Let $\alpha = \sqrt{2} + \sqrt{3}$ and $K \subseteq \mathbb{R}$ such that $\mathbb{Q} \subset K \subset \mathbb{Q}[\alpha]$

Let $\alpha = \sqrt{2} + \sqrt{3}$ and $K \subseteq \mathbb{R}$ such that $\mathbb{Q} \subset K \subset \mathbb{Q}[\alpha]$.

I want to find the degree of the minimum polynomial of $\alpha$ on K.

I know that the degree of the minimum polynomial of $\alpha$ on $\mathbb{Q}$ is four, I think that means that $[\mathbb{Q} : \mathbb{Q}[\alpha]]$ = 4. To find the degree of the minimum polynomial, I think I need to use the following: $$[\mathbb{Q} : K] \times [K : \mathbb{Q}[\alpha] ] = [\mathbb{Q} : \mathbb{Q}[\alpha]] = 4$$

I think I need to find this term: $[K : \mathbb{Q}[\alpha] ]$, so I need to take a look at $[\mathbb{Q} : K]$. There are three possibilities: 1 2 and 4. However, I am struggling to find which one to esclude.

Any tips?

• Does $\subset$ mean $\subsetneq$ or $\subseteq$? Jun 20 '18 at 13:45
• It means proper subset. Jun 20 '18 at 13:46

Let $m=[K:\mathbb{Q}]$ and $n=[\mathbb{Q}[\alpha]:K]$. Then $mn=4$.
Since the inclusions in $\mathbb{Q} \subset K \subset \mathbb{Q}[\alpha]$ are strict, we have $m>1$ and $n>1$.
Therefore, $m=n=2$.