I've just started some reading and doing exercises on field theory with Galois theory in scope, and have had some trouble with this exercise. I think I have simply misunderstood some of the definitions, and would like someone to set this straigth to me.

If $K$ is an extension field of $\mathbb{Q}$ such that $[K:\mathbb{Q}] = 2$, prove that $K = \mathbb{Q}(\sqrt{d})$ for some square-free integer $d$.

Now, I understand that since the extension is finite-dimensional, so it has to be algebraic. So in particular if I take any element $u \in K$ not in $\mathbb{Q}$ then it must be algebraic. Since the basis of $K$ over $\mathbb{Q}$ is of size 2, the set $\{1, u, u^2\}$ must be linearly dependant and with it I could construct a polynomial of degree two with $u$ as a root.

If the polynomial is $f(x) = x^2 + ax + b$, then I know $u = -a/2 + \sqrt{a^2/4 -b}$, where $t = \sqrt{a^2/4 -b}$ cannot be a square or else $u \in \mathbb{Q}$. I can see why $\mathbb{Q}(u) = \mathbb{Q}(t)$.

In this way I get the chain of fields $\mathbb{Q} \subset \mathbb{Q}(t) \subset K$, but because $[K:\mathbb{Q}]= 2$ and certainly $\mathbb{Q} \neq \mathbb{Q}(t)$ then $\mathbb{Q}(t) = K$. Now, my problem lies in proving that the field $\mathbb{Q}(t)$ actually can be represented by $\mathbb{Q}(\sqrt d)$ where $d$ is square-free.

What bothers me is the following. The polynomial $f(x) = x^2 - 2/3$ has a root in $\sqrt{2/3}$ and is certainly irreducible in $\mathbb{Q}$. But then the field $\mathbb{Q}(\sqrt{2/3})$ has dimension 2. How is this field equal to some field $\mathbb{Q}(\sqrt d)$ where $d$ is square-free?

EDIT: Thanks for the help in the comments. Obviously if $n/m$ is a rational number in reduced form then $nm$ is square-free and $\mathbb{Q}(\sqrt{n/m}) = \mathbb{Q}(\sqrt{nm})$. Feel free to close the question.

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    $\begingroup$ The square root of $2/3$ and that of $2 \cdot 3$ generate the same field. $\endgroup$ – franz lemmermeyer Mar 25 '12 at 11:30
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    $\begingroup$ $\mathbb Q(\sqrt{2/3}) = \mathbb Q(3\sqrt{2/3}) = \mathbb Q(\sqrt{6})$ $\endgroup$ – marlu Mar 25 '12 at 11:33
  • $\begingroup$ @franz Oh, of course. Thank you very much. $\endgroup$ – Bartosz Malman Mar 25 '12 at 11:51
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    $\begingroup$ @Barre: On the contrary, now that you see the solution, you should feel free to write an answer to your own question - this is explicitly encouraged on StackExchange sites. $\endgroup$ – Zev Chonoles Mar 25 '12 at 14:30
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    $\begingroup$ A question gets closed only when it's a duplicate, off-topic, a duplicate, etc. Your question is a perfectly good one for the site. If you feel that it has been answered to your content, choose the best answer and accept it. It may even be your own answer. $\endgroup$ – lhf Mar 26 '12 at 11:55

From the discussion in the original post it follows that if $[K:\mathbb{Q}] = 2$ then $K = \mathbb{Q}(t)$ where $t$ is the square root of some rational number in reduced form, say $t = \sqrt{n/m}$. Then $\gcd(n,m) = 1$. Then the integer $nm$ is clearly square-free and what is left to show is that $\mathbb{Q}(\sqrt{n/m}) = \mathbb{Q}(\sqrt{nm})$.

Since $\mathbb{Q}(\sqrt{n/m}) = \{a+b\sqrt{n/m} \ | \ a,b \in \mathbb{Q}\}$, letting $a = 0, b = m$ we get that $\sqrt{mn} \in \mathbb{Q}(\sqrt{n/m})$ and since $\mathbb{Q}(\sqrt{nm})$ is the smallest field containing this element, we ge the inlusion $\mathbb{Q}(\sqrt{nm}) \subset \mathbb{Q}(\sqrt{n/m})$.

By a similar argument, $\mathbb{Q}(\sqrt{n/m}) \subset \mathbb{Q}(\sqrt{nm})$ since $\mathbb{Q}(\sqrt{nm}) = \{a+b\sqrt{nm} \ | \ a,b \in \mathbb{Q}\}$ and letting $a=0, b = 1/m$ we get that $\mathbb{Q}(\sqrt{n/m}) \subset \mathbb{Q}(\sqrt{nm})$.

So in conclusion $\mathbb{Q}(\sqrt{n/m}) = \mathbb{Q}(\sqrt{nm})$ and every field extension of the rationals of degree 2 is on form $\mathbb{Q}(\sqrt{d})$ where $d$ is square-free.


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