Let be squarefree. Find a basis for where is the conjugate of .
Let . Evidently, is a root of ; the other root is . If , then this polynomial has no rational roots, and so is irreducible over . In this case, is the minimal polynomial of , and so .
Note that if , then cannot be a basis for as it contains only one element. Thus we may assume that .
We claim that is a basis for so long as . To that end, suppose first that is a basis; if , then we have and . Now , with . So , a contradiction since is not rational. Thus .
Conversely, suppose . To show that is a basis, it suffices to show that it generates , as we know that this field has degree 2 over . To achieve this, we will show that and are linear combinations of and . Indeed, evidently we have and , as desired.
In particular, is a basis for over .