quadratic fields that are not isomorphic

Within this entry, $S$ denotes the set of all squarefree integers not equal to $1$.

Theorem.

Let $m\mathrm{,}n\mathrm{\in }S$ with $m\mathrm{\ne }n$. Then $\mathrm{Q}\mathit{}\mathrm{(}\sqrt{m}\mathrm{)}$ and $\mathrm{Q}\mathit{}\mathrm{(}\sqrt{n}\mathrm{)}$ are not isomorphic (http://planetmath.org/FieldIsomorphism).

Proof.

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

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

This yields an obvious corollary:

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.∎

Note that the above corollary could have also been obtained by using the result regarding Galois groups of finite abelian extensions of $\mathbb{Q}$ (http://planetmath.org/GaloisGroupsOfFiniteAbelianExtensionsOfMathbbQ). On the other hand, using this result to prove the above corollary can be likened to “using a sledgehammer to kill a housefly”.