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?)