# Q(sqrt(2)) and Q(sqrt(3)) not isomorphic?

• geor
In summary, the conversation discusses the exercise of proving that Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic. The person is confused because they initially thought the two vector spaces should be isomorphic, but they realize that it is not true for fields. It is then mentioned that this holds for any square free integers, and a proof is suggested by showing that an isomorphism cannot exist between the fields. The conversation ends with the clarification that this proof can be generalized to other square free integers.

#### geor

Hello all,

I am studying Algebra and in the chapter where Galois theory is introduced, I
see the following exercise:

"Prove that Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic"

Well, It seems that I am a bit behind because I really don't get it... :(
I mean, I'm sure that this is the case, since it is a question in the book
(and surely 'not' is not a typo!), but these are vector spaces over Q,
both of dimension 2, so shouldn't they be isomorphic by sending
sqrt(2) to sqrt(3) and any rational number to itself?!

What do I miss here?

Last edited:
Ooops! I think I see it now..
They are isomorphic as vector spaces but not as fields, right?
The isomorphism I said above does not respect the product..

That's it, right?!

Right. (Have you yet shown there isn't a field isomorphism?)

What more is true, is that given any square free intgers m and n, Q(sqrt(m)) and Q(sqrt(n)) are nonisomorphic. Intution serves right when you say that it "does not respect the product" but being more rigorous, show that no ismorphism can possibly exist between the two fields by first showing that any isomorphism fixes Q and that sqrt 2 (in this specific case) cannot be sent to any rational number, ie. sqrt 2 is sent to a+b*sqrt 3 for some nonzero rational b. This proof easily generalizes to square free m and n.

Thanks a lot!