$\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ iff $n=m$

Question

Page 105 of D. Burton's A First Course in Rings and Ideals reads

It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ if and only if $n=m$.

Well, it is getting difficult for me, any help or would be appreciated.

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My progress so far: Suppose $\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ is an isomorphism. Then $\phi(u)=u$ for every $u \in \mathbb{Q}$, so that $\phi(\sqrt{n})^2=\phi(n)=n$. If $\phi(\sqrt{n})=a+b\sqrt{m}$ for some $a,b\in\mathbb{Q}$, then $(a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n$ implies $$a^2+b^2m = n$$ and $$2ab=0.$$

Then $b=0$ cannot happen since that would imply that $\sqrt{n}\in\mathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$ If $\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ were an isomorphism then there would exist some $s\in\mathbb{Q}$ such that $$s^2m=n,$$ then $s=b$ or $s=-b$. Therefore the only isomorphisms from $\mathbb{Q}(\sqrt{n})$ to $\mathbb{Q}(\sqrt{n})$ are $\phi(u+v\sqrt{n})=u+vs\sqrt{m}$ and $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.

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I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :)

Answer 1

Everything up to where you get $$b^2m = n$$ is fine. After that what you want to say is that because $n$ is square-free $b = \pm 1$ and hence $m = n$.

I will also say that it is not immediately obvious that $$\mathbf{Q}(\sqrt d) = \{a + b\sqrt d : a, b \in \mathbf{Q}\}.$$ This isn't hard to show either but perhaps it would be good to mention that this was shown somewhere before.

Answer 2

A somewhat stupid way using algebraic number theory: any isomorphism $K = \mathbb{Q}(\sqrt{n}) \rightarrow L = \mathbb{Q}(\sqrt{m})$ must preserve integral closures of $\mathbb{Z}$, and hence the splitting data of primes. The choice of a squarefree integer $n$ is equivalent to a choice of finitely many prime numbers as well as a choice of sign. We can conclude that for $n \neq m$ squarefree, the splitting data of primes in $\mathcal O_K$ and $\mathcal O_L$ are different based on the following standard result:

1 . The discriminant of $\mathbb{Q}(\sqrt{n})$ (whose prime divisors are exactly the ramified primes) is $n$ if $n \equiv 1 \pmod{4}$, and otherwise $4n$.

In particular, $\mathbb{Q}(\sqrt{n})$ and $\mathbb{Q}(\sqrt{-n})$ are never isomorphic for $n$ odd, because $2$ ramifies in exactly one of them. Moreover, we see that $\mathbb{Q}(\sqrt{n})$ and $\mathbb{Q}(\sqrt{m})$ are not isomorphic if there exists an odd prime divisor of $n$ which does not divide the other.

The final possibility to consider is that $n$ and $m$ share all their odd prime divisors, but one of them, say $n$, is even (if they are both even, then we must have $n = \pm m$, and we are done by the above paragraph). The three cases are $n = 2m, n =-2m$, and $n = -m$, both even.

Since $2$ ramifies in $\mathbb{Q}(\sqrt{m})$, it would have to ramify in $\mathbb{Q}(\sqrt{n})$, so we can conclude that the ring of integers of $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ are respectively $\mathbb{Z}[\sqrt{m}]$ and $\mathbb{Z}[\sqrt{n}]$. Thus a prime $p$ splits in $\mathbb{Q}(\sqrt{m})$ (resp. $\mathbb{Q}(\sqrt{n})$) if and only if $m$ (resp. $n$) is a square mod $p$. Now either $n = 2m$ or $n = -2m$. Using the Legendre symbol, we have in the first case:

$$(\frac{n}{p}) = (\frac{2}{p})(\frac{m}{p}) = (-1)^{\frac{p^2-1}{8}}(\frac{m}{p})$$

and in the second:

$$(\frac{n}{p}) = (\frac{-1}{p})(\frac{2}{p})(\frac{m}{p}) = (-1)^{\frac{(p-1)(p^2-1)}{16}}(\frac{m}{p})$$

and in the third:

$$(\frac{n}{p}) = (\frac{-1}{p})(\frac{m}{p}) = (-1)^{\frac{p-1}{2}}(\frac{m}{p})$$

Looking at the residues of primes modulo $16$, we can choose one which splits in $K$ but not in $L$.