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.
Answer
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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