Originally Posted by

**ulysses123** That doesn't seem like its sufficient,

as a counter example take √3 and Q(√(2))

Using [E:G]=[E:F][F:G]

where E=Q(√(2))

F=Q(√(3))

G=Q

the dimension of [Q(√(2)):Q(√(3))]=1

Uh? How exactly is $\displaystyle \mathbb{Q}(\sqrt{2}}$ a vector space over $\displaystyle \mathbb{Q}(\sqrt{3})$??

Either one is a subfield of the

other, and the multiplication of vectors by scalars is obvious, or else you must define this product.

Tonio

But still √(3) is not actually contained in Q(√(2)

Since [Q(√(2))(√(3)):Q]=4

and each of root 2 and root 3 have dimension 2 over Q, this then says that there is no unique way to write root 3 in Q(√(2), otherwise if there was the dimension of [Q(√(2))(√(3)):Q]would be two.