Let $p(x)=x^2+rx+s$ and $q(x)=x^2+tx+u$ be irreducible polynomials in $\mathbb{Q}[x]$. If $p(\alpha)=q(\beta)=0$ for some $\alpha, \beta \notin \mathbb{Q}$, I need to find necessary and sufficient conditions on the coefficients that guarantee $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ are isomorphic.
My work
Since $p(x)$ and $q(x)$ are irreducible polynomials, then we have
$\mathbb{Q}(\alpha) \cong \mathbb{Q}[x]/(p(x))$ and $\mathbb{Q}(\beta) \cong \mathbb{Q}[x]/(q(x))$
If $\mathbb{Q}(\alpha) \cong \mathbb{Q}(\beta)$ then $\mathbb{Q}[x]/(p(x)) \cong \mathbb{Q}[x]/(q(x))$.
Can I conclude that $p(x)=q(x)$? if not, is there any relation between the two polynomials?
First note that $\Bbb Q(\alpha)=\Bbb Q(\sqrt{d_1})$ where $d_1=r^2-4s.$ Similarly, $\Bbb Q(\beta)=\Bbb Q(\sqrt{d_2})$ where $d_2=t^2-4u.$
If these two fields are isomorphic then $\Bbb Q(\sqrt{d_2})$ contains a square root of $d_1,$ i.e.
$$\exists a,b\in\Bbb Q,\quad d_1=(a+b\sqrt{d_2})^2=a^2+b^2d_2+2ab\sqrt{d_2}$$ i.e. $d_1=a^2+b^2d_2$ and $ab=0.$ Since $d_1$ is not a square in $\Bbb Q,$ $b$ cannot be $0.$ Therefore, $a=0$ and $$d_1=b^2d_2.$$
We thus proved that if $\Bbb Q(\alpha)\cong\Bbb Q(\beta)$ then $d_1=b^2d_2$ for some $b\in\Bbb Q.$ The converse being obvious, we can conclude:
$$\Bbb Q(\alpha)\cong\Bbb Q(\beta)\iff\exists b\in\Bbb Q,\quad\frac{r^2-4s}{t^2-4u}=b^2.$$