# Show that the extension has degree 9 [duplicate]

Let $$F=\mathbb{Q}(\sqrt{2}, \sqrt{3})$$. Show that over $$\mathbb{Q}$$ this is an extension of degree 9.

I see that this is equivalent of showing that $$\sqrt{2}\notin \mathbb{Q}(\sqrt{3})$$. I don't know any other way of showing this except going through $$\sqrt{2}=a+b\sqrt{3}+c\sqrt{9}$$ which doesn't lead me anywhere.

1/ If you would like to proceed that way then by contradiction, suppose that $$K=\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{3})$$. Consider the splitting field $$\mathbb{Q}(\sqrt{2}, \omega)$$ of $$x^3-2$$ (where $$\omega$$ is the 3-th root of 1), then this is also the splitting field of $$x^3-3$$ (since we just adjoin $$\omega$$ to $$K$$).
The automorphism $$\sigma$$ such that $$\sigma(\sqrt{3}) = \omega \sqrt{3}, \sigma(\omega \sqrt{3}) = \omega^2 \sqrt{3}, \sigma(\omega^2 \sqrt{3}) = \sqrt{3}$$ also maps $$\sqrt{2}$$ to a solution of $$x^3-2$$, i.e, $$\sigma(\sqrt{2})$$ is either $$\sqrt{2},\omega \sqrt{2}$$ or $$\omega^2 \sqrt{2}$$.
Now assume that $$\sqrt{2}=a+b\sqrt{3}+c\sqrt{9}$$, by taking $$\sigma$$ both sides and considering case by case, you should get a contradiction.
2/ If you know about trace, however, there is a nicer proof in my opinion and this would help handle more general situation. Suppose that $$K=\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{3})$$ as before. Then you can check that $$\mathrm{Trace}_{K/\mathbb{Q}}(\sqrt{2}) = \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt{4}) = \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt{3})= \mathrm{Trace}_{K/\mathbb{Q}}(\sqrt{9}) = 0$$ If we assume that $$\sqrt{2}=a+b\sqrt{3}+c\sqrt{9}$$ then by taking trace both sides, we get $$0=3a$$ (since $$a\in \mathbb{Q}$$), hence $$a=0$$. Now multiplying both sides of $$\sqrt{2}=b\sqrt{3}+c\sqrt{9}$$ by $$\sqrt{3}$$ and $$\sqrt{9}$$ and taking trace, we get $$b=c=0$$ as well. (Notice that $$\mathrm{Trace}_{K/\mathbb{Q}}(\sqrt{6}) = 0$$ as well).