The degree of the extension Q(sqrt2,sqrt6) over Q(sqrt3) is either 1 or 2. If it is 1, then sqrt2 + sqrt6 can be written as a linear combination of 1 and sqrt3 with rational coefficients. However, a calculation shows the square of sqrt2 + sqrt6 contains a sqrt3 term, which implies the degree is 2. Therefore, the basis of Q(sqrt2,sqrt6) over Q(sqrt3) has 2 elements.

1. find the degree and basis for Q(sqrt2,sqrt6) over Q(sqrt3)
Solution
letting a = sqrt(2) + sqrt(6) for short, calculation shows that
a^2 = 2 + 2 sqrt(12) + 6 = 8 + 4 sqrt(3).
This implies that a is a root of the polynomial p in Q(sqrt(3))[x] given by
p(x) = x^2 - (8 + 4sqrt(3))
This implies that the degree of the extension Q(a) over Q(sqrt(3)) is either 2 or 1 (2 in the case
that p is irreducible over Q(sqrt(3)), by the above theorem, and 1 in the case that it factors, as we
can then apply the theorem to one of the linear factors of p).
Suppose it is 1. Then the field Q(a) is equal to the field Q(sqrt(3)), and in particular, a is in
Q(sqrt(3)). Since {1, sqrt(3)} is a basis for Q(sqrt(3)) over Q, there must be rationals p and q
with a = p + q sqrt(3). But then a short calculation shows that a^2 = (p^2 + 3 q^2) + 2 pq sqrt(3).