1. ## field extension question

So i need to prove that the extension field:
Q(sqrt(2)+sqrt(3)) = the set {a + b.sqrt(2) + c.sqrt(3) + d.sqrt(6) ; with a,b,c,d are rational numbers}

Since Q(sqrt(2)+sqrt(3)) is the SMALLEST field that contains all rationals and sqrt(2)+sqrt(3) hence Q belongs to the set

However, I dont know how to prove the other way around, i.e. the set belongs to Q(sqrt(2)+sqrt(3)).

2. I wonder if you can use something like synthetic division -
a + b rt 2 + c rt 3 + d rt 6 =
(m + n rt 2)(p + q rt 3) + r + s rt 2 + t rt 3

3. Originally Posted by pc31
So i need to prove that the extension field:
Q(sqrt(2)+sqrt(3)) = the set {a + b.sqrt(2) + c.sqrt(3) + d.sqrt(6) ; with a,b,c,d are rational numbers}
I am not sure what you are trying to prove.
I assume it is $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$.

Trivially, $\mathbb{Q}(\sqrt{2}+\sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2},\sqrt{3})$

Now $\frac{1}{\sqrt{2}+\sqrt{3}} = \sqrt{2} - \sqrt{3} \in \mathbb{Q}(\sqrt{2}+\sqrt{3})$.
Thus, $\tfrac{1}{2}[(\sqrt{2}-\sqrt{3}) + (\sqrt{2}+\sqrt{3}) ]= \sqrt{2} \in \mathbb{Q}$.
Similarly, $\sqrt{3}\in \mathbb{Q}$.

Thus. $\mathbb{Q}(\sqrt{2},\sqrt{3})\subseteq \mathbb{Q}(\sqrt{2}+\sqrt{3})$

Putting this together we get $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$.