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

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!

• what exactly is your definition for $T_\alpha$? – Pink Panther Jul 28 '18 at 16:36
• Sorry for the lack of details and thanks for the comment! I added that info to the problem. – user110320 Jul 28 '18 at 16:40
• so first you can take two minimal polynomials $m_\alpha$ and $m_{T_\alpha}$ and then you see that $m_{T_\alpha}(\alpha)=m_{T_\alpha}(1\cdot\alpha)=0$, since we assume that $m_{T_\alpha}(T_\alpha(x))=m_{T_\alpha}(\alpha\cdot x)=0$ for all $x\in F$ (if i recall correctly, please tell me if i'm wrong). Then $\deg(m_{T_\alpha})\geq\deg(m_\alpha)$, since $m_{T_\alpha}$ is already a candidate for $m_\alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_\alpha(x)-m_{T_\alpha}(x)\neq 0$ for some $x\in F$. I'll let you try from here. Hope this helps! – Pink Panther Jul 28 '18 at 17:15

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 $f(T_\alpha)=T_{f(\alpha)}$.