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:

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”.

Title	quadratic fields that are not isomorphic
Canonical name	QuadraticFieldsThatAreNotIsomorphic
Date of creation	2013-03-22 16:19:44
Last modified on	2013-03-22 16:19:44
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	9
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11R11