Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$?

Let $K=\Bbb Q(\sqrt3)$. Then for $\alpha$, $\beta\in K$, $\sqrt\beta\in K(\sqrt\alpha)$ if and only if $\beta$ or $\alpha\beta$ is a square in $K$.

In this example. $\alpha=3+\sqrt3$ and $\beta=3-\sqrt3$. Then $\alpha$ is not a square in $K$, since its norm to $\Bbb Q$ is $6$ which is not a square in $\Bbb Q$. Also $\alpha\beta=6$ and that is not a square in $K$: the rational numbers which are squares in $K$ are $a^2$ and $3a^2$ for $a\in\Bbb Q$.

In this case, $\sqrt\beta\not\in\Bbb Q(\sqrt\alpha)$.

Use the tracial method, for example. The trace map on a number field $Q(\alpha)$ assigns to each $\beta \in \mathbb Q(\alpha)$ the quantity $\sum_{\sigma} \sigma(\beta)$ where $\sigma(\beta)$ are all the conjugates of $\beta$.

Suppose that $\sqrt 2 \in L$. Then $\sqrt 3 \in L$ implies that $\mathbb Q(\sqrt{3+\sqrt 3}) = \mathbb Q(\sqrt 2 ,\sqrt 3)$.

Write $\sqrt{3 + \sqrt 3} = a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$. Taking the trace of both sides gives $2a = 0$(note that $\sqrt{3+\sqrt 3}$ has minimal polynomial $x^4 - 6x^2+6$ hence has trace zero) hence $a = 0$.

Next multiply by $\sqrt 2$ to get $2b + c \sqrt 6 + 2d \sqrt 3 = \sqrt{6+2\sqrt 3}$. Once again the trace of the RHS is seen to be zero(find the minimal polynomial, it is easy!), so $b = 0$ is obtained on taking trace.

We are left with $\sqrt{3+\sqrt 3} = c \sqrt 3 + d \sqrt 6$. Multiplying by $\sqrt 3$ gives $\sqrt{9+3\sqrt 3} = 3c + 2d \sqrt 2$. Again the trace of the LHS is zero(easy again!) so we get $c = 0$.

Finally we are left with $\sqrt{3 + \sqrt{3}} = d \sqrt 6$, here squaring gives $\sqrt 3 = 6d^2 - 3$, an obvious problem. Hence, we are done.

Note that $\sqrt{3 - \sqrt 3}$ is a conjugate of $\sqrt{3+\sqrt 3}$, therefore the above shows that this extension is not normal, hence not Galois.

Let $\alpha, \beta$ be $\sqrt{3+\sqrt{3}}$ and $\sqrt{3-\sqrt{3}}$ respectively; they clearly have the same minimum polynomial on $\mathbb{Q}$ which is $f(x) = x^4-6x^2+6$, by direct calculation and using Eisenstein Criterion.

Now let $E$ be the splitting field of $f$ on $\mathbb{Q}$ and let $G$ be its Galois group. The order of $G$ may be $4,8,12,24$; I'll prove that it is exactly 8. In this way it can be shown that $\beta$ cannot be in $\mathbb{Q}(\alpha)$ because otherwise $E = \mathbb{Q}(\alpha)$ which has order 4 on $\mathbb{Q}$.

By Galois Correspondence Theorem there is at least one subgroup $H$ of $G$ with index 4 in it: that's the subgroup consisting of all $\mathbb{Q}$-automorphisms of E which fix the elements of $\mathbb{Q}(\alpha)$. Since the elements of H send roots of $f$ in roots of $f$, it's clear that the only possibility for $H$ is to be of the form $H = \{id, \sigma\}$ where $\sigma$ is the $\mathbb{Q}$-automorphism of E that sends $\beta$ to $-\beta$ and fixes $\alpha$ (and consequently fixes $-\alpha$).

Therefore $G$ contains a subgroup having 2 elements and index 4 in $G$; by Lagrange's Theorem follows that $G$ has order 8.