I was originally trying to show this isomorphism: $$Aut_K(K(x)) \cong PGL(2, K) \cong GL(2, K)/Scal(2, K), \text{ where } K \text{ is a field, and } Scal(2,K) \text{ is the group of scalar 2x2 matrices over } K.$$
I have already succeeded in producing a group homomorphism of this form: $$\Psi : GL(2, K) \to Aut_K(K(x)) \text{ such that } \Psi(A) = \phi_{A^{-1}}$$ and
$$\phi_{A} : K(x) \to K(x), \quad \phi_A(x) = \frac{ax+b}{cx+d}, \text{ where } A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in GL(2,K)$$
I have proven that each $\phi_A$ is a K-algebra automorphism of $K(x)$; hence, this group homomorphism makes sense. I have also proved the fact that $$\ker{\Psi} = Scal(2, K)$$
So, if this group homomorphism is surjective, then the first isomorphism theorem gives us the required isomorphism. While proving this is surjective, I have to prove that each K-algebra automorphism is of the form $\phi_A$ that has been described.
I was hinted by a text online that this particular theorem (or lemma?) should be of help, but I have not seen a proof of this in the text. Hence, I started with proving this seemingly simple lemma and am unable to complete the same for quite a few hours:
Let $K$ be a field, $K[t]$ be the polynomial ring of a single variable over K, and $Frac(K[t]) = K(t)$ be its field of fractions. Let $u \in K(t) \setminus K$ such that $u = \frac{f(t)}{g(t)}$ for non-zero polynomials $f(t), g(t) \in K[t]$, and $gcd(f(t), g(t)) = 1$, which we can define as $K[t]$ is an Euclidean Domain.
- Show that $h(x) = f(x) - ug(x) \in K(u)[x]$ is irreducible over $K(u)$ and $t \in K(t)$ is a root of the polynomial.
2. Show that $\deg(h) = \max{\left[\deg(f(t)), \deg(g(t))\right]}$ and hence conclude $[K(t) : K(u)] = \max{\left[\deg(f(t)), \deg(g(t))\right]}.$
The part that $t \in K(t)$ is the root of $h(x)$ is pretty apparent, and also, the degree of $h(x)$ must be the maximum of the degrees of $f$ and $g$, as per addition is defined in $K(u)[x]$. I also can see that if this is the case, $h(x)$ is the minimal polynomial of the extension $K(u,x)/K(u)$ and hence the degree of the extension is equal to the degree of the polynomial.
However, I am unable to prove that the polynomial is irreducible. As the gcd of the coefficients is 1, Eisenstein's Criterion fails to help - and for Gauss's Lemma to work, the problem is equivalent to showing that
$g(u)f(x) - f(u)g(x) \in (K[u])[x]$ is irreducible, which in turn would let me say that the polynomial is also irreducible in $K(u)[x]$.
As of now, I am unable to prove this fact even after careful consideration of the information available. Any help is appreciated. If there are any different methods to approach this, that would also help.
The polynomial $h=f(x)-ug(x)$ lies in fact in $K[u,x]=K[x][u]$. Since $K[x]$ is a UFD, and since $h$ is primitive as a polynomial in $u$ (this is because $f$ and $g$ are coprime), $h$ is irreducible in $K[x][u]$ if and only if it is irreducible in $K(x)[u]$, which is clear since it has degree $1$ in $u$.
Hence $h$ is irreducible in $K[x,u]=K[u][x]$. Now if we can show that $h$ is also primitive as a polynomial in $x$, we could conclude as before that $h$ is irreducible in $K(u)[x]$.
Write $f=a_nx^n+...+a_0,g=b_nx^n+...+b_0$, where $n=max (deg(f);deg(g))$, so one of $a_n$ or $b_n$ may be zero, but not both.
Then $h=(a_n- b_nu)x^n+...+(a_0-b_0 u)$. If $b_n=0$, then a gcd of the cofficients is invertible and $h$ is primitive. If $b_n\neq 0$, a gcd of the coefficients of $h$ is non invertible if and only if there is a $\lambda_i\in K$ such that $(a_i-b_iu)=\lambda_i (a_n-b_nu)$ for all $i$, where we set $\lambda_n=1$. Multiplying by $x^i$ and summing yields $f-u g=Q . (a_n -b_n u)$ for some $Q\in K[x]$ which is monic of degree $n$ . We then get $f=a_n Q, g=b_n Q$ , so $f$ and $g$ are not coprime, contradiction.
Maybe there is a shorter argument, but it does not pop into my mind right now.