Let \(R\) be the number ring corresponding the the number field \(K\). Then it can be easily shown that every unit \(u\) satisfies \(N(u) = \pm 1\). Also, if \(a \in R\) satisfies \(N(a) = 1\), then \(1 = n(a)^{n/d} = a(a^{n/d - 1}(\sigma_2(a)...\sigma_n(a))^{n/d})\) thus \(a\) is indeed factor of unity in \(R\) since all conjugates of \(a\) must also be algebraic integers.

Hence an element \(a \in R\) is a unit if and only if \(N(a) = {\pm 1}\). For example, for squarefree \(m < 0\) we have that the only units in the number ring of \(\mathbb{Q}[\sqrt{m}]\) are \({\pm 1}\) if \(m \ne {-1},{-3}\). If \(m = {-1}\) we also have \(i, -i\), and if \(m = {-3}\), we have the primitive sixth roots of unity. On the other hand \(\mathbb{Z}[\sqrt{2}]\) has infinitely many units generated by the powers of \((1+\sqrt{2})\) (which correspond to the solutions of \(a^2 - 2b^2 = {\pm 1}\) over the integers).

For \(a \in R\), if \(N(a)\) is prime (in \(\mathbb{Z}\)) then clearly \(a\) is irreducible. For example, \(9 + \sqrt{10}\) is irreducible in \(\mathbb{Z}[\sqrt{10}]\).

Note \(x^2 + 5 y^2\) cannot be 2 or 3 if \(x,y\) are integers so all elements of \(\mathbb{Z}[\sqrt{-5}]\) of norm 6 are irreducible. Hence in \(\mathbb{Z}[\sqrt{-5}]\) we have \(2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\) as an example of nonunique factorization.

The trace can be used to show that certain elements are not contained in certain fields. For example, consider the field \(\mathbb{Q}[\alpha]\) where \(\alpha = \root{4}{2}\). We wish to determine if \(\sqrt{3}\) is contained in this field. If it is, we have the equation \(\sqrt{3} = a + b \alpha + c \alpha^2 + d \alpha^3\). Now \(T(\sqrt{3}) = T(\alpha) = T(\alpha^2) = T(\alpha^3) = 0\), and \(T(a) = 4a\), thus we must have \(a=0\). Dividing both sides of the equation by \(\alpha\) gives \(\root{4}{9/2} = b + c \alpha + d \alpha + a(\alpha^3)/2\), and taking traces now shows \(b = 0\). Similarly, dividing by \(\alpha\) again shows \(c = 0\). We can divide again to show \(d = 0\) and thus obtain a contradiction, or simply say that \((d \alpha^3)^2 = 2 d^2 \sqrt{2} \ne \sqrt{3}\).