[proofplan]
Write the rational map in lowest terms as $P/Q$. If its degree is $1$, then both numerator and denominator are linear-or-constant polynomials, so they have the form $az+b$ and $cz+d$. The only point to check is that lowest terms and nonconstancy force $ad-bc \ne 0$, which is precisely the determinant condition for a [Möbius transformation](/page/M%C3%B6bius%20Transformation). Conversely, a Möbius transformation has such a fractional linear formula with nonzero determinant; the determinant condition prevents a common factor, so the reduced rational map has degree $1$.
[/proofplan]
[step:Rewrite a degree-one rational map as a fractional linear formula]
Assume first that $\deg R=1$. By the definition of degree for a rational map in lowest terms,
\begin{align*}
\max\{\deg P,\deg Q\}=1.
\end{align*}
Therefore there exist coefficients $a,b,c,d \in \mathbb{C}$ such that
\begin{align*}
P(z)=az+b
\end{align*}
and
\begin{align*}
Q(z)=cz+d.
\end{align*}
The polynomial $Q$ is not the zero polynomial, and since $R$ is nonconstant, $P$ and $Q$ are not both constant. Thus at least one of $a$ and $c$ is nonzero.
[guided]
We start from the hypothesis $\deg R=1$ and unpack exactly what that means. The rational map is already written in lowest terms as $R=P/Q$, so the degree convention gives
\begin{align*}
\deg R=\max\{\deg P,\deg Q\}=1.
\end{align*}
This equality says two things at once. First, neither $P$ nor $Q$ has degree greater than $1$. Second, at least one of them has degree exactly $1$. Hence both polynomials are linear-or-constant. We may therefore choose coefficients $a,b,c,d \in \mathbb{C}$ such that
\begin{align*}
P(z)=az+b
\end{align*}
and
\begin{align*}
Q(z)=cz+d.
\end{align*}
The denominator polynomial $Q$ is not the zero polynomial because it represents the denominator of the rational expression. Also, $P$ and $Q$ cannot both be constant, since then $P/Q$ would define a constant rational map on $\widehat{\mathbb{C}}$, contrary to the hypothesis that $R$ is nonconstant. Equivalently, at least one of the leading coefficients $a$ and $c$ is nonzero.
[/guided]
[/step]
[step:Use coprimality to force the determinant condition]
We claim that
\begin{align*}
ad-bc \ne 0.
\end{align*}
Suppose instead that $ad-bc=0$. Then the coefficient vectors $(a,b)$ and $(c,d)$ in $\mathbb{C}^2$ are linearly dependent. Since neither $P$ nor $Q$ is the zero polynomial, there is a nonzero scalar $\lambda \in \mathbb{C}$ such that either $P=\lambda Q$ or $Q=\lambda P$. Because at least one of $P$ and $Q$ has degree $1$, this gives a nonconstant common factor of $P$ and $Q$, contradicting that the rational expression is in lowest terms. Hence $ad-bc \ne 0$.
By the definition of a Möbius transformation, the formula
\begin{align*}
z \mapsto \frac{az+b}{cz+d}
\end{align*}
with $ad-bc \ne 0$, interpreted with its usual extended value at the pole and at $\infty$, defines a Möbius transformation of $\widehat{\mathbb{C}}$. Therefore $R$ is a Möbius transformation.
[/step]
[step:Show that every Möbius transformation has reduced degree one]
Conversely, assume that $R$ is a Möbius transformation. Then there exist $a,b,c,d \in \mathbb{C}$ with
\begin{align*}
ad-bc \ne 0
\end{align*}
such that, on finite points where the denominator is nonzero,
\begin{align*}
R(z)=\frac{az+b}{cz+d}.
\end{align*}
Define polynomials $P_0,Q_0 \in \mathbb{C}[z]$ by
\begin{align*}
P_0(z)=az+b
\end{align*}
and
\begin{align*}
Q_0(z)=cz+d.
\end{align*}
The determinant condition implies that $P_0$ and $Q_0$ are not both constant: if $a=c=0$, then $ad-bc=0$.
We now check that $P_0$ and $Q_0$ are coprime. Since $\mathbb{C}$ is algebraically closed, any nonconstant common factor of two linear-or-constant polynomials would have a common zero $\alpha \in \mathbb{C}$. If such an $\alpha$ existed, then
\begin{align*}
a\alpha+b=0
\end{align*}
and
\begin{align*}
c\alpha+d=0.
\end{align*}
Thus the nonzero vector $(\alpha,1) \in \mathbb{C}^2$ would lie in the kernel of the [linear map](/page/Linear%20Map) $A: \mathbb{C}^2 \to \mathbb{C}^2$ defined by $A(u,v)=(au+bv,cu+dv)$. This is impossible because the determinant of $A$ is $ad-bc \ne 0$. Hence $P_0$ and $Q_0$ are coprime.
Since $P_0$ and $Q_0$ are coprime and at least one has degree $1$, the lowest-terms degree of $R$ is
\begin{align*}
\deg R=\max\{\deg P_0,\deg Q_0\}=1.
\end{align*}
This proves the converse implication and completes the proof.
[/step]
