I'm working on this problem:
Find infinitely many quadratic extension fields of $\mathbb{Q}$ that are pairwise non isomorphic, and prove that they are indeed pairwise non isomorphic.
So far, I found that there are infinitely many $\mathbb{Q}(\sqrt{n})$ , where $n$ is a prime number. Now I'm stuck at a showing $\phi : \mathbb{Q}(\sqrt{n}) \rightarrow \mathbb{Q}(\sqrt{m})$ is not an isomorphism, for any distinct prime $n$ and $m$.
Thanks for you help!
Let $p$ and $q$ be distnct primes. Suppose $\phi \colon {\Bbb Q}(\sqrt{p})\to {\Bbb Q}(\sqrt{q})$ is an isomorphism. Suppose $\phi(\sqrt{p})=a+b\sqrt{q}$. Then, since $\phi$ is identity on $\Bbb Q$, it follows that $p=\phi(p)=(a+b\sqrt{q})^2$. You can easily get a contradiction from this.