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.

  • $\begingroup$ So, the minimal polinomial of $T$ in $L$ is $m(x)=x-\alpha$. Is $\alpha$ in $F$? $\endgroup$ Oct 29 '18 at 1:32
  • $\begingroup$ Since I was given that $\alpha \in L$, I would assume it would be since $L$ is the field extension? $\endgroup$
    – dmsj djsl
    Oct 29 '18 at 1:36
  • $\begingroup$ 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. $\endgroup$ Oct 29 '18 at 1:42
  • $\begingroup$ I guess the exercise wants the minimal polynomial of $T$ in $F$, that is the minimal polynomial $m(x)\in F[x]$ such that $m(T)=0$. $\endgroup$ Oct 29 '18 at 1:44
  • $\begingroup$ I am given that $m \in F[x]$ $\endgroup$
    – dmsj djsl
    Oct 29 '18 at 1:45

Follow these steps:

(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 $m(x)=m_\alpha(x)$.

  • $\begingroup$ sorry, what is $\alpha Id$? $\endgroup$
    – dmsj djsl
    Oct 29 '18 at 1:59
  • 1
    $\begingroup$ $\alpha Id$ is the multiplication of the scalar $\alpha$ with the linear operator identity $Id$. $\endgroup$ Oct 29 '18 at 2:00
  • $\begingroup$ It's another way of viewing $T$. The definition $T(x)=\alpha x$, means that $T=\alpha Id$. $\endgroup$ Oct 29 '18 at 2:02

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