Why is Q(sqrt(2)) not isomorphic with Q(sqrt(3))

[Gtay]

Homework Statement

why is Q[$$\sqrt{2}$$] is not isomorphic to Q[$$\sqrt{3}$$]?

Homework Equations

The Attempt at a Solution

I do not know where to start?

 

[eok20]

Assume you have an isomorphism $$f$$ from $$\mathbb{Q}[\sqrt{2}]$$ to $$\mathbb{Q}[\sqrt{3}]$$. Think about what $$\sqrt{2}$$ can be mapped to. Hint: you can deduce that $$f(\sqrt{2})^2 = 2$$

 

[Gtay]

Thanks for the help.
If I assume there is an isomorphism from $$\mathbb{Q}[\sqrt{2}]$$ to $$\mathbb{Q}[\sqrt{3}]$$ then since $$\sqrt{2}$$ is not rational, it can only be mapped to $$\sqrt{3}$$. This says that (a+b$$\sqrt{2}$$) is mapped to (a+b$$\sqrt{3}$$) but the map here is not a homomorphism since it fails this requirement: f(ab) = f(a)f(b).
I am just curious if there is some characteristic that $$\mathbb{Q}[\sqrt{2}]$$ but $$\mathbb{Q}[\sqrt{3}]$$ does not? ( I guess not?)