Showing two fields are not isomorphic

**AlexP** · August 5th 2011, 05:14 PM

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

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

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

**HallsofIvy** · August 5th 2011, 05:36 PM

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

**Drexel28** · August 5th 2011, 06:11 PM

Originally Posted by AlexP

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

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

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

Suppose that $f:\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{3})$ is an isomorphism then $f(\sqrt{2})^2-2=f\left(\sqrt{2}^2-2\right)=0$ and so there exists some $y\in\mathbb{Q}(\sqrt{3})$ such that $y^2-2=0$ .

**AlexP** · August 6th 2011, 07:36 PM

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 $\mathbb{Q}(\sqrt{3})$ , but can you explain the $f(\sqrt{2})^2-2 = f(\sqrt{2}^2-2)$ part?

**Drexel28** · August 6th 2011, 08:05 PM

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 $\mathbb{Q}(\sqrt{3})$ , but can you explain the $f(\sqrt{2})^2-2 = f(\sqrt{2}^2-2)$ part?

Since we are dealing with fields every morphism is unital, i.e. if $f:\mathbb{Q}(2)\to\mathbb{Q}(3)$ is an isomorphism then $f(1)=1$ and so $f(2)=f(1+1)=f(1)+f(1)=2$ thus $f\left(\sqrt{2}^2-2\right)=f\left(\sqrt{2}^2\right)-f(2)=f\left(\sqrt{2}\right)^2-2$ .

**AlexP** · August 7th 2011, 02:37 PM

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

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