) 0 with equality when z = 0. Also, from our knowledge of complex numbers we know that v(zw) = v(z)v(w). It is easy to show that if v(z) = 1, then zis a unit, and that the only units of Z\\V/3 4] are 1 and 1, We claim that 4 has two distinct factorizations into irreducible elements: 4=2-2=(1- V3i)(14+ V3). We must show that each of these factors is an irreducible element in Z[,/3 4]. If 2 is not irreducible, then 2 = zw for elements z, w in Z[./3 i] where v(z) = v(w) = 2. However, there does not exist an element in z in Z[./3 i] such that v(z) = 2 because the equation a? + 3b? = 2 has no integer solutions. Therefore, 2 must be irreducible. A similar argument shows that both 1 /3i and1+ V3iare irreducible. Since 2 is not a unit multiple of either 1 V3ior1+ /3i, 4 has at least two distinct factorizations into irreducible elements