How to prove this splitting field isomorphy criterium?

Question

Given two irreducible polynomials $f(x) = x^2+a, g(x) = x^2+b \in \Bbb Q [x]$, the task is to prove that splitting fields of $f$ and $g$ ($\Bbb Q[\sqrt{-a}]$ and $\Bbb Q[\sqrt{-b}]$ actually) are isomorphic iff $\frac ab = q^2$ for some $q\in \Bbb Q$.

I have some thoughts about going left direction but I can't write them right. Actually my guess is that $\sqrt{-a}$ and $\sqrt{-b}$ got both the same irrational or complex part and are equal up to rational factor, but it could be more general I guess.

I appreciate your help, or some kind of hint. Thank you in advance!

Answer 1

Well, if $\;\phi:\Bbb Q(\sqrt{-a})\to\Bbb Q(\sqrt{-b})\;$ is a isomorphism, then

$$\exists r,s\in\Bbb Q\;\;s.t.\;\;\phi(r+s\sqrt{-a})=\sqrt{-b}\implies r+s\phi(\sqrt{-a})=\sqrt{-b}$$

But we know it must be that $\;\phi(\sqrt{-a})=\pm\sqrt{-a}\;$ , so

$$r\pm s\sqrt{-a}=\sqrt{-b}$$

and now: if $\;a>0\;$ , and thus also $\;b>0\;$ , then the above line is an equality in the complex numbers and real and imaginary parts in both sides are corr. equal, thus

$$r=0\;\;\text{and}\;\;s\sqrt a=\sqrt b\implies\frac ab=\frac1{s^2}\in\Bbb Q$$

If $\;a,b <0\;$ then the same deduction above is possible, this time taking into account that $\;\{1,\sqrt x\}\;,\;0<x\in\Bbb Q\;$ is a linearly independent set in $\;\Bbb R_{\Bbb Q}\;$

Answer 2

Another approach based on a theorem about quadratic field extensions of the rationals:

A general fact proved here is that for $d,d'$ square-free, $\mathbb{Q}(\sqrt{d}) \cong \mathbb{Q}(\sqrt{d'})$ as fields iff $d=d'$.

In your case $\alpha = -a$ and $\beta = -b$ are not required to be square-free, so write $\alpha = r^2 j$ and $\beta = s^2 k$ where $j$ and $k$ are square-free integers. Then $\sqrt{\alpha} = r\sqrt{j}$ and $\sqrt{\beta} = s\sqrt{k}$ and so $\mathbb{Q}(\sqrt{\alpha}) = \mathbb{Q}(\sqrt{j})$ and $\mathbb{Q}(\sqrt{\beta}) = \mathbb{Q}(\sqrt{k})$. Thus the mentioned theorem gives that $\mathbb{Q}(\sqrt{\alpha}) \cong \mathbb{Q}(\sqrt{\beta})$ iff $j = k$. Using our equations, this is equivalent to $\alpha/r^2 = \beta/s^2$, again equivalent to $\alpha / \beta = q^2$ where $q = r/s$, which is what you wanted to prove.