Showing the minimal polynomial for an element in an extension field is the same as the minimal polynomial of a linear transformation.

Question

This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:

Let $K$ be a finite extension of $F$ of degree $n$. Let $\alpha$ be an element of $K$. Prove that the minimal polynomial for $\alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_{\alpha}$. In this problem, $T_{\alpha}$ is an $F$-linear transformation of $K$ that arises from $\alpha$ acting by left multiplication on $K$.

I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!

Answer 1

Hints:

1. Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.

2. We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.

3. Hence we have to prove $f(\alpha)=0 \iff f(T_\alpha) =0$ for any polynomial $f\in F[x]$.

   For this, observe $f(T_\alpha)=T_{f(\alpha)}$.