Suppose we are an injective rational function $f(x) = \frac{p(x)}{q(x)}$, from $\mathbb{C} \backslash \{\text{roots of q}\} \rightarrow \mathbb{C}$.
I would like to prove that $p$ and $q$ have degree at most 1, ie that $f$ is actually a mobius transformation.
This post answers the question but I believe the proof is faulty. Given $p(x) – tq(x)$ has only one root, how do we conclude that $p$ and $q$ are linear? It could be that $p(x) – tq(x) = (x-a)^n$. Here is my attempt at fixing this.
Suppose that $n = \max(\deg(p), \deg(q)) \geq 2$. Let $t_0$ be the ratio of the lead coefficients of $p$ and $q$. If $t \neq t_0$, then $p(x) – tq(x)$ is a degree $n$ polynomial, and by injectivity, we know it has a single root. Hence $p(x) – tq(x) = b(x-a)^n$. The $b$ and $a$ both depend on $t$, so we can write $p(x) – tq(x) = b(t)(x-a(t))^n$ as elements in the polynomial ring $F(t)[x]$, where $F(t)$ is the ring of functions $\text{Im}(f)\backslash\{t_0\} \rightarrow \mathbb{C}$. Hence, we expand out and compare term by term.
$$p(x) – tq(x) = (c_n -t d_n)x^n + … + (c_0 – td_0) =b(t) x^n + … + b(t)(-1)^na(t)^n = b(t)(x-a(t))^n $$
Note that the assumption of $t\neq t_0$ means that $c_n – td_n \neq 0$ for all $t$, but it is possible that $c_n =0$ or $d_n = 0$. Regardless, we get the following equations:
$$c_m – td_m = b(t){n \choose m}(-1)^{n-m} a(t)^{n-m}$$
and in particular, $b(t) = c_n – td_n \neq 0$. We consider two cases: $d_n = 0$ and $d_n \neq 0$. In the first case, $b(t) = c_n$ is a constant, and we look at the next equation, $$c_{n-1} – td_{n-1} = -b(t) {n \choose n-1} a(t)$$
We can rescale to get that $a(t) = c_{n-1}' -td_{n-1}'$, and again we consider two cases. First, if $a(t)$ is a constant, then so is $a(t)^m$, and we don't get any powers of $t$ on the RHS of the original equation, meaning that $tb_m = 0$ for all $m$, so $q$ is the 0 polynomial, a contradiction. Otherwise $a(t)$ is a genuine linear function, and so $a(t)^m$ are polynomials of degree $m$ in $t$. However, we don't see any higher powers of $t$, so it must be that $q(x) = d_{n-1}x^{n-1}$. Furthermore, it is obvious that $p(x)$ has only one distinct root, so $f(x) = \alpha\frac{(x-\beta)^n}{x^{n-1}}$. I am not sure how to derive a contradiction from this.
In the second case, we get that $b(t) a(t)^m$ is linear, so $a(t)^m$ must be a mobius transformation. Looking again at $a(t)^1$, we also get that $a(t)^m$ is a power of a mobius transformation. More specifically, $$ a(t)^m = \frac{c_m' – td_m'}{b(t)} = \left(\frac{c_{n-1}' – td_{n-1}'}{b(t)}\right)^m$$
I am also unsure of how to derive a contradiction from here. I would prefer not to use complex analysis such as Cauchy integral formula, nor to lift the problem to the Riemann sphere $\hat{\mathbb{C}}$. I am aware that it is possible to solve this problem by using the fact that the automorphisms of $\mathbb{P}^1(\mathbb{C})$ are exactly the mobius transformations, but I would prefer to use more elementary means to solve this problem.
For context, this problem came up in describing the injective morphisms $\mathbb{C} \backslash \{x_1, … x_n\} \rightarrow \mathbb{C}$ as an affine variety, so using a bit of affine algebraic geometry is fine.
Best Answer
This is much easier than you're making it. All we need is the fundamental theorem of algebra and a little trick. Write $p,q$ in lowest terms. Consider $p(x)-tq(x)$. Except at possibly one value of $t$, this is a polynomial of degree $\max(\deg p,\deg q)$ and except at finitely many more values of $t$, it has distinct roots (these values correspond exactly to the roots of the discriminant, a nonzero polynomial in $t$). Pick a $t=t_0$ which isn't one of the above finite values to avoid. Then $p(x)-t_0q(x)$ has $\max(\deg p,\deg q)$ distinct roots, which correspond to solutions of $p/q=t_0$. If $f$ is to be injective, then $\max(\deg p,\deg q) \leq 1$ and you have your result.