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Let $K = F(a)$ be a finite extension of $F$. For $\alpha \in K$, let $L_{\alpha}$ be the map from $K$ to $K$ defined by $L_{\alpha}(x) = \alpha x$. Show that $L_{\alpha}$ is an $F$-linear transformation. Also show that $\det(xI - L_{\alpha})$ is the minimal polynomial $\min(F,a)$ of $a$. For which $\alpha \in K$ is $\det(xI-L_{\alpha})=\min(F,\alpha)$?

I showed that $L_{\alpha}$ is a $F$-linear transformation. How I show that $\det(xI - L_{\alpha}) = \min(F,a)$? I like any hint, no solutions. Thanks for the advance!

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Hint: $P=det(xI-L_\alpha)(L_\alpha)=0$ Cayley-Hamilton, $P(L_\alpha)(1)=0$, deduce that $\alpha$ is a root of $P$, if $P=QR$ show that $\alpha$ is a root of $Q$ or $R$, deduce that $[K:F]$is the degree of $Q$ or $R$. Deduce that $Q=P$ or $R=P$ up to a scalar.

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  • $\begingroup$ In the question, $det(xI-L_{\alpha}) = \min(F, a)$, i.e, the minimal polynomial is associated to $a$, not $\alpha$. If I understand correctly your solution, you showed that $\min(F,\alpha)$ is $\det$. Can it be a typo of the question? Or did I misunderstand your solution correctly? $\endgroup$ – Corrêa Apr 5 '18 at 21:32
  • $\begingroup$ @LucasCorrêa: I suspect it's a typo in the question, since there are examples where $\min(F, \alpha) \ne \det$; note that $\deg(\det(x - L_{\alpha})) = [F(a):F]$ since $\alpha$ is a linear map on all of $F(a)$; so if we find any $\alpha$ with $\min(F, \alpha) < [F(a):F]$, we are done. Now consider $\Bbb Q(\omega)$ where $\omega^7 = 1$; we have $[\Bbb Q(\omega):\Bbb Q] = 6$, but $\alpha = \omega^6 + \omega$ satisfies a cubic. See math.stackexchange.com/questions/2722850/… $\endgroup$ – Robert Lewis Apr 7 '18 at 23:15
  • $\begingroup$ @LucasCorrêa: Also, the fact that the OP asks which $\alpha$ satisfy $\det = \min(F, \alpha)$ hints that perhaps not all $\alpha$ do! Cheers! $\endgroup$ – Robert Lewis Apr 7 '18 at 23:18
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    $\begingroup$ @RobertLewis, thank you very much! $\endgroup$ – Corrêa Apr 9 '18 at 22:48

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