Extension fields

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September 19th 2010, 05:15 AM
ulysses123

Extension fields

is √7 ∈ Q((5+√7)^(1/2)

Using:

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

This gives me that the dimension of Q((5+√7)^(1/2) as a vector space over Q(√7 ) is 2.
But i don't know if the irreducible polynomial of degree 4 over Q((5+√7)^(1/2), is still irreducible over Q(√7 ).Is it sufficient to state that a basis is [1,5+√7)^(1/2)] and
√7 cannot be written as a linear combination of this basis?
September 19th 2010, 03:35 PM
tonio

Quote:

Originally Posted by ulysses123

is √7 ∈ Q((5+√7)^(1/2)

Using:

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

This gives me that the dimension of Q((5+√7)^(1/2) as a vector space over Q(√7 ) is 2.

Stop here! This already tells you that $\sqrt{7}\notin\mathbb{Q}(\sqrt{5+\sqrt{7}})$ , otherwise you'd

get that $dim_{\mathbb{Q}(\sqrt{7})}\mathbb{Q}(\sqrt{5+\sqrt {7}})=1$ ...

Tonio

But i don't know if the irreducible polynomial of degree 4 over Q((5+√7)^(1/2), is still irreducible over Q(√7 ).Is it sufficient to state that a basis is [1,5+√7)^(1/2)] and
√7 cannot be written as a linear combination of this basis?

.
September 19th 2010, 07:11 PM
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
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.
September 19th 2010, 07:15 PM
tonio

Quote:

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 $\mathbb{Q}(\sqrt{2}}$ a vector space over $\mathbb{Q}(\sqrt{3})$ ??
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.

.