How to show that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 5)$ are non-isomorphic?

Question

• How to show that $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 5)$ are non-isomorphic as a ring?

  All I could manage to show is that,

  for any isomorphism $\phi:\mathbb(\sqrt 2)\to\mathbb(\sqrt 5),$ $\phi(1)=1.$

  Since $\phi$ is a homomorphism, $\phi(a+b\sqrt 2)=\phi(a)+\phi(b)\phi(\sqrt 2)=a+b.\phi(\sqrt 2)$ (Since $\phi(\dfrac{p}{q})=\dfrac{p}{q}\phi(1)$ for $p,q(\neq 0)\in\mathbb Z$) $\forall~a,b\in\mathbb Q.$

Answer 1

Hints without many words: supose there is such an isomorphism, then

$$\phi(\sqrt 2)=a+b\sqrt 5\implies \phi(2)=a^2+2ab\sqrt 5+5b^2$$

But also

$$\phi(2)=2\phi(1)=2$$

Deduce now your contradiction....

Answer 2

We know that $\sqrt{2}\times \sqrt{2} - 1 - 1 =0$. Hence $\phi(\sqrt{2})\times\phi(\sqrt{2})-\phi(1)-\phi(1)=\phi(0)$. Hence $x^2=2$ has a solution in $\mathbb{Q}(\sqrt{5})$.

Answer 3

Another example of equation having sol-n in $Q[\sqrt2]$ but not in $Q[\sqrt5]$: take $(x+(a+b\sqrt2))\cdot(x+(a-b\sqrt2))=x^2 +2xa+(a^2-2b^2)$ (the coefficients belong to $Q[\sqrt5]$, since belong to $Q$)