# Basis for an extension field

• Prometheos
In summary, the conversation is about finding the basis of a field adjoined with an element, specifically Q(sqrt(2)+sqrt(3)). The degree of this element is 4, so the basis should have 4 dimensions. The actual basis is {1, sqrt(2), sqrt(3), sqrt(6)}, but it is not clear how sqrt(6) can be obtained as a Q-linear combination of sqrt(2) and sqrt(3). The key to understanding this concept is to show that sqrt(6) is independent from sqrt(2) and sqrt(3) over Q, which will result in a basis of {1, sqrt(2), sqrt(3), sqrt(6)}.

#### Prometheos

I am having no luck understanding how to find the basis of a field adjoined with an element.

For example
Q(sqrt(2)+sqrt(3))
I know that if i take a=sqrt(2)+sqrt(3) that i can find a polynomial (1/4)x^4 - (5/2)x^2 + 1/4 that when evaluated at a is equal to zero.

So, from that I know the degree is 4 and the basis should have 4 dimensions.

{1, sqrt(2), sqrt(3), Y} where Y is the part I don't understand.
{1, sqrt(2), sqrt(3), sqrt(6)} is the actual basis, but how do you get sqrt(6) as Y?

I don't see how it is a linear combination of sqrt(2) and sqrt(3) as defined in the book I'm using. Since it should be a additive combination of scalars from Q times sqrt(2) and sqrt(3)... but the only way to obtain sqrt(6) is sqrt(2)*sqrt(3), in which neither are scalars as elements of Q.

Any help understanding this concept is greatly appreciated.

Remember, you should not be able to get sqrt(6) as a Q-linear combination of sqrt(2) and sqrt(3) for otherwise the set {1,sqrt(2), sqrt(3), sqrt(6)} would be dependent and therefore not a basis. You know that sqrt(6) is in your field however (being the product of sqrt(2) and sqrt(3)) so if you can show it is independent from sqrt(2) and sqrt(3) over Q you will have your basis (since you already know the dimension is 4).

So by putting sqrt(2) and sqrt(3) adjoined to the field, then it has to have multiplicative closure so sqrt(6) has to be there... so simple now that I think about it.

Thanks!