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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 $K(f(x))$?

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    $\begingroup$ As written, this seems to be a transcendental extension. Maybe you need to clarify $\endgroup$ Nov 22 '20 at 10:46
  • $\begingroup$ Can you say how $x$ acts on $K(f(x))$? $\endgroup$ Nov 22 '20 at 11:55
  • $\begingroup$ Did you really mean $K[x]/(f(x))$? $\endgroup$ Nov 22 '20 at 11:56
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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.

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  • $\begingroup$ Thank you. I may have overlooked something, but for now I'm convinced. If I repeat the division by f (x), left side seems to become a polynomial of degree n or less on K. $\endgroup$
    – Kazsugi
    Nov 22 '20 at 12:37
  • $\begingroup$ @Kazsugi Essentially keep on this process would simply show that all $s_i$ must all be zero polynomials, and considering where they come from would show you that all $p_i$'s are zero polynomials, and therefore $P(t)=0$ and contradiction. I thought I should have seen a proof of this somewhere but I couldn't find any, so I just used brute force :( sorry if it's more complicated than need be. $\endgroup$
    – August Liu
    Nov 22 '20 at 17:31

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