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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.


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 $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.


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 :)

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  • $\begingroup$ Once you have $b^2m=n$, you automatically get that $b=1$ since $n$ is squarefree, so you obtain $n=m$. The other implication ($n=m \implies \mathbb{Q}(\sqrt{n})=\mathbb{Q}(\sqrt{m})$) is trivial, and you do not need to introduce $\psi$ unless you are trying to find the automorphisms of $\mathbb{Q}(\sqrt{n})$. $\endgroup$ – tristan Jun 26 '17 at 1:34
  • $\begingroup$ See the answers here, and here $\endgroup$ – Dietrich Burde Jun 26 '17 at 8:19
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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.

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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 $L$.

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    $\begingroup$ I feel like this question will be viewed by many people who don't know what the words "integral" or "splitting" or "ramify" mean. If you feel like it would take too much space to explain this to people unfamiliar with algebraic number theory then perhaps you could add a preface to your answer saying that it is meant for readers who are familiar with such devices. $\endgroup$ – Trevor Gunn Jun 26 '17 at 3:49

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