Cfrgtkky

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 $phi :mathbb{Q}(sqrt{n})tomathbb{Q}(sqrt{m})$ is an isomorphism. Then $phi(u)=u$ for every $u in mathbb{Q}$, so that $phi(sqrt{n})^2=phi(n)=n$. If $phi(sqrt{n})=a+bsqrt{m}$ for some $a,binmathbb{Q}$, then $(a+bsqrt{m})^2=a^2+b^2m + 2absqrt{m}=n$ implies
$$a^2+b^2m = n$$ and $$2ab=0.$$

Then $b=0$ cannot happen since that would imply that $sqrt{n}inmathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$
If $psi:mathbb{Q}(sqrt{n})tomathbb{Q}(sqrt{m})$ were an isomorphism then there would exist some $sinmathbb{Q}$ 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 $phi(u+vsqrt{n})=u+vssqrt{m}$ and $psi(u+vsqrt{n})=u-vssqrt{m}$.

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 :)

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 $$mathbf{Q}(sqrt d) = {a + bsqrt d : a, b in mathbf{Q}}.$$ This isn't hard to show either but perhaps it would be good to mention that this was shown somewhere before.

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 $L$.