Determining Subfields

**demode** · October 12th 2010, 04:12 AM

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

Attempt:
For the first part I think $\mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}$ . 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 $\mathbb{Q} (\sqrt{-3})$ has complex numbers whereas the other one does not...

**tonio** · October 12th 2010, 05:13 AM

Originally Posted by demode

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

Attempt:
For the first part I think $\mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}$ . 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 $\mathbb{Q} (\sqrt{-3})$ has complex numbers whereas the other one does not...

As linear spaces, $\dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2$ 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

**demode** · October 12th 2010, 02:43 PM

Originally Posted by tonio

As linear spaces, $\dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2$ 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 $\mathbb{Q}$ such that $\mathbb{Q}(\sqrt{7})$ is spanned by them?

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

Here's an alternative approach:

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

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

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

Is this correct?

**tonio** · October 12th 2010, 03:31 PM

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 $\mathbb{Q}$ such that $\mathbb{Q}(\sqrt{7})$ is spanned by them?

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

Here's an alternative approach:

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

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

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

Is this correct?

Yes.

Tonio

**demode** · October 12th 2010, 09:22 PM

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...

**tonio** · October 13th 2010, 04:44 AM

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

**demode** · October 14th 2010, 12:20 AM

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 $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{7})$ ?

**tonio** · October 14th 2010, 05:34 AM

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 $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{7})$ ?

Yes...

Tonio

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