Let $K$ be a field and $f(x)$ be a one-variable monic polynomial on $K$.
Is the minimal polynomial on $K(f(x))$ of $x$ (with $t$ as a variable) $f(t) - f(x)$?
I know that the minimal polynomial of $x$ I want to know is a divisor of $f(t)-f (x)$.
I expect the minimal polynomial to be $f(t)-f(x)$. So, in general, is $f(t)-f(x)$ irreducible on $K(f(x))$?
I think the answer is yes, and hopefully this is a valid proof.
If there's a polynomial $P(t)\in K(f(x))[t]$ that has $x$ as a root with degree less than $\deg f$, then we can write
$$P(t)=\sum_{i=0}^n\frac{p_i(f(x))}{q_i(f(x))}t^i$$
where $p_i$, $q_i$ are polynomials with coefficient in $K$ and $n<\deg f$. Then plugging in $x$ we can write
$$\sum_{i=0}^ns_i(f(x))x^i=0$$
where $s_i$'s are polynomials with coefficient in $K$. Considering this equation $\operatorname{mod} f(x)$ implies that all the $s_i$'s have $0$ constant term. Now divide the equation by $f(x)$ and repeat until you reach a contradiction.