Every time I see an example of a basis for $\mathbb{Q}(\sqrt{d})$ in my book, $d$ is prime. Is there a reason for this?
Let $d=p_1\times p_2$ where $p_1$ and $p_2$ are different primes, and $\sqrt{d} \not \in \mathbb{Q}$. Is the basis for $\mathbb{Q}(\sqrt{d})=\{a+ b\sqrt{d} \mid a,b \in \mathbb{Q}\}$?
Thanks in advance
Edit: I meant: the basis for $\mathbb{Q}(\sqrt{d})=\{1, \sqrt{d}\}$ as pointed out in the answers. Thanks!
$\{a + b\sqrt{d} \mid a, b \in \Bbb Q\}$ is not a basis for $\Bbb Q(\sqrt{d})$. It is not linearly independent. It contains $0$, for example.
One example for a $\Bbb Q$-basis for $\Bbb Q(\sqrt{d})$ is $\{1, \sqrt{d}\}$. This works not only for $d = p_1p_2$ (with $p_1 \neq p_2$) but also for any $d \in \Bbb Q$ which is not the square of a rational number. (How?)