Vector Space/Fields Q(sqrt(2),sqrt(3))

[magicarpet512]

Homework Statement

Q(sqrt(2),sqrt(3)) is a field generated by elements of the form a + b*sqrt(2) + c*sqrt(3); where a,b,c are in Q. We are supposed to show that this is a vector space with dimension 4.

Homework Equations

Given above.

The Attempt at a Solution

So showing this is a vector space with dimension 4 will be easy for me to do, however I am not understanding why the problem above doesn't have a term d*sqrt(6)? For instance in a previous problem, I showed that
Q(sqrt(2),sqrt(3)) = {a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6) | a,b,c,d are in Q} is a field.
In Q(sqrt(2),sqrt(3)) it is easy to show that the minimum polynomial is
p(x) = x^4 - 10x^2 + 1, so the vector space would have dimension 4 and a basis would be
{1, sqrt(2), sqrt(3), sqrt(6)}; so I would think that a d*sqrt(6) term should be included in the original problem? Might it be the case that the term c*sqrt(3) is of the form (c + d*sqrt(2))*sqrt(3)?
So could someone help me understand the difference in saying:
1.Q(sqrt(2),sqrt(3)) is a field generated by elements of the form a + b*sqrt(2) + c*sqrt(3); where a,b,c are in Q.
and
2.Q(sqrt(2),sqrt(3)) = {a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6) | a,b,c,d are in Q} is a field.

 

[Sethric]

Because you don't actually need the sqrt(6) term to generate the field - since any sqrt(6) term can be created from a combination of sqrt(3) and sqrt(2) terms. That is the key difference in the the 2 ways you state the field. One uses a set of elements to generate a field, one gives the form of the field generated by such a set. Consequently, the form of the field generated is what makes it obvious that it is a dimension 4 vector space.

 

[Hurkyl]

To emphasize further, it would even be correct to say that \mathbb{Q}(\sqrt{2}, \sqrt{3}) is generated (as a field) by the two-element set \{ \sqrt{2}, \sqrt{3} \}. Of course, to generate it as a Q-vector space, you need four Q-linearly independent elements.