I am currently studying elementary Algebraic Number Theory and came across the following statement:

Any Number Field $K$ such that $[K:\mathbb{Q}] = 2$ is equal to $\mathbb{Q}(\sqrt{d})$ for a unique square-free integer $d\neq1$.

Unfortunately I can't quite understand the proof for this given in the book and was wondering if anyone has a simple proof for it?

  • $\begingroup$ Pick an element of $K$ which is not rational. Show that it's the root of a quadratic polynomial. Then complete the square. $\endgroup$ – Qiaochu Yuan Apr 14 '15 at 6:00
  • $\begingroup$ Can you include what you did not understand in their proof $\endgroup$ – Elaqqad Apr 14 '15 at 6:10
  • $\begingroup$ So, this is what I have done so far, Since the number field has degree $2$, then the minimal polynomial of any element $\alpha$ in K which isn't rational has degree $1$ or $2$. If it is 1 then obviously $\alpha$ will be in $\mathbb{Q}$, so therefore $\alpha$ will be the root of a quadratic polynomial but after completing the square I can't quite see how that would imply the above statement. $\endgroup$ – user1314 Apr 14 '15 at 6:14

Let $ \mathbb{Q} \subset K$ be an extension of fields. Pick an element $u \in K- \mathbb{Q}$ such that $u$ is algebraic over $\mathbb{Q}$ and since $[K:Q]=2$ we know that $\mathbb{Q}(u)=K$. Here I am recaling the fact that if $[K:F]=p$ where $p$ is prime then there exists no proper subfield $L$ with $F \subset L \subset K$ where $L$ is not equal to $F$ or $K$. Continuing, since $\mathbb{Q}(u) = K$ we know that the degree of $u$ is $2$ over $\mathbb{Q}$; hence it satisfies a unique monic irreducible polynomial of degree $2$. Denote this polynomial as $g(x) = x^2+bx+c$. Now using the quadratic formula we have that $g(u) = 0 \Rightarrow$,

$$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2} \Rightarrow \mathbb{Q}(u) = \mathbb{Q}\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2}\right) = \mathbb{Q}\left(\sqrt{b^2-4ac}\right)$$

If $b^2-4ac$ is square free then $d:=b^2-4ac$. Otherwise we can write $\sqrt{b^2-4ac} = c\sqrt{k}$ where $k$ is now square free. For example, $\sqrt{8} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$. With this second case that I showed the example for we have,

$$\mathbb{Q}(c\sqrt{k}) = \mathbb{Q}(\sqrt{k})$$

Hence, $k$ is square free and just let $k=d$. I will also make that remark that we are able to reduce the field when we have elements that we know are in the corresponding field, we say that the field absorbs these elements. For example, $\mathbb{Q}(c) = \mathbb{Q}$ if $c \in \mathbb{Q}$. An extension is suppose to give you something new, so adjoining elements already in the fields does nothing basically. $[K:F]=p$ where $p$ is prime. Let $u \in K-F$ be algebraic over $F$. Then we must have that $F(u) = K \Rightarrow \text{deg}(m_{u,F}(x))=p$.

$\textbf{Answer to question in comment}$

Take $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ which has degree $3$. Pick an element $u \in \mathbb{Q}(\sqrt[3]{2})$. If $\text{deg}(m_{u,\mathbb{Q}(\sqrt[3]{2})}(x))<3$ namely $2$ then we would get $\mathbb{Q} \subset \mathbb{Q}(u) \subset \mathbb{Q}(\sqrt[3]{2})$. But applying the tower theorem we must have $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}] \Rightarrow 2m=3$, where $m =[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(u)]$.

  • $\begingroup$ That is extremely helpful! Just a small comment regarding the second line. When we say $\mathbb{Q}(u) = K$, can we only say that when the extension is of prime degree? I feel like I have read that statement somewhere before: $L/K$ is a simple extension if its degree is prime? $\endgroup$ – user1314 Apr 14 '15 at 6:26
  • $\begingroup$ Well if this is the answer you are looking for check it so others will know it is complete. I made the remark that if $K/F$ is an extension which is prime then if you take any algebraic extension, if must be $K$. $\endgroup$ – Mr.Fry Apr 14 '15 at 6:33
  • $\begingroup$ Ok, so going by that premise,suppose we have $K/F$ which is an extension of prime degree $p$, and suppose we pick an element $\alpha$ in $K - F$, then we know that the minimal polynomial of $\alpha$ will have degree {$2...p-1$}. Then as we said $F(\alpha) = K$ but wouldn't this mean $K$ will have different degrees because the minimal polynomial of $\alpha$ can have any degree in that range? $\endgroup$ – user1314 Apr 14 '15 at 6:44
  • $\begingroup$ $[K:F]=p$ where $p$ is prime. Let $u \in K-F$ be algebraic over $F$. Then $F(u) = K \Rightarrow \text{deg}(m_{u,F}(x))=p$. $\endgroup$ – Mr.Fry Apr 14 '15 at 6:48
  • $\begingroup$ oh that is quite interesting, didn't realise that till now even though I have been working with field extensions! Thanks a lot $\endgroup$ – user1314 Apr 14 '15 at 6:53

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