I read the proof that $\mathbb{Q}(\sqrt2)$ is isomorphic to $\mathbb{Q}(\sqrt3)$ and it makes since. However I cannot understand what is wrong with the following map $\varphi$.
$$ \varphi:\mathbb{Q}(\sqrt2) \to\mathbb{Q}(\sqrt3) $$ defined by $$ \varphi(a+b\sqrt2)\mapsto a+b\sqrt3. $$ and this is a 1-1 map and has the properties $$ \varphi(a+b)=\varphi(a)+\varphi(b) \\ \varphi(ab)=\varphi(a)\varphi(b) $$ My error will be appreciated.
On
The map $\varphi$ is a $\mathbb{Q}$-vector space isomorphism, but not a ring homomorphism.
Under a ring isoomorphism, corresponding elements should have the same minimal polynomial over $\mathbb{Q}$, but $\sqrt{2}$ and $\sqrt{3}$ don't. So, not only $\varphi$ is not a ring homomorphism, but there can be no isomorphism, because no element in $\mathbb{Q}(\sqrt{2})$ has $X^2-3$ as its minimal polynomial.
$$2=\varphi(2)=\varphi(\sqrt2\cdot\sqrt2)=\varphi(\sqrt2)\varphi(\sqrt2)=\sqrt3\cdot\sqrt3=3$$