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Theorem   Let $m,n \in S$ with $m \neq n$ . Then $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are not isomorphic.

Proof. Suppose that $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are isomorphic. Let $\varphi \colon \mathbb{Q}(\sqrt{m}) \to \mathbb{Q}(\sqrt{n})$ be a field isomorphism. Recall that field homomorphisms fix prime subfields. Thus, for every $x \in \mathbb{Q}$ , $\varphi(x)=x$ .

Let $a,b \in \mathbb{Q}$ with $\varphi(\sqrt{m})=a+b\sqrt{n}$ . Since $\varphi(a)=a$ and $\varphi$ is injective, $b \neq 0$ . Also, $m=\varphi(m)=\varphi((\sqrt{m})^2)=(\varphi(\sqrt{m}))^2=(a+b\sqrt{n})^2=a^2+2ab\sqrt{n}+b^2n$ . If $a \neq 0$ , then $\displaystyle \sqrt{n}=\frac{m-a^2-b^2n}{2ab} \in \mathbb{Q}$ , a contradiction. Thus, $a=0$ . Therefore, $m=b^2n$ . Since $m$ is squarefree, $b^2=1$ . Hence, $m=n$ , a contradiction. It follows that $K$ and $L$ are not isomorphic.

Corollary   There are infinitely many distinct quadratic fields.

Proof. Note that there are infinitely many elements of $S$ . Moreover, if $m$ and $n$ are distinct elements of $S$ , then $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are not isomorphic and thus cannot be equal.

quadratic fields that are not isomorphic is owned by Warren Buck.