1. ## 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?

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

Originally Posted by AlexP
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 $\mathbb{Q}\cup \{\sqrt{2}\}$ .

3. ## 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.

4. ## 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?

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

Originally Posted by FernandoRevilla
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!

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 $F$ 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 $F = \mathbb{Q}(\sqrt{2}).$

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

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