Obtain a normal extension of containing and and express this extension in the form
clearly Q(√3) contains 3+√3. if we then take Q(√3,√5) this contains 3+√3 and √15 = √3√5.
note that Q(√3) is a normal extension of Q, since all roots of x2 - 3 are contained in Q(√3) (it is a splitting field of that polynomial).
in much the same vein, Q(√3,√5) = (Q(√3)(√5) is a splitting field of x2 - 5 over Q(√3), so Q(√3,√5) is normal over Q(√3). we will see later than Q(√3,√5) is actually normal over Q.
but first, let's find an element u in Q(√3,√5) such that Q(√3,√5) = Q(u). i claim we can take u = √3+√5.
it is obvious that Q(√3+√5) is a subfield of Q(√3,√5). so to show that √3+√5 actually generates Q(√3,√5) it suffices to show that √3 and √5 are contained in Q(√3+√5).
if we compute (√3+√5)3 = 18√3 + 14√5, it becomes obvious that (1/4)((√3+√5)3 - 14(√3+√5)) = √3, which shows that √3 is in Q(√3+√5).
therefore √5 = (√3+√5) - √3 is likewise in Q(√3+√5), so Q(√3,√5) = Q(√3+√5).
as promised we will now show Q(√3+√5) is normal by exhibiting an irreducible polynomial in Q[x] that splits completely in Q(√3+√5):
f(x) = (x + √3+√5)(x - √3+√5)(x + √3-√5)(x - √3-√5)
= (x2 - (8+2√15))(x2 - (8-2√15))
= x4 - ((8+2√15)+(8-2√15))x2 + (8+2√15)(8-2√15)
= x4 - 16x2 + (64-60)
= x4 - 16x2 + 4
convince yourself that if f(x) = (x2+ax+b)(x2+cx+d) that we must have c = -a, and a(d-b) = 0, and that a = 0 leads to b2 - 16b +4 = 0, while b = d leads to a2 = 12, neither of which can occur. note as well that f(x) is a quadratic in x2, and there is no x2 in Q that is a root, since the discriminant 256 - 4(4) = 240 is not a perfect square. thus this has no rational roots x (since x2 is surely rational if x is), and so conclude that f(x) is irreducible over Q.
but this means Q(u) is a splitting field of an irreducible polynomial in Q[x], and is thus a normal extension. in fact, since f(x) is separable, it is a galois extension.