# Proof verification. Question about finite extension, minimal polynomial.

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 $\det(xI-L_{a})$ is the minimal polynomial $\min(F,a)$ of $a$. For which $\alpha \in K$ is $\det(xI-L_{\alpha})=\min(F,\alpha)$?

$\bf{For\,the\,first\,part}.$ Take $x,y \in K$ and $\beta \in F$, so $$L_{\alpha}(\beta x + y) = \alpha(\beta x + y) = \alpha \beta x + \alpha y = \beta L_{\alpha}(x) + L_{\alpha}(y).$$

If $\beta \in F$, $L_{\alpha}(\beta) = L_{\alpha}(\beta 1) = \beta L_{\alpha}(1) = 1$, since $L_{\alpha}$ is surjective. Then $L_{\alpha}(F) = F$.$\square$

$\bf{For\,the\,second\,part}.$ By Cayley-Hamilton Theorem, $\det(xI-L_{a}) = \chi_{a}(x)$ where $\chi_{a}(x)$ is the characteristic polynomial of $xI-L_{a}$ and $\chi_{a}(L_{a}) = \chi_{a}(a)= 0$, so $a$ is a root of $\chi_{a}$. The minimal polynomial of $xI-L_{a}$ divides $\chi_{a}$, but the minimal polynomial of $xI-L_{a}$ is the $\min(F,a) = p(x)$. If $\chi_{a}(x) = p(x)q(x)$, $\deg(\chi_{a}) \geq \deg(p) = n$. Since $[K:F] = n$, $\chi_{a}(x) = \min(F,a) = \det(xI-L_{a})$.$\square$

I need help with the last part:"For which $\alpha \in K$ is $\det(xI-L_{\alpha})=\min(F,\alpha)$?"

I'm not sure about the my solution for the second part.

• One would think that $\det(xI-L_\alpha)$ was the minimal polynomial for $\alpha$ precisely when $K=F(\alpha)$. – Lubin Apr 10 '18 at 3:57

The equality $\chi_a(L_a) = \chi_a(a)$ is not really correct. $\chi_a(L_a)$ is a map $K \to K$ while $\chi_a(a)$ is an element of $K$. But what could you evaluate the map $\chi_a(L_a)$ at in order to get $\chi_a(a)$? I.e., what can you evaluate $L_a$ at in order to get $a$?
I don't quite follow your conclusion that $\chi_a = p$. You have that $p$ divides $\chi_a$ and $n = \deg(p) \leq \deg(\chi_a) = n$, so they have the same degree. This shows that $q$ must have degree $0$, but this just means $q$ is a scalar, not necessarily $1$. So why couldn't $\chi_a = -p$ or $\chi_a = 2p$?
For the last part, compare the degrees of $\det(xI - L_\alpha)$ and $\min(F,\alpha)$. (This does not give you an "explicit" answer, but I think it's the best one can do.) It might also help to consider an example, say $F = \mathbb{Q}$, $a = \sqrt{2}$ and $\alpha = \sqrt{2}$. What do you get for $\det(xI - L_\alpha)$ and $\min(F,\alpha)$ in this case?
Notice that the argument is this, $\deg(\chi_a)=[K:F]=\deg(\min(F,a))$ and $\min(F,a)|\chi_a$
For a general $\alpha$, we still get that $\min(F,\alpha)|\chi_\alpha$ and that $\deg(\chi_\alpha)=[K:F]$. So the result will hold iff $[K:F]=\deg(\min(F,\alpha))$ which is true iff $K = F(\alpha)$