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Let $d\ne 0,1$ be a square-free integer, and let $K=\mathbb{Q}(\sqrt{d})$ be the associated quadratic field. ThenThe pair $1,{\omega }_d$ is an integral basis. The field discriminant is $d$ in the first case and $4d$ in the second. This statement requires $d$ to be a square-free integer: for example, $\mathbb{Q}(\sqrt{8})=\mathbb{Q}(\sqrt{2})$ but $\mathbb{Z}[\sqrt{8}]$ omits $\sqrt{2}$.^[ref]

For $d=2$, the integral basis is $1,\sqrt{2}$, and the field discriminant is $8$. For $d=5$, it is $1,\varphi$ with $\varphi =(1+\sqrt{5})/2$, and the field discriminant is $5$. The identity ${\varphi }^2=\varphi +1$ makes $\mathbb{Z}[\varphi ]$ closed under multiplication. Since $\sqrt{5}=2\varphi -1$, the smaller order in a number field $\mathbb{Z}[\sqrt{5}]$ consists precisely of $a+b\varphi$ with even $b$. This computes its index two and shows concretely why adjoining a square root does not always supply every algebraic integer.

In this second integral basis, the field norm isIn particular, $N(\varphi )=-1$ and ${\varphi }^{-1}=\varphi -1$: the extra algebraic integer is also a unit. The units of the quadratic field Q square root of five describe all its powers.