Prove $\sqrt{2} \notin \mathbb{Q}(\sqrt{3+\sqrt{3}})$ using field trace I would like to prove that $\sqrt{2} \notin \mathbb{Q}(\sqrt{3+\sqrt{3}})$. I would like to do so using field trace, since it seems like a pretty useful tool. I have not seen it in my lectures, so I am still pretty dubious when it comes to applying it.
I have discovered that the Galois closure of $\beta \equiv \sqrt{3 + \sqrt{3}}$ is given by $\mathbb{Q}(\beta, \sqrt{2})$, which I now need to prove is an extension of $\mathbb{Q}$ of degree 8. To prove it, is is sufficient to see that $\mathbb{Q}(\beta, \sqrt{2}) | \mathbb{Q}(\sqrt{2})$ has degree $2$. Since $\sqrt{3} \in \mathbb{Q}(\beta, \sqrt{2})$, I think this is equivalent to proving that $\beta \notin \mathbb{Q}(\sqrt{2}, \sqrt{3})$.
Suppose $ \beta \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Then, there would exist $a_0, a_1, a_2, a_3 \in \mathbb{Q}$ such that:
$$\beta = a_0 + a_1 \sqrt{2} + a_2 \sqrt{3} + a_3 \sqrt{6}$$
If we define $T \equiv Tr_{\mathbb{Q}(\sqrt{2}, \sqrt{3})/Q}$ the trace operator we have that:
$$T(\beta) = 0 = a_0 + a_1 T(\sqrt{2}) + a_2 T(\sqrt{3}) + a_3 T (\sqrt{6})$$
and I think all that traces are $0$, since $T(\beta) = \beta - \beta + \gamma - \gamma = 0$ (where $\gamma = \sqrt{3-\sqrt{3}}$) and $T(\sqrt{2}), T(\sqrt{3}), T(\sqrt{6}) $ are $0$. Therefore $a_0 = 0$.
I am not really sure about how to go on. I have seen other similar posts and usually the solution is just multiplying by some appropriate element, like $\sqrt{3}$. But I don't really know how to calculate the trace of elements like $\beta \sqrt{3}$. Can you give me any hint about how to go on?
Other proofs are ok, but please use reasonably elementary Galois properties. Thanks in advance.
EDIT: I may have found a solution.
We multiply by $\sqrt{3}$ and we have:
$$\beta \sqrt{3} = a_1 \sqrt{6} + 3 a_2  + 9 a_3 \sqrt{2}$$
and all the traces of all those terms but $T (3 a_2)$ are zero (right?). So $a_2 = 0$.
If we repeat the process multiplying by $\sqrt{2}$ and take traces again we get that $a_1 = 0$. So $\beta$ is a rational multiple of $\sqrt{6}$ which is absurd. Is this ok? I am quite sure I am calculating traces wrong.
 A: I think that looks ok, since the sum of roots of a quadratic without a linear term, or a quartic without a cubic term is zero (giving a zero trace). Alternatively, squaring the linear combination for $\beta$ yields $$3+\sqrt3=A_0+(2a_0a_1+3a_2a_3)\sqrt2+A_2\sqrt3+(2a_0a_3+a_1a_2)\sqrt6$$ where $A_0,A_2\in\Bbb Q$ are unimportant coefficients. The $\sqrt6$ coefficient is zero by a standard argument so $a_1=-2a_0a_3/a_2$. The $\sqrt2$ coefficient is also zero and substituting the expression for $a_1$ gives $4a_0^2=3a_2^2$ which is absurd since $a_0\in\Bbb Q$.
A: This is a general technique used to deal with real radicals and I learnt it on mathse.
Lemma: Let $F$ be a field with $F\subseteq \mathbb {R} $ and let $a$ be a real number and $m$ a positive integer such that $a\notin F, a^m\in F$. Then $\text{tr} _F^{F(a)} (a) =0$.
Note that if $K$ is an extension field of $F(a) $ of finite degree then we get $\text{tr} _F^K(a) =0$ as well.
We proceed in two steps. First we show that $\sqrt {2}\notin\mathbb {Q} (\sqrt {3})$. Let $$a=\sqrt{2},b=\sqrt{3},c=ab,F=\mathbb {Q} $$ Then we know that none of $a, b, c$ lie in $F$ but their squares lie in $F$. We can assume on the contrary that $a\in F(b) $ so that $$a=p+qb$$ for some $p, q\in F$. Let $K=F(a, b) $ be of degree $n$ over $F$. Applying trace $\text{tr} _F^K$ on above equation we get $$0=np+q\cdot 0$$ ie $p=0$ or $a=qb$ or $$2=a^2=qab=qc$$ Applying trace again we get $2n=q\cdot 0$ which is absurd. Thus we have $a\notin F(b) $.
Next we establish the result asked in question. Let $$a=\sqrt{2},b=\sqrt{3+\sqrt{3}},c=ab,F=\mathbb {Q} (\sqrt{3})$$ Then we can observe that $b, c$ are of degree $4$ over field of rationals and hence they don't lie in $F$. We have already established in first part that $a\notin F$. But let us note that $a^2,b^2,c^2$ all lie in $F$. Also observe that $\mathbb{Q}(b)=F(b) $ and $F(b) $ is of degree $2$ over $F$.
Let us assume on the contrary that $a\in F(b)$. Then we can  write $$a=p+qb$$ for some $p, q\in F$. Using $K=F(a, b) $  and applying trace $\text{tr} _F^K$ on it we get $$0=np+q\cdot 0$$ ie $p=0$. Then we have $$2=a^2=qc$$ and applying trace again we get $2n=q\cdot 0$ which is absurd. Thus $a\notin F(b) $.
The argument in both steps is exactly the same, we just change the base field $F$. The procedure can be used to deal with nested real radicals in a similar manner.
A: This is an argument using norm instead of trace.
Consider the Galois extension $\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q(\sqrt 2)$ whose Galois group is $S_2 = \{1, \sigma\}$ where $\sigma$ swaps $\pm\sqrt 3$. If $\beta\in\mathbb Q(\sqrt 2, \sqrt 3)$, its norm for the extension is $\beta \cdot\sigma(\beta)\in\mathbb Q(\sqrt 2)$, where $\sigma(\beta)$ is among $\pm \sqrt{3\pm \sqrt 3}$.
However, $\beta \cdot \pm \sqrt{3+\sqrt 3} = \pm (3 + \sqrt 3)\not\in\mathbb Q(\sqrt 2)$ and $\beta \cdot \pm \sqrt{3-\sqrt 3} = \pm \sqrt 6\not\in\mathbb Q(\sqrt 2)$.
A: We'll use the approach of this answer, which states: incommensurable radicals are linearly independent.
As basic field we choose $F= \mathbb{Q}(\sqrt{3})\subset \mathbb{R}$. Our radicals are $1$, $\sqrt{2}$,  and $\sqrt{3 + \sqrt{3}}$. Let's check for instance that
$$\frac{\sqrt{3 + \sqrt{3}}}{\sqrt{2}} \not \in \mathbb{Q}(\sqrt{3})$$
Assume the contrary, so $\frac{3+ \sqrt{3}}{2} = (a + b\sqrt{3})^2$, and from here
$$\frac{9-3}{4} = N^{\mathbb{Q}(\sqrt{3})}_{\mathbb{Q}}(\frac{3+ \sqrt{3}}{2}) = (a^2 - 3 b^2)^2 $$
a contradiction.
One checks similarly for the other two ratios.
We conclude
$$\sqrt{2} \not = \alpha\cdot 1 + \beta\cdot  \sqrt{3+ \sqrt{3}}$$
for any $\alpha$, $\beta\in \mathbb{Q}(\sqrt{3})$.
