Proving that Minimal Polynomial is Monic

I am given that $$F \subseteq L$$ and $$L$$ is an extension. Now, I am also given $$T: L \to L$$, $$x \mapsto \alpha x$$ and $$m(x)\in F[x]$$ to be the minimal polynomial of $$T$$ in $$F$$.

I have to demonstrate that $$m$$ is also the monic polynomial in $$F[x]$$ of minimal degree for which $$m(\alpha)=0$$, which means showing that $$m$$ is minimal polynomial of $$\alpha$$ over $$F$$.

Now, I think that we are usually familiar with showing and defining the minimal polynomial by what I said on the third sentence. This question is asking you to prove the converse, and I am not exactly sure how to even start this. Any tips as to how should I start? Help would be appreciated.

Thank you.

• So, the minimal polinomial of $T$ in $L$ is $m(x)=x-\alpha$. Is $\alpha$ in $F$? Oct 29 '18 at 1:32
• Since I was given that $\alpha \in L$, I would assume it would be since $L$ is the field extension? Oct 29 '18 at 1:36
• Sorry, so I didn't get it. Since $T(x)=\alpha x$, we have that $x-\alpha$ is the minimal polynomial of $T$ in $L$. And if $x-\alpha$ is the minimal polynomial of $\alpha$ in $F$, then it means that $\alpha\in F$, which is not necessarily true. Oct 29 '18 at 1:42
• I guess the exercise wants the minimal polynomial of $T$ in $F$, that is the minimal polynomial $m(x)\in F[x]$ such that $m(T)=0$. Oct 29 '18 at 1:44
• I am given that $m \in F[x]$ Oct 29 '18 at 1:45

(1) Verify that $$\alpha$$ is a root of $$m(x)$$.
Indeed, since $$T=\alpha Id$$, and $$m(T)=0$$, $$\begin{array}{rcl} m(x)=a_nx^n+...+a_1x+a_0 & \Rightarrow & 0=m(T)=a_n\alpha^nId+...+a_1\alpha Id+a_0Id \\ & \Rightarrow & (a_n\alpha^n+...+a_1\alpha+a_0)Id = 0 \\ & \Rightarrow & m(\alpha) = a_n\alpha^n+...+a_1\alpha+a_0 = 0. \end{array}$$ (2) Since $$\alpha$$ is a root of $$m(x)$$, the minimal polynomial of $$\alpha$$ in $$F$$, $$m_\alpha(x)\in F[x]$$, divides $$m(x)$$.
(3) But then, since $$m(x)$$ is irreducible, the fact that $$m_\alpha(x)$$ divides $$m(x)$$ implies that $$m(x)=m_\alpha(x)$$.
• sorry, what is $\alpha Id$? Oct 29 '18 at 1:59
• $\alpha Id$ is the multiplication of the scalar $\alpha$ with the linear operator identity $Id$. Oct 29 '18 at 2:00
• It's another way of viewing $T$. The definition $T(x)=\alpha x$, means that $T=\alpha Id$. Oct 29 '18 at 2:02