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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.

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    $\varphi(\sqrt{2}^2)=\varphi(2)=2$ or $\varphi(\sqrt{2}^2)=\varphi(\sqrt{2})^2=\sqrt{3}^2=3$. – 2017-01-19
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    You say it has the property $\phi(ab)=\phi(a)\phi(b)$, but it doesn't - that's the problem. – 2017-01-19
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    They are isomorphic _as vector spaces over $\mathbb Q$_, but not as fields. – 2017-01-19
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    For further insight see the prior question: [Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})?$](http://math.stackexchange.com/q/9188/242) – 2017-01-19

2 Answers 2

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$$2=\varphi(2)=\varphi(\sqrt2\cdot\sqrt2)=\varphi(\sqrt2)\varphi(\sqrt2)=\sqrt3\cdot\sqrt3=3$$

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    must i test more things then just homomorphism and 1-1? – 2017-01-19
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    @DanielHouston: Asaf's point is that _it is not a homomorphism_: it fails to satisfy $\varphi(ab)=\varphi(a)\varphi(b)$ when $a=b=\sqrt2$. – 2017-01-19
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    @HenningMakholm I understand now thank you. – 2017-01-19
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    @Daniel: You should continue checking that $\varphi$ is an isomorphism only if you agree that $2=3$, in which case, since $0=1$, many more things can be proved. The point is that your homomorphism might be additive, but not multiplicative. – 2017-01-19
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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.