# Degree of extension field over $\mathbb{Q}$

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!

• 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 $\alpha$. – basket Mar 13 '16 at 1:52
• 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. – basket Mar 13 '16 at 1:59

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 $K=Q(\sqrt d)$.