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Math Help - Subfields of Q(sqrt(2))

  1. #1
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    Subfields of Q(sqrt(2))

    I am correct in thinking that the only subfields of \mathbb{Q}(\sqrt{2}) are \mathbb{Q} and \mathbb{Q}(\sqrt{2}) itself? How do we prove that these are the only ones?

    Is it fair to say that there are no infinite fields 'smaller' than \mathbb{Q} since \mathbb{Q} is the field of fractions for \mathbb{Z} so that no 'smaller' field can contain \mathbb{Z}, which we know it must?
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  2. #2
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    Re: Subfields of Q(sqrt(2))

    Quote Originally Posted by AlexP View Post
    I am correct in thinking that the only subfields of \mathbb{Q}(\sqrt{2}) are \mathbb{Q} and \mathbb{Q}(\sqrt{2}) itself? How do we prove that these are the only ones?
    Hint: By definition, \mathbb{Q}(\sqrt{2}) is the smallest field which contains \mathbb{Q}\cup \{\sqrt{2}\} .
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  3. #3
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    Re: Subfields of Q(sqrt(2))

    another way is to apply the galois's correspondence theorem. first note that every subfield of  \mathbb{Q}(\sqrt{2}) contains \mathbb{Q}. now let G be the galois group of \mathbb{Q}(\sqrt{2}) over \mathbb{Q}. then |G|= [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2 and so G has only two subgroups. thus the number of subfields of \mathbb{Q}(\sqrt{2}) is also two.
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  4. #4
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    Re: Subfields of Q(sqrt(2))

    Every member of Q(\sqrt{2}) is of the form a+ b\sqrt{2} where a and b are rational numbers. Taking b= 0 gives the subfield Q. What if you take a= 0?
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    Re: Subfields of Q(sqrt(2))

    Quote Originally Posted by FernandoRevilla View Post
    Hint: By definition, \mathbb{Q}(\sqrt{2}) is the smallest field which contains \mathbb{Q}\cup \{\sqrt{2}\} .
    So combining the reasoning I included in my post and your hint, do we know that \mathbb{Q} contains no subfields, and \mathbb{Q}(\sqrt{2}) contains no subfields but \mathbb{Q} because they could not include \sqrt{2} and thus must be the rationals (but in this case, how do we know this)?

    NonCommAlg: how do we know that every subfield must contain \mathbb{Q}? Your method is beyond what I know, but it does make me look forward to Galois theory!
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  6. #6
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    Re: Subfields of Q(sqrt(2))

    ok, well, if you don't know galois theory, then you have to solve the problem directly. let F be a subfield of \mathbb{Q}(\sqrt{2}). then 1 \in F and thus \mathbb{Z} \subset F. since F is a subfield, it contains the inverse of all non-zero elements and so \mathbb{Q} \subseteq F.
    now suppose that \mathbb{Q} \neq F. then there exists x = a+b\sqrt{2} \in F \setminus \mathbb{Q}, \ a,b \in \mathbb{Q}. thus b \neq 0. let y=a-b \sqrt{2}. then ...

    try to finish the proof yourself and if you couldn't, see the rest:

    Spoiler:
    xy=a^2-2b^2 \in \mathbb{Q} \subset F and so y=\frac{a^2-2b^2}{x} \in F.


    can you finish it now? if not, see below

    Spoiler:
    hence 2b \sqrt{2}=x-y \in F and therefore \sqrt{2} \in F because b \neq 0. thus F = \mathbb{Q}(\sqrt{2}).
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  7. #7
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    Re: Subfields of Q(sqrt(2))

    ok, there's an easier way. let F be a subfield of \mathbb{Q}(\sqrt{2}). i showed in my previous post that \mathbb{Q} \subseteq F. now

    2=[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}):F][F:\mathbb{Q}] \ \ \ \ \ \ \ (*).

    so if F \neq \mathbb{Q}, then [F:\mathbb{Q}] > 1 and hence, by (*), [F:\mathbb{Q}]=2 and [\mathbb{Q}(\sqrt{2}):F]=1. thus F=\mathbb{Q}(\sqrt{2}).
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  8. #8
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    Re: Subfields of Q(sqrt(2))

    ok, I feel good about this now. Thank you.
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