Prove that $\sqrt{2} \notin \mathbb{Q}(\sqrt{3})$.

Question

Suppose there exists an isomorphism $\Phi \colon \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$. Then, of course, it must be the case that $\Phi(1) = 1$. Hence \begin{align*} 2 &= 1+1 = \Phi(1) + \Phi(1) = \Phi(1 + 1) \\ &= \Phi(2) = \Phi(\sqrt{2} \sqrt{2}) = \Phi(\sqrt{2}) \Phi(\sqrt{2}) = \Phi(\sqrt{2})^2. \end{align*} In other words, $\Phi(\sqrt{2}) = \pm\sqrt{2}$. I don't know how to proceed further.

Answer 1

Suppose $ p =a + b\sqrt{3}$ (with $a,b \in \mathbb{Q}$), which is a general element of $\mathbb{Q}(\sqrt{3})$, is such that $p^2 = 2$.

And $p^2 = (a^2 + 3b^2) + 2ab\sqrt{3} = 2$. So $2ab = 0$ and $a^2 + 3b^2 = 2$, hence...

Answer 2

If $(a+b\sqrt3)^2=2$, then $$a^2+3b^2+2ab\sqrt3=2, $$ so $a^2+3b^2=2$ and $2ab=0$. You have either $a=0$ or $b=0$, giving $3b^2=2$ or $a^2=2$, both impossible.

Answer 3

An other way. Suppose $\sqrt 2\in \mathbb Q(\sqrt 3)$, then $\mathbb Q(\sqrt 3)=\mathbb Q(\sqrt 2)=K$. Let $\sqrt 2=a+b\sqrt 3$ with $a,b\in \mathbb Q$. You have that $$[\times \sqrt 3]=\begin{pmatrix}0&3\\1&0\end{pmatrix}$$ refer to the basis $\{1,\sqrt 3\}$ and $$[\times \sqrt 2]=\begin{pmatrix}0&2\\ 1&0\end{pmatrix}$$ refer to the basis $\{1,\sqrt 2\}$.

Therefore, $$0=\text{Tr}_{K/\mathbb Q}(\sqrt 2)=a\text{Tr}_{K/\mathbb Q}(\sqrt 3)+2b\implies 2b=0\implies b=0$$ and thus $$\sqrt 2 =a\sqrt 3\implies a=\frac{\sqrt 2}{\sqrt 3}\in \mathbb R\backslash \mathbb Q$$ which is a contradiction.