How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3).$ [duplicate]

Solution 1:

Consider the polynomial $$p(x)=x^{2}-2$$

Since any isomorphism $$\varphi:\,\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{3})$$ is the identity when restricted to $\mathbb{Q}$ we get that if $q\in\mathbb{Q}(\sqrt{2})$ then $$\varphi(p(q))=\varphi(q^{2})-\varphi(2)=\varphi^{2}(q)-2$$

Since $p$ have a root $r\in\mathbb{Q}(\sqrt{2})$ we get that $$0=\varphi(0)=\varphi(p(r))=\varphi^{2}(r)-2$$

but $\varphi(r)\in\mathbb{Q}(\sqrt{3})$ and there is no element $\alpha$ in $\mathbb{Q}(\sqrt{3})$ that satisfies the equation $\alpha^{2}-2=0$ since this would imply $\alpha=\sqrt{2}\in\mathbb{Q}(\sqrt{3})$.

Note: It is true in general that if $\mathbb{F}_{1}\cong\mathbb{F}_{2}$ and $\sum_{i=0}^{n}a_{n}x^{n}=p(x)\in\mathbb{F}_{1}[x]$ have a root $r$ then the polynomial $\sum_{i=0}^{n}\varphi(a_{n})x^{n}=q\in\mathbb{F}_{2}[x]$ have a root $\varphi(r)\in\mathbb{F}_{2}$.

Solution 2:

Isomorphic quadratic number fields have the same discriminant. But $\mathbb{Q}(\sqrt{2})$ has discriminant $8$, and $\mathbb{Q}(\sqrt{3})$ has discriminant $12$, see here.

Solution 3:

Let $\phi:\sqrt 2\mapsto p.$ Then $2\mapsto p^2$ whence $p^2=2.$ But $x^2-2=0$ has no solution in $\mathbb Q(\sqrt 3)$ since any element of $\mathbb Q(\sqrt 3)$ is of the form $ax+b+\langle x^2-3\rangle$ and $(ax+b+\langle x^2-3\rangle)^2-(2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies(a^2x^2+2abx+b^2-2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies b=0.$ So $p$ is of the form $ax+\langle x^2-3\rangle$ and $a^2x^2+\langle x^2-3\rangle=2+\langle x^2-3\rangle\implies\dfrac{2}{a^2}=3$ a contradiction to $a\in\mathbb Q$