Let $K = \mathbb{Q_3}$ and $L=K(\alpha)$ where $f := \min_K(\alpha) = x^4 - 3x^2 + 18$.
The roots of $f$ are given by $\pm \alpha$ and $\pm \alpha'$ where $$ \alpha' = \frac{(2 \alpha^2 -3) u}{\alpha} $$ and $u = \sqrt{-2/7} \in K$.
Consider the Möbius Transformation $$ \phi(x) = \frac{a x + b}{c x + d} $$ with coefficients $a,b,c,d \in L$ such that $\phi(0) = \alpha$, $\phi(1) = -\alpha$ and $\phi(\infty) = \alpha'$.
In fact, this is possible for $a = \alpha' c$, $c = -\frac{2\alpha d}{\alpha+\alpha'}$, $b = \alpha d$ and $d \in L$ arbitrary.
I was told that because $\alpha'$ is a root of $f$ and $\phi(\infty) = \alpha'$, we obtain $$ f(\phi(x)) = \frac{1}{(c x + d)^4} f_1(x) $$ where $f_1$ is a polynomial of degree $3$ (and not $4$), i.e. the coefficient corresponding to the monomial $x^4$ disappears!
Question: Without computing that by hand, can someone explain why this happens?
I could compute all of this by hand but this would not give me any insight on how the Möbius Transformation works here.
Let $g(x)=f(\phi(x))$. We have $g(\infty)=f(\phi(\infty))=f(\alpha')=0$. If a rational function vanishes at $\infty$, that means its numerator has lower degree than its denominator. How you prove this depends on how you define evaluation at $\infty$. For instance, one way of defining it is that $g(\infty)=h(0)$ where $h$ is the rational function given by $h(x)=g(1/x)$. If $g(x)=p(x)/q(x)$ for polynomials $p$ and $q$ of degree $m$ and $n$, we then have $$h(x)=\frac{p(1/x)}{q(1/x)}=\frac{x^{-m}r(x)}{x^{-n}s(x)}=x^{n-m}\frac{r(x)}{s(x)}$$ where $r(x)$ and $s(x)$ are polynomials not divisible by $x$. So, $h(0)=0$ iff $n>m$.
In your case, since the denominator has degree $4$, the numerator must have degree at most $3$. (And in this case, the numerator actually has degree exactly $3$, since $g$ has $3$ zeroes at finite points, namely at $\phi^{-1}(\alpha)$, $\phi^{-1}(-\alpha)$, and $\phi^{-1}(-\alpha')$.)