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Let $K$ a field, $p,n\in \mathbb{N}$, $B\in \mathcal{M}_p({K})$ and let denote $S_B=\{X\in \mathcal{M}_p(K) \ \mid \ X^n=B\}$.

If $X\in S_B$, I have to prove that $\mu_X$ (the minimal polynomial of $X$) divides $\mu_B(\xi^n)$.

If $X\in S_B$ then we can deduce that $X^n-B=0$ but after I do not know how to link this with $\mu_X$ or $\mu_B$...

Thanks in advance !

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Hint: It suffices to note that if $X \in S_B$, then $\mu_B(X^n) = 0$

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  • $\begingroup$ Does it come from the fact that $\mu_B(B)=0$ by definition ? $\endgroup$ – Maman Jan 11 at 19:15
  • $\begingroup$ @Maman exactly. $\endgroup$ – Omnomnomnom Jan 11 at 19:16
  • $\begingroup$ Then I think we have : $\mu_{X^n}(X^N)=0$ ? What is the link with $\xi^n$ ? $\endgroup$ – Maman Jan 12 at 21:27
  • $\begingroup$ If $p(A)=0$, then the minimal polynomial of $A$ divides $p$. $\endgroup$ – Omnomnomnom Jan 13 at 1:49
  • $\begingroup$ @Maman $\xi$ is used to emphasize the polynomial itself. That is, $p(\xi) = \mu_B(\xi^n)$ is a polynomial satisfying $p(X) = 0$ $\endgroup$ – Omnomnomnom Jan 13 at 16:54

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