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.