I'm trying to solve the following problem:
Let $k$ be a field, $K=k(x)$ the field of fractions of $k[x]$ and $u \in K$. Show that $K= k(u)$ if and only if $u=\frac{ax+b}{cx+d}$ for some $a,b,c,d \in k$ such that $ad-bc \neq 0$.
For $\Leftarrow$ side, we can compute $x$ in terms of $u$ and prove that $x$ is a fraction of $u$ over $k$.
But in order to prove $\Rightarrow$, I'm facing some difficulties.
Assume $K = k(u) = k(x)$ then we can write $u = \frac{P(x)}{Q(x)}$ and $x = \frac{R(u)}{S(u)}$ for some $P,Q \in k[x]$, $R,S \in k[u]$ non-zero polynomial (since $u$ and $x$ are non-zero). The problem now is to reduce these polynomials to linear ones. My first idea is to write $$ u=\frac{P\left(\frac{R(u)}{S(u)}\right)}{Q\left(\frac{R(u)}{S(u)}\right)} $$ then note that the denominator shoud be equal to 1 (up to the choice of good $P$ and $Q$), moreover we should have $\left( P \circ \frac{R}{S} \right) (u) = u$.
At this point how can I go further ? Or is it a wrong path ? Should we use the Euclidean division algorithm to reduce the degree of $P$ and $Q$ ?
Edit: $u \notin k$ and the polynomial $uQ(y) + P(y) \in k(u)[y]$ has $x$ has a root, ie $x$ is algebraic over $K(u)$ thus there exists a irreducible monic polynomial $\mu_u \in k(u)[y]$ such that $\mu_u(x)=0$ and $\mu_u$ divides $f_u(y) := uQ(y)+P(y)$.
Then, the argument remains obscure to me...
One point is that $\mu_u$ is monic ie : $\mu_u(y) = y^n + \dfrac{P_{n-1}(u)}{Q_{n-1}(u)}y^{n-1} + \dots \dfrac{P_{0}(u)}{Q_{0}(u)}$.
The other point is that $\mu_u$ is an irreducible factor of $f_u(y)$, namely there exists $g_u(y) \in k(u)[y]$ such that $$ f_u(y) = g_u(y) \mu_u(y) $$
The Gauss' Lemma tells to us that (since k[u] is a PID, hence factorial) $$ \mu_u \in k[u][y] \text{ monic irreducible} \quad \Longleftrightarrow \quad \mu_u \in k(u)[y] \text{ irreducible.} $$
$ \bullet$ Can we deduce from $\gcd(P,Q) =1$ that $f_u(x)$ is irreducible (w.r.t $x$) and then $f_u = \mu _u$ ? Does it follow from the Gauss' Lemma ? But $f_u$ do not have to be monic in the variable $y$, isn't it ?
I mean since we have the minimal polynomial of $x$ over $k(u)[y]$ we can build a field out of the quotient by the ideal $(\mu_u)$: $$ \dfrac{k(u)[y]}{(\mu_u)} \simeq k(u,x). $$ Next, we get $[k(u,x): k(u)] = \deg_y(\mu_u)$ and here would use the equality $\mu_u = f_u$ to see on the right hand side $\deg_y(\mu_u) = \max\deg(P,Q)$, and one the left hand side $[k(u,x),k(u)] =1$ from the assumption $k(u,x)=k(u)=k(x)$.
