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

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

Is it fair to say that there are no infinite fields 'smaller' than $\displaystyle \mathbb{Q}$ since $\displaystyle \mathbb{Q}$ is the field of fractions for $\displaystyle \mathbb{Z}$ so that no 'smaller' field can contain $\displaystyle \mathbb{Z}$, which we know it must?

2. ## Re: Subfields of Q(sqrt(2))

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

3. ## Re: Subfields of Q(sqrt(2))

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

4. ## Re: Subfields of Q(sqrt(2))

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

5. ## Re: Subfields of Q(sqrt(2))

Originally Posted by FernandoRevilla
Hint: By definition, $\displaystyle \mathbb{Q}(\sqrt{2})$ is the smallest field which contains $\displaystyle \mathbb{Q}\cup \{\sqrt{2}\}$ .
So combining the reasoning I included in my post and your hint, do we know that $\displaystyle \mathbb{Q}$ contains no subfields, and $\displaystyle \mathbb{Q}(\sqrt{2})$ contains no subfields but $\displaystyle \mathbb{Q}$ because they could not include $\displaystyle \sqrt{2}$ 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 $\displaystyle \mathbb{Q}$? Your method is beyond what I know, but it does make me look forward to Galois theory!

6. ## Re: Subfields of Q(sqrt(2))

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

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

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

can you finish it now? if not, see below

Spoiler:
hence $\displaystyle 2b \sqrt{2}=x-y \in F$ and therefore $\displaystyle \sqrt{2} \in F$ because $\displaystyle b \neq 0$. thus $\displaystyle F = \mathbb{Q}(\sqrt{2}).$

7. ## Re: Subfields of Q(sqrt(2))

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

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

so if $\displaystyle F \neq \mathbb{Q},$ then $\displaystyle [F:\mathbb{Q}] > 1$ and hence, by $\displaystyle (*)$, $\displaystyle [F:\mathbb{Q}]=2$ and $\displaystyle [\mathbb{Q}(\sqrt{2}):F]=1.$ thus $\displaystyle F=\mathbb{Q}(\sqrt{2}).$

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# determine all subfields of q(square root of 2)

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