# Determining Subfields

• Oct 12th 2010, 04:12 AM
demode
Determining Subfields
(a) Determine all subfields of $\displaystyle \mathbb{Q}(\sqrt{7})$. (b) Prove that $\displaystyle \mathbb{Q}(\sqrt{5}) \ncong \mathbb{Q}(\sqrt{-3})$.

Attempt:
For the first part I think $\displaystyle \mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}$. So, I know about the "Subfield Test", but how would I find the possible subfields to be tested? Or is there a simpler way of dealing with this problem?

For the second part to show they're not isomorphic do I just need to show there is not a one to one correspondance? I think $\displaystyle \mathbb{Q} (\sqrt{-3})$ has complex numbers whereas the other one does not...
• Oct 12th 2010, 05:13 AM
tonio
Quote:

Originally Posted by demode
(a) Determine all subfields of $\displaystyle \mathbb{Q}(\sqrt{7})$. (b) Prove that $\displaystyle \mathbb{Q}(\sqrt{5}) \ncong \mathbb{Q}(\sqrt{-3})$.

Attempt:
For the first part I think $\displaystyle \mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}$. So, I know about the "Subfield Test", but how would I find the possible subfields to be tested? Or is there a simpler way of dealing with this problem?

For the second part to show they're not isomorphic do I just need to show there is not a one to one correspondance? I think $\displaystyle \mathbb{Q} (\sqrt{-3})$ has complex numbers whereas the other one does not...

As linear spaces, $\displaystyle \dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2$ and since any subfield would be a linear subspace of

the last one, there are no many subfields left out there, right?

About the non-isomorphism: you must know by now that two quadratic extensions of a field are isomorphic

iff they're generated by elements hich are roots of the same irreducible quadratic pol. over the field, so...

Tonio
• Oct 12th 2010, 02:43 PM
demode
Quote:

Originally Posted by tonio
As linear spaces, $\displaystyle \dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2$ and since any subfield would be a linear subspace of

the last one, there are no many subfields left out there, right?

So since the dimension is 2, there are 2 elements in its basis. But the question doesn't ask how many there are, it's asking to identify them! Do I somehow need to find two elements in $\displaystyle \mathbb{Q}$ such that $\displaystyle \mathbb{Q}(\sqrt{7})$ is spanned by them?

Quote:

About the non-isomorphism: you must know by now that two quadratic extensions of a field are isomorphic

iff they're generated by elements hich are roots of the same irreducible quadratic pol. over the field, so...

Tonio
So I must show that generators of $\displaystyle \mathbb{Q}(\sqrt{-3})$ and $\displaystyle \mathbb{Q}(\sqrt{5})$ are not the roots of the same polynomial. How do I find them? I'm very confused...

Here's an alternative approach:

Suppose that $\displaystyle \phi: \mathbb{Q}(\sqrt{-3}) \to \mathbb{Q}(\sqrt{5})$ is an isomorphism. Since $\displaystyle \phi(1) = 1$ (is this right?), we have $\displaystyle \phi(-3)=-3$. Then

$\displaystyle -3=\phi(-3)=\phi(\sqrt{-3} \sqrt{-3})= (\phi(\sqrt{-3}))^2$

This is impossible, since $\displaystyle \phi(\sqrt{-3})$ is a real number.

Is this correct?
• Oct 12th 2010, 03:31 PM
tonio
Quote:

Originally Posted by demode
So since the dimension is 2, there are 2 elements in its basis. But the question doesn't ask how many there are, it's asking to identify them! Do I somehow need to find two elements in $\displaystyle \mathbb{Q}$ such that $\displaystyle \mathbb{Q}(\sqrt{7})$ is spanned by them?

So I must show that generators of $\displaystyle \mathbb{Q}(\sqrt{-3})$ and $\displaystyle \mathbb{Q}(\sqrt{5})$ are not the roots of the same polynomial. How do I find them? I'm very confused...

Here's an alternative approach:

Suppose that $\displaystyle \phi: \mathbb{Q}(\sqrt{-3}) \to \mathbb{Q}(\sqrt{5})$ is an isomorphism. Since $\displaystyle \phi(1) = 1$ (is this right?), we have $\displaystyle \phi(-3)=-3$. Then

$\displaystyle -3=\phi(-3)=\phi(\sqrt{-3} \sqrt{-3})= (\phi(\sqrt{-3}))^2$

This is impossible, since $\displaystyle \phi(\sqrt{-3})$ is a real number.

Is this correct?

Yes.

Tonio
• Oct 12th 2010, 09:22 PM
demode
Quote:

Originally Posted by tonio
Yes.

Tonio

Do you mean my answer to part (b) is correct? Then what about part (a)? I'm really stuck on part this part...
• Oct 13th 2010, 04:44 AM
tonio
Quote:

Originally Posted by demode
Do you mean my answer to part (b) is correct? Then what about part (a)? I'm really stuck on part this part...

The ONLY part you did is (b), so what did I mean by "yes"?

Anyway, what kind of subspaces has a vector space of dimension 2? Then...?

You must have your linear algebra sharp as a butcher's knife BEFORE you want to mess with fields, extensions

and stuff, let alone Galois Theory.

Tonio
• Oct 14th 2010, 12:20 AM
demode
Quote:

Originally Posted by tonio
The ONLY part you did is (b), so what did I mean by "yes"?

Anyway, what kind of subspaces has a vector space of dimension 2? Then...?

You must have your linear algebra sharp as a butcher's knife BEFORE you want to mess with fields, extensions

and stuff, let alone Galois Theory.

Tonio

for any subspace v, the dimension is defined to be the number of elements in a basis for v. Subspaces with dimension 2 have 2 elements in their basis (for example a plane through the origin has dim=2).

So, are the two subfields $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Q}(\sqrt{7})$?
• Oct 14th 2010, 05:34 AM
tonio
Quote:

Originally Posted by demode
for any subspace v, the dimension is defined to be the number of elements in a basis for v. Subspaces with dimension 2 have 2 elements in their basis (for example a plane through the origin has dim=2).

So, are the two subfields $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Q}(\sqrt{7})$?

Yes...(Rock)

Tonio