Suppose we have $K$ is an algebraically closed field, $x$ is transcendental over $K$ and $f$ is a polynomial in $K[x]$. Consider the tower of field extensions $$K\subseteq K(f(x))\subseteq K(x).$$ We also have $[K(x):K(f(x))]=\deg f$.
First, I consider $g\in K(f(x))[t]$, the ring of polynomials over $K(f(x))$, and $$g(t)=\dfrac{1}{f(x)}f(t)-1.$$ It is clear to see $g(x)=0$. However, I can't prove $g$ is irreducible. I wonder if this statement is true?
Consider the monic polynomial $f(t)-f(x)$, and let $z=f(x)$. Since $t$ is transcendental over $K[z]$ and $z$ is transcendental over $K$, $z$ is transcendental over $K[t]$. Suppose $f(t)-z=u(t)v(t)$ for some $u,v\in K[z][t]$. Note that $f(t)-z$ is a linear polynomial in $z$. Then we may assume $u(t)=a(t)$ and $v(t)=b(t)z+c(t)$ with $a,b,c\in K[t]$. Comparing the coefficients we obtain $a(t)b(t)=-1$. Now $u(t)=a(t)$ is a constant, so $f(t)-z$ is irreducible over $K[z]$. By Gauss's Lemma, that is irreducible over $K(z)$.