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

  • Thread starter geor
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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.
  • #1
geor
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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?

Thanks a lot in advance..
 
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  • #2
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?!
 
  • #3
Right. (Have you yet shown there isn't a field isomorphism?)
 
  • #4
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.
 
  • #5
Thanks a lot!
 

1. Why are Q(sqrt(2)) and Q(sqrt(3)) not isomorphic?

The two fields Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic because they have different algebraic properties. While both are extensions of the rational numbers, Q(sqrt(2)) is a quadratic extension while Q(sqrt(3)) is a cubic extension. This means that the elements in Q(sqrt(2)) can be expressed as a linear combination of 1 and sqrt(2), while the elements in Q(sqrt(3)) can be expressed as a linear combination of 1, sqrt(3), and sqrt(3^2).

2. Can Q(sqrt(2)) and Q(sqrt(3)) be isomorphic under a different mapping?

No, Q(sqrt(2)) and Q(sqrt(3)) cannot be isomorphic under any mapping. This is because isomorphisms preserve the algebraic properties of a field, and since Q(sqrt(2)) and Q(sqrt(3)) have different algebraic properties, they cannot be isomorphic.

3. Do Q(sqrt(2)) and Q(sqrt(3)) have the same cardinality?

Yes, Q(sqrt(2)) and Q(sqrt(3)) have the same cardinality as both fields have an infinite number of elements. However, this does not mean that they are isomorphic, as cardinality only refers to the number of elements in a set and does not take into account their algebraic properties.

4. Are there any similarities between Q(sqrt(2)) and Q(sqrt(3))?

Yes, there are some similarities between Q(sqrt(2)) and Q(sqrt(3)). Both fields are extensions of the rational numbers and contain infinitely many elements. They also both have a multiplicative identity of 1 and an additive identity of 0. However, as mentioned before, their algebraic properties are different, making them non-isomorphic.

5. What implications does the non-isomorphism of Q(sqrt(2)) and Q(sqrt(3)) have?

The non-isomorphism of Q(sqrt(2)) and Q(sqrt(3)) has implications in algebra and number theory. It means that these fields have distinct algebraic structures and cannot be used interchangeably. This has consequences in areas such as Galois theory, where isomorphisms play a crucial role in understanding the structure of fields and their extensions.

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