How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3).$ How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$ 
My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when restricted to $\mathbb Q.$
Ya! I can see that: $\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle$ and $\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.$ Now if $\phi$ be such an isomorphism then $\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.$ Hence the restriction of $\phi$ over $\{q+\langle x^2-2\rangle:q\in\mathbb Q\}$ remains the identity.
Now what?
 A: 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.
A: 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 $\varphi(r)\in\mathbb{F}_{2}$.
A: 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 $a\in\mathbb Q$
A: Well, we use the following facts: 


*

*$\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a,b \in \mathbb{Q}\}$

*$\mathbb{Q}(\sqrt{3}) = \{a + b\sqrt{3} : a,b \in \mathbb{Q}\}$

*$\sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$


Suppose they are isomorphic, and let $\phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ be an isomorphism. Then $\phi(a + b\sqrt{2}) = a + b \phi(\sqrt{2})$ (by the fact that $\phi$ restricted to $\mathbb{Q}$ is the identity. But notice $\phi(\sqrt{2})^2 = \phi(2) = 2$, so $\phi(\sqrt{2}) = \pm \sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$.
