I've been struggling trying to understand the Jacobson's Basic Algebra vol. II proof of the Luröth's theorem.
Let $K$ be a field, $K(X)$ the field of rational fonctions and take $L$ to be a sub-extension of $K(X):K$ such that $L \neq K$. Let $F \in L - K$. Then some lemma states that $X$ is algebraic over $K(F)$, it follows $X$ algebraic over $L$. Then let, $$ \Phi = T^n + f_1T^{n-1} + f_{2}T^{n-2} + \cdots + f_0 \in L[T] $$ be the minimal polynomial of $X$ over $L$. Now comes the strange assertion made in Jacobson's book :
There is a polynomial $P_n \in K[X]$ of least degre such that, $$ P_n \Phi = P_nT^{n} + P_{n-1}T^{n-1} + \cdots + P_0 \in (K[X])[T] $$ and such that $P_n \Phi$ is primitive. I suppose we can take $P_n$ to be the least common multiple of the denominators of the $f_i \in L \subset K(X)$ but then I can't see why the greatest common divisor of the $\{ P_0, \ldots P_{n}\}$ is $1$.
So the question is why is it true ?
From the way Jacobson's states this assertion, it seems easy to see this but I tried very hard and failed.
Some insight would be much appreciated.
You are on the right track. By letting $P$ to be that least common multiple of the denominators of the $f_i$s you get a polynomial $P(X)$ such that $P\Phi\in K[X][T]$. It may initially happen that the coefficients $P_0(X),P_1(X),\ldots,P_n(X)$ have a non-constant common divisor $d(X)$. But the $P(X)=P_n(X)$ is also a multiple of $d(X)$, and we can use $\tilde{P}(X)=P(X)/d(X)$ instead of $P(X)$ to begin with. After all, the coefficient of $\tilde{P}\Phi$ are the rational functions $P_i(X)/d(X)$ that we now know to be polynomials.
Mind you, I think that it follows from this that we actually must have had $\gcd\{P_0,P_1,\ldots,P_n\}=1$ all the time.