# Minimal polynomial of $x$ over $K(f(x))$ [duplicate]

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 $$K(f(x))$$?

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

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.

• 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. Nov 22 '20 at 12:37
• @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. Nov 22 '20 at 17:31