Assume you have an isomorphism [tex]f[/tex] from [tex]\mathbb{Q}[\sqrt{2}][/tex] to [tex]\mathbb{Q}[\sqrt{3}][/tex]. Think about what [tex]\sqrt{2}[/tex] can be mapped to. Hint: you can deduce that [tex]f(\sqrt{2})^2 = 2[/tex]
Thanks for the help.
If I assume there is an isomorphism from [tex]\mathbb{Q}[\sqrt{2}][/tex] to [tex]\mathbb{Q}[\sqrt{3}][/tex] then since [tex]\sqrt{2}[/tex] is not rational, it can only be mapped to [tex]\sqrt{3}[/tex]. This says that (a+b[tex]\sqrt{2}[/tex]) is mapped to (a+b[tex]\sqrt{3}[/tex]) 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 [tex]\mathbb{Q}[\sqrt{2}][/tex] but [tex]\mathbb{Q}[\sqrt{3}][/tex] does not? ( I guess not?)
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