Isomorphisms

• May 18th 2009, 07:24 AM
TheEmptySet
Isomorphisms
I need to prove that $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

So here is my thoughts on this:

Since $\displaystyle \mathbb{Q}$ is the prime subfield is must be fixed by any isomorphism from $\displaystyle \tau : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$

Since $\displaystyle \{1, \sqrt{2} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \{1, \sqrt{3} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{3})$

The isomorphism must be of the form

$\displaystyle \tau(a+b\sqrt{2})=a+b{\sqrt{3}}$ or $\displaystyle \tau(a+b\sqrt{2})=a-b{\sqrt{3}}$

Both of these only work for the addative group structure, but don't preserve multiplication. So they are not isomorphisms.

Therefore $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

I feel unsatisfied with the above proof. Another thought that I had was since \mathbb{Q}(\sqrt{2}) has only two automorphism showing that neither of those could be made into an isomorphism to $\displaystyle \mathbb{Q}(\sqrt{3})$ but that came up empty.

Any thoughts or comments would be awsome.

Thanks

TES
• May 18th 2009, 08:31 AM
Swlabr
This came up in a thread a couple of days ago, I believe. Take $\displaystyle \phi: \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ and look at where $\displaystyle \sqrt{2}$ maps to. Square it, and then in a few lines you can see that the thing you're mapping to must be outwith $\displaystyle \mathbb{Q}(\sqrt{3})$.

• May 18th 2009, 09:52 PM
Isomorphism
sah_mat has a solution to the problem:Try this
• May 18th 2009, 10:15 PM
TheEmptySet
Thanks I have something similar. I guess the reason I feel unsatified is the proof is out of place in my text. In Dummit and Foote this is in the beginning of chapter 14(the beginning of Galois theory), but I didn't use anything from the current chapter. I used only tools from the previous chapter 13. Of course this is not a reason to think a proof is incorrect, but it does make me suspect that I may have missed an applicaiton of a more recent idea.
• May 23rd 2009, 06:05 PM
ThePerfectHacker
Quote:

Originally Posted by TheEmptySet
I need to prove that $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

So here is my thoughts on this:

Since $\displaystyle \mathbb{Q}$ is the prime subfield is must be fixed by any isomorphism from $\displaystyle \tau : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$

Since $\displaystyle \{1, \sqrt{2} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \{1, \sqrt{3} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{3})$

The isomorphism must be of the form

$\displaystyle \tau(a+b\sqrt{2})=a+b{\sqrt{3}}$ or $\displaystyle \tau(a+b\sqrt{2})=a-b{\sqrt{3}}$

Both of these only work for the addative group structure, but don't preserve multiplication. So they are not isomorphisms.

Therefore $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

I feel unsatisfied with the above proof. Another thought that I had was since \mathbb{Q}(\sqrt{2}) has only two automorphism showing that neither of those could be made into an isomorphism to $\displaystyle \mathbb{Q}(\sqrt{3})$ but that came up empty.

Any thoughts or comments would be awsome.

Thanks

TES

It is not hard to show $\displaystyle \tau (r) = r, r\in \mathbb{Q}$.

Notice that $\displaystyle \tau(\sqrt{2})^2 = \tau(\sqrt{2}^2) = \tau(2) = 2$. Therefore, $\displaystyle \tau(\sqrt{2}) = \pm \sqrt{2}$.
This means, $\displaystyle \tau(a+b\sqrt{2}) = a \pm b\sqrt{2}$.
Now argue that $\displaystyle \sqrt{3} \not = c + d\sqrt{2}$ for any $\displaystyle c,d\in \mathbb{Q}$.