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

  • Thread starter Gtay
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In summary, the question is asking why Q[\sqrt{2}] is not isomorphic to Q[\sqrt{3}]. The attempt at a solution suggests that if there were an isomorphism between the two, it would imply that \sqrt{2} could only be mapped to \sqrt{3}, which would violate the requirement for a homomorphism. Therefore, there is no characteristic that Q[\sqrt{2}] has but Q[\sqrt{3}] does not.
  • #1
Gtay
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Homework Statement



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

Homework Equations


The Attempt at a Solution



I do not know where to start?
 
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  • #2
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]
 
  • #3
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?)
 

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

Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic because they have different algebraic properties. Specifically, Q(sqrt(2)) has elements of order 2, while Q(sqrt(3)) does not. This means that there is no isomorphism between the two fields that preserves the order of elements, making them not isomorphic.

2. Can Q(sqrt(2)) and Q(sqrt(3)) be isomorphic in any other way?

No, Q(sqrt(2)) and Q(sqrt(3)) cannot be isomorphic in any other way. This is because isomorphism is a bijective mapping that must preserve the algebraic structure of the fields, including the order of elements and the operations defined on them. As mentioned before, Q(sqrt(2)) and Q(sqrt(3)) have different algebraic properties, making it impossible for them to be isomorphic.

3. Are there any other examples of fields that are not isomorphic?

Yes, there are many examples of fields that are not isomorphic. For instance, the fields Q and R (the set of rational and real numbers, respectively) are not isomorphic because Q is countable while R is uncountable. This means that there is no bijective mapping that can preserve the order and operations of both fields.

4. Can we prove that Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic?

Yes, we can prove that Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic using the fundamental theorem of field isomorphisms. This theorem states that two fields are isomorphic if and only if there exists a bijective mapping between them that preserves the algebraic structure. Since Q(sqrt(2)) and Q(sqrt(3)) have different algebraic properties, there can be no such mapping, proving that they are not isomorphic.

5. How does the non-isomorphism of Q(sqrt(2)) and Q(sqrt(3)) affect their mathematical properties?

The non-isomorphism of Q(sqrt(2)) and Q(sqrt(3)) means that they are fundamentally different mathematical objects with distinct algebraic properties. This affects how operations and computations are carried out in each field, and also has implications for other mathematical concepts that involve these fields, such as Galois theory and algebraic number theory.

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