# Field Extension of the Rationals

So I'm considering a Field $\mathbb{F}$, such that $\mathbb{Q}$ is a subset of $\mathbb{F}$ and when it's considered a vector space over $\mathbb{Q}$, it has dimension 2. I want to show two things:

1) There exists an element a of $\mathbb{F}$ that's not in $\mathbb{Q}$ such that it satisfies the equation $a^2-n=0$ for some $n\in\mathbb{Z}$.

2) That $\mathbb{F}$ is isomorphic to $\mathbb{Q}[\surd(n)]$ and further that $n$ is square-free.

So, showing there's an element in the complement of $\mathbb{F}$ and $\mathbb{Q}$ I've managed, but I can't show that it satisfies the equation in question. For number 2, I'm lost.

(Trying LaTeX, hope I didn't screw it up too bad.)

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There are many elements in $\mathbb F \setminus \mathbb Q$. Not all of them satisfy $a^2-n=0$. The point is that given some $b \in \mathbb F \setminus \mathbb Q$ you can find a related $a$ that works. – Erick Wong Sep 21 '12 at 2:29

A start: Let $\alpha\in \mathbb{F}$, with $\alpha\not\in \mathbb{Q}$. Argue that by dimensionality considerations, $\alpha^2$, $\alpha$, and $1$ are linearly dependent over the rationals.
This shows that $\alpha$ is the root of a quadratic with rational (or equivalently integer) coefficients.
Complete the square, or use the Quadratic Formula, to produce the requisite $n$ (think discriminant).
The discriminant is definitely not zero. The quadratic will have two solutions. I will assume you got to there are integers $a$, $b$, $c$ with $a\ne 0$ such that $a\alpha^2+b\alpha+c=0$. Let $b^2-4ac=k^2n$ where $n$ is squae-free (this is for later, not needed now). Argue that a linear combination of $\alpha$ and $1$ has square equal to $n$. – André Nicolas Sep 21 '12 at 2:58
@AsinglePANCAKE: To help thinking, pretend the quadratic turned out to be $2x^2-4x-9$. Then $n$ turns out to be $22$. – André Nicolas Sep 21 '12 at 3:01
No, the discriminant is (say) $b^2-4ac$ as usual. Let $k^2$ be largest square that divides the discriminant, and define $n$ by $b^2-4ac=k^2n$. Or else, but you will have to change your mind in a later part of the question, just let $n=b^2-4ac$. In the numerical example I suggested, suppose the linear relation turned out to be $2\alpha^2-4\alpha-9=0$. Find $\alpha$ (one of the roots). Discriminant is $88$. Then it turns out that $\mathbb{F}$ is isomorphic to $\mathbb{Q}(\alpha)$, which is $\mathbb{Q}(\sqrt{22})$. – André Nicolas Sep 21 '12 at 3:29