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.
Best Answer
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 $a_0\in\Bbb Q$.