If $u\in K$, show that $K=k(u)$ if and only if $u=(ax + b)/(cx + d)$ where $a,b,c,d\in k$ with $ad-bc\neq 0$ Let $k$ be a field, and let $K = k(x)$ be the rational function field in one variable over $k$.
I proved that if $u=(ax + b)/(cx + d)$ where $a,b,c,d\in k$ with $ad-bc\neq 0$ then $K=k(u)$ but idk how to prove the other part.
I considered $u=f(x)/g(x)$ for some $f,g\in K$, but i have nothing more.
 A: Let $\deg$ be the usual degree function on $k[x]$ with $\deg(0)=-\infty$.

Extend $\deg$ to $k(x)$ by $\deg(p/q)=\deg(p)-\deg(q)$ for nonzero $p,q\in k[x]$.

The following properties of $\deg$ are easily verified:

*

*If $s,t\in k(x)$, then $\deg(st)=\deg(s)+\deg(t)$.$\\[4pt]$

*If $s,t\in k(x)$ with $t\ne 0$, then $\deg(s/t)=\deg(s)-\deg(t)$.$\\[4pt]$

*If $s,t\in k(x)$ with $\deg(s) > \deg(t)\ge 0$, then $\deg(s+t)=\deg(s)$.$\\[4pt]$

*If $s,t\in k(x)$ with $\deg(s)\ne 0$ and $t$ non-constant,
then $\deg(s\circ t)=\deg(s)\deg(t)$.$\\[4pt]$

*If $s\in k(x)$ with $\deg(s)\le 0$,
there is a unique constant $c\in k$ such that $\deg(s-c) < 0$.

Now suppose $u\in k(x)$ is such that $k(u)=k(x)$.

Necessarily $u$ is non-constant.

Without loss of generality, we can assume $\deg(u)\le 0$, else replace $u$ by $1/u$.

Let $c\in k\;$be the unique constant such that $\deg(u-c) < 0$, and let $v=1/(u-c)$.

Then we have $\deg(v) > 0$ and $k(v)=k(u)$, so $k(v)=k(x)$.

Then $x=p(v)/q(v)$ for some nonzero $p,q\in k[x]$ with $q$ monic and $\gcd(p,q)=1$.

Since $\gcd(p,q)=1$, there exist $f,g\in k[x]$ such that $fp+gq=1$.
\begin{align*}
\text{Then}\;\;&
fp+gq=1
\\[4pt]
\implies\;&
fxq+gq=1
\\[4pt]
\implies\;&
(fx+g)q=1
\\[4pt]
\end{align*}
so $q$ is a unit element of $k[x]$/

Hence since $q$ is monic, it follows that $q=1$.
\begin{align*}
\text{Then}\;\;&
x=p(v)/q(v)
\\[4pt]
\implies\;&
x=p(v)
\\[4pt]
\implies\;&
\deg(p\circ v)=1
\\[4pt]
\implies\;&
\deg(p)\deg(v)=1
\\[4pt]
\implies\;&
\deg(p)=1
\\[4pt]
\end{align*}
hence $p=ax+b$ for some $a,b\in k$ with $a\ne 0$.
\begin{align*}
\text{Then}\;\;&
x=p(v)
\\[4pt]
\implies\;&
x=av+b
\\[4pt]
\implies\;&
v=\frac{x-b}{a}
\\[4pt]
\implies\;&
u-c=\frac{a}{x-b}
\\[4pt]
\implies\;&
u=\frac{a}{x-b}+c
\\[4pt]
\implies\;&
u=\frac{cx+(a-bc)}{x-b}
\\[4pt]
\end{align*}
Hence we have
$$
u=\frac{Ax+B}{Cx+D}
$$
where $A=c,\;B=a-bc,\;C=1,\;D=-b$, and
$$
AD-BC
=
(c)(-b)-(a-bc)(1)
=
a
\ne
0
$$
so $u$ has the claimed form.
