# quadratic fields that are not isomorphic

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

###### Theorem.

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

###### 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, $m=\varphi(m)=\varphi((\sqrt{m})^{2})=(\varphi(\sqrt{m}))^{2}=(a+b\sqrt{n})^{2}% =a^{2}+2ab\sqrt{n}+b^{2}n$. If $a\neq 0$, then $\displaystyle\sqrt{n}=\frac{m-a^{2}-b^{2}n}{2ab}\in\mathbb{Q}$, 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:

###### 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”.

Title quadratic fields that are not isomorphic QuadraticFieldsThatAreNotIsomorphic 2013-03-22 16:19:44 2013-03-22 16:19:44 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Theorem msc 11R11