I think I have the later parts of this proof worked out pretty well but what's really stumping me is how to go from knowing $[K:\mathbb{Q}]=2$ to knowing that $K = \mathbb{Q}[x]/a_2x^2 + a_1x + a_0$.

I mean all I know from $[K:\mathbb{Q}]=2$ is that every element of $K$ can be written in the form $bk_1 + ck_2$ for $b,c\in \mathbb{Q}$. As far as I can tell I don't yet have any theorems at my disposal that say if $[K:\mathbb{Q}]$ is finite than $K$ must be algebraic over $\mathbb{Q}$, or anything like that. How do I go from this premise about $K$ as a 2-dimensional vector space over $\mathbb{Q}$ to knowing something about elements of $K$ as roots of polynomials in $\mathbb{Q}[x]$? Thanks.

  • $\begingroup$ You know that a basis for a subspace ($\mathbb Q$ in this case) can be extended to a basis for $K$. A suitable basis for $\mathbb Q$ is $\{1\}$. Say you extend it by $\alpha$. Can you show that $\alpha$ must satisfy a quadratic equation? Can you massage $\alpha$ into a $\beta$ of the form you want? $\endgroup$ – anthonyquas Jun 29 '12 at 22:11
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    $\begingroup$ If $K$ is $n$-dimensional, you get a relation between $1,\alpha,\ldots,\alpha^n$ by vector space stuff. $\endgroup$ – anthonyquas Jun 29 '12 at 22:12
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    $\begingroup$ Use the quadratic formula, dude. $\endgroup$ – Zarrax Jun 29 '12 at 22:15
  • $\begingroup$ actually, this is true for any field $F$ s.t $char(F)\neq2$ (hint: consider what Zarrax wrote) $\endgroup$ – Belgi Jun 29 '12 at 22:20
  • $\begingroup$ I'm vexed by Zarrax's and Dylan's comments, but going off of anthonyquas if I can show $\alpha^2$ is an element of the rationals than $\alpha$ is a solution to the equation $x^2 - \alpha^2$ which is in $\mathbb{Q}[x]$. Is this the right direction? $\endgroup$ – Thoth Jun 29 '12 at 22:34

Let $\alpha \in K - \mathbb{Q}$. Since $1 , α, α^2$ are linearly dependent over $\mathbb{Q}$, $aα^2 + bα + c = 0$, where $a, b, c \in \mathbb{Q}$ and not all of $a, b, c$ are zero. By multiplying a suitable nonzero integer, we can assume $a, b, c \in \mathbb{Z}$. If $a = 0$, we get $bα + c = 0$. Since $b$ or $c$ is not zero, this can't happen. Hence $a \neq 0$. Hence $\mathbb{Q}(\alpha) = \mathbb{Q}((b^2 - 4ac)^{1/2})$. Since 1$ , α$ are linearly independent over $\mathbb{Q}$, [$\mathbb{Q}(\alpha) : \mathbb{Q}$] = 2. Hence $K = \mathbb{Q}(\alpha)$ and we are done.

  • $\begingroup$ Hello,can you explain please way $a \neq 0$ follows that $\mathbb{Q}(\alpha) = \mathbb{Q}((b^2 - 4ac)^{1/2})$ and why it means that [$\mathbb{Q}(\alpha) : \mathbb{Q}$] = 2 ? thank you. $\endgroup$ – Bar B Jun 6 '17 at 18:48

This is easy enough once you know the primitive element theorem. Since any finite extension of $\Bbb{Q}$ is separable it follows that $K = \Bbb{Q}(\alpha)$ for some $\alpha\notin \Bbb{Q}$ that has degree 2 over $\Bbb{Q}$. Suppose without loss of generality that the minimal polynomial of $\alpha$ over $\Bbb{Q}$ is

$$x^2 + bx + c.$$

Then by the quadratic formula, $\alpha = \frac{-b\pm \sqrt{b^2 - 4c}}{2}.$ Now $b^2 - 4c$ cannot be a square in $\Bbb{Q}$ for then this would contradict $\alpha$ not being a rational number. Furthermore because $b$ is a rational number we have

$$\Bbb{Q}(\alpha)= \Bbb{Q}(\sqrt{b^2 -4c})$$

so finally at the end we can say that $K = \Bbb{Q}(\sqrt{e})$ where $e = b^2 - 4c$ that is not a square in $\Bbb{Q}$. Now write $e = \alpha/\beta$ where $\alpha,\beta$ are integers with $\alpha,\beta \neq 0$. We may assume that $e$ is in its lowest terms otherwise we can cancel factors off top and bottom. Then

$$\sqrt{e} = \sqrt{\alpha}/\sqrt{\beta} = \sqrt{\alpha\beta}/\beta$$

and consequently $K = \Bbb{Q}(\sqrt{e}) = \Bbb{Q}(\sqrt{\alpha\beta})$ because $\beta$ is an integer. Now if we write $d = \alpha\beta$, we have shown that $K = \Bbb{Q}(\sqrt{d})$, where $d$ is not the square of any integer.

$$\hspace{5in} \square$$

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    $\begingroup$ The full primitive element theorem is somewhat overkill here, since you are assuming that $[K:\mathbf{Q}]=2$. Any element $\alpha$ of $K\setminus\mathbf{Q}$ will be such that $\mathbf{Q}(\alpha)=K$. $\endgroup$ – Keenan Kidwell Jun 29 '12 at 23:08
  • $\begingroup$ @KeenanKidwell :D $\endgroup$ – user38268 Jun 30 '12 at 0:09
  • $\begingroup$ You've got \mathbb Q(\sqrt{d}) for $d$ not a square in $\mathbb Q$, but you haven't quite shown it's a non-square integer... (upvoting anyway because it's really not a serious problem) $\endgroup$ – Ben Millwood Jun 30 '12 at 1:23
  • $\begingroup$ @BenMillwood Write $d = a/b$. Then $\sqrt{d} = \sqrt{a}/\sqrt{b} = \sqrt{ab}/b$. And then $\Bbb{Q}(\sqrt{d}) = \Bbb{Q}(\sqrt{ab})$. Is this better? $\endgroup$ – user38268 Jun 30 '12 at 1:44
  • $\begingroup$ Hmm. I'm not sure you've convinced me that your method of getting an integer preserves square-free-ness (it clearly doesn't if you don't choose $a$, $b$ coprime, for example) but possibly you can just say "if it turns out that $ab$ is not square-free, just take out any square factors". But this is really just pedantry and might be obscuring the real point of your argument, so I'm not picky :) $\endgroup$ – Ben Millwood Jun 30 '12 at 1:51

First, we know that if $K/\Bbb Q$ is quadratic. Then $K/\Bbb Q$ simple extension. Let $K=\Bbb Q(\alpha).$ Let $f$ be minimal polynomial of $\alpha$. Then $K=\Bbb Q(D(f)).$


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