1
$\begingroup$

Suppose $|K:\mathbb{Q}|= 2$.Then how can we prove the existence of $a\in \mathbb{Q}$ such that $K = \mathbb{Q}[\sqrt{a}]?$ Is there some kind of uniqueness of $a$? Here is my proof: clearly $\sqrt{a}\in K$, Then there exist a minimal polynomial $x^2-a$ in $K$ as the degree of extension is 2. So $a$ must be in $\mathbb{Q}$ and this $a$ is unique. Am I correct? Thanks!

$\endgroup$
  • $\begingroup$ The extension is algebraic since it is finite. Let $\alpha$ be an element of $K$ of not in $\mathbb{Q}$. It's minimal polynomial is degree 2 (why?) and this shows it is the square root of some rational (why?). Lastly, by considering the degree of the extension, $K$ must be a simple extension generated by $\alpha$. $\endgroup$ – basket Mar 13 '16 at 1:52
  • 1
    $\begingroup$ Your proof is circular. It begins with "Clearly $\sqrt{a} \in K$..." but what is $a$?. The conclusion about uniqueness is also unjustified and false. $\endgroup$ – basket Mar 13 '16 at 1:59
1
$\begingroup$

Let $x\in K$ and $x$ is not in $Q$, $1,x$ generated $K$, so $x^2=ax+b$ where $a,b\in Q$. Thus $(x-a/2)^2 =a^2/4+b.$ Write $a^2/4+b=d$. You have $x-a/2=\sqrt d$ or $x-a/2=-\sqrt d$. This implies $K=Q(\sqrt d)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.