I have the following theorem
The remarks state that it is obvious that the caley hamilton theorem follows from this as $f_s$ is the minimal polynomial. However, I do not see why it must be that. I think that under this decomposition I can see that in the right hand side ($f_s$) will disappear, this corresponds to $f_s(\alpha)$, so I can deduce that $f_s(\alpha)=0$ (?). Therefore the minimal polynomial divides it. However I don't see how this implies that this is actually the minimal polynomial. Could someone clarify the situation for me?
Question: Could someone clarify why $f_s$ is the minimal polynomial here?
Clarification: Here we have
$f_s$ will be the minimal polynomial because: if $g\in\mathcal{F}[X]$ acts as zero on $V_\alpha$, by the isomorphism of modules it must act as zero over every single one of the $\mathcal{F}[X]/(f_i)$. But that’s true if and only if it is divisible by every single $f_i$, which is true if and only if it is divisible by $f_s$.