Thread: 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)).

Can you please help me? Thank you.

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

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 .

Trivially,

Now .
Thus, .
Similarly, .

Thus.

Putting this together we get .