# Thread: Showing two fields are not isomorphic

1. ## Showing two fields are not isomorphic

I am trying to show that $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ are not isomorphic.

My question is this... Isomorphisms preserve subfields (including, in this case, $\displaystyle \mathbb{Q}$), and I WANT to say something like '$\displaystyle \mathbb{Q}$ must be preserved so the mapping acts as the identity on $\displaystyle \mathbb{Q}$ but then it must act as the identity on the whole field, but then clearly it is not an isomorphism so $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ cannot be isomorphic.'

But, is any of this true, or even a useful way of thinking about it? Sadly, I suspect not...

2. ## Re: Showing two fields are not isomorphic

why is $\displaystyle f(a+ b\sqrt{2})= a+ b\sqrt{3}$ not an isomorphism?

3. ## Re: Showing two fields are not isomorphic

Originally Posted by AlexP
I am trying to show that $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ are not isomorphic.

My question is this... Isomorphisms preserve subfields (including, in this case, $\displaystyle \mathbb{Q}$), and I WANT to say something like '$\displaystyle \mathbb{Q}$ must be preserved so the mapping acts as the identity on $\displaystyle \mathbb{Q}$ but then it must act as the identity on the whole field, but then clearly it is not an isomorphism so $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ cannot be isomorphic.'

But, is any of this true, or even a useful way of thinking about it? Sadly, I suspect not...
Suppose that $\displaystyle f:\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{3})$ is an isomorphism then $\displaystyle f(\sqrt{2})^2-2=f\left(\sqrt{2}^2-2\right)=0$ and so there exists some $\displaystyle y\in\mathbb{Q}(\sqrt{3})$ such that $\displaystyle y^2-2=0$.

4. ## Re: Showing two fields are not isomorphic

HallsofIvy: it is not a homomorphism. When you multiply the square roots cancel each other so you end up with a 2 on one side and a 3 on the other.

Drexel28, I understand the point of showing that there is no number that squares to 2 in $\displaystyle \mathbb{Q}(\sqrt{3})$, but can you explain the $\displaystyle f(\sqrt{2})^2-2 = f(\sqrt{2}^2-2)$ part?

5. ## Re: Showing two fields are not isomorphic

Originally Posted by AlexP
HallsofIvy: it is not a homomorphism. When you multiply the square roots cancel each other so you end up with a 2 on one side and a 3 on the other.

Drexel28, I understand the point of showing that there is no number that squares to 2 in $\displaystyle \mathbb{Q}(\sqrt{3})$, but can you explain the $\displaystyle f(\sqrt{2})^2-2 = f(\sqrt{2}^2-2)$ part?
Since we are dealing with fields every morphism is unital, i.e. if $\displaystyle f:\mathbb{Q}(2)\to\mathbb{Q}(3)$ is an isomorphism then $\displaystyle f(1)=1$ and so $\displaystyle f(2)=f(1+1)=f(1)+f(1)=2$ thus $\displaystyle f\left(\sqrt{2}^2-2\right)=f\left(\sqrt{2}^2\right)-f(2)=f\left(\sqrt{2}\right)^2-2$.

6. ## Re: Showing two fields are not isomorphic

Ah, right. I didn't see how f(2)=2 but that explains it. Thanks.