I am trying to determine a basis of $\mathbb{Q}(\sqrt{3+ \sqrt{2}})$ as a field extension over $\mathbb{Q}$, and my attempt is to choose an arbitrary element of $\mathbb{Q}(\sqrt{3+ \sqrt{2}})$ then find a basis through doing the multiples of that element, and I get:
$x^0 = 1$
$x^1 = a + b \sqrt{3+ \sqrt{2}}$
$x^2 = (a^2 + 3b^2) + 2b^2 \sqrt{2} + 2ab \sqrt{3+ \sqrt{2}}$
$\cdots$
But through this way I don't know whether this process will end by getting a set which is multiplicatively closed or not. So can someone tell me what the right way to do this problem is, or give me some hint on my way of doing it? Thanks!
It is not true that $1,\sqrt{2},\sqrt{3+\sqrt{2}}$ have to be in the basis. Bases are fairly arbitrary, only $\Bbb Q$-linear dependence is needed.
There is a canonical way to find a basis. Find a primitive element... well, that's easy - take $\alpha=\sqrt{3+\sqrt{2}}$. Find the degree of the minimal polynomial. "Clearly" it should be four; explicitly, $(\alpha^2-3)^2-2=0$ and $x^4-6x^2+7$ is irreducible over $\Bbb Q$, e.g. reduce modulo $2$. Then $1,\alpha,\alpha^2,\alpha^3$ is a basis!