I am having trouble proving (or finding a counterexample but I believe it to be true) the following. Prior to this I did some problems such as: Show that $\mathbb Q (-3+i\sqrt{2},2-\sqrt{8})=\mathbb Q (i,\sqrt{2})$ and I didn't really encounter any difficulty (providing I did them correctly). My idea was to apply the same strategy I used before on the following:
Prove or give a counterexample: For all nonzero complex numbers $z$ and all nonzero rational numbers $a$ and $b, \mathbb Q (az+b)=\mathbb Q(z)$ and $\mathbb Q (az^2+b)=\mathbb Q(z)$
What I did doesn't feel right. I just copied the same procedure as the other problems but I have no means to determine if this was appropriate or not. Here's what I did for $\mathbb Q (az+b)=\mathbb Q(z)$:
We need to show two things:
(1) $\mathbb Q (az+b)\subseteq \mathbb Q(z)$, and
(2) $\mathbb Q (az+b)\supseteq \mathbb Q(z)$
For (1): Since $\mathbb Q (az+b)$ is the smallest subfield of $\mathbb C$ containing the subfield $\mathbb Q$, and the element $az+b$, it suffices to show that $\mathbb Q (z)$ is a subfield of $\mathbb C$ which contains $\mathbb Q$ and $az+b$.
$\mathbb Q (z)$ is a subfield of $\mathbb C$ and $\mathbb Q \subseteq \mathbb Q(z)$ and by definition $a,z,$ and $b$ are all in $\mathbb Q(z)$. Since $\mathbb Q(z)$ is a subfield, $az+b$ is in $\mathbb Q(z)$ as well. Thus, since $\mathbb Q (az+b)$ is the smallest subfield of $\mathbb C$ containing the subfield $\mathbb Q$ and $az+b$, $\mathbb Q (az+b)\subseteq \mathbb Q(z)$.
For (2): Since $\mathbb Q(z)$ is the smallest subfield of $\mathbb C$ containing the subfield $\mathbb Q$, and the element $z$, it suffices to show that $\mathbb Q (az+b)$ is a subfield of $\mathbb C$ containing $\mathbb Q$ and $z$. By definition, $\mathbb Q(az+b$ is a subfield of $\mathbb C$ which contains $\mathbb Q$.
Furthermore, since $a,b,$ and $z$ are in $\mathbb Q (az+b)$, $z=((az+b)-b)(a)^{-1}$. Thus, it follows that since $\mathbb Q(z)$ is the smallest subfield of $\mathbb C$ containing $\mathbb Q$ and $az+b$, $\mathbb Q (az+b)\supseteq \mathbb Q(z)$.
I did the same thing for $\mathbb Q(az^2+b)$. Can anyone verify if this makes any sense?