the minimal polynomial is the characteristic polynomial of a suitable map How would I have to go about proving that the minimal polynomial $p_\beta$ of a field extension $F\subseteq G$ coincides (modulo the sign) with the characteristic polynomial of the linear mapping $f:F(\beta)\rightarrow F(\beta), \ v \mapsto \beta v$ between the $F$-spaces $F(\beta)$ ?
My attempt was trying to express $f$ in a basis; and since $B:=(1,\beta,\ldots,\beta^{d-1})$, where $d$ is the degree of $p_\beta$, is a basis, this seemed natural; if we assume that $$p_\beta=X^d-a_{d-1}X^{d-1}-\cdots-a_1 X-a_0, $$ for some $a_i\in K$, and compute the characteristic polynomials $c_\beta $ of $f$ with the help of the above basis, we get $$  (-1)^{2d}(-X)^d + (-1)^{2d}a_{d-1}(-X)^{d-1}+(-1)^{2d-1}\beta^{d-1}a_{d-2}(-X)^{d-2}+\cdots+(-1)^{d+2}\beta^2\cdots\beta^{d-1}a_1(-X)+(-1)^{d+1}\beta^1 \beta^2\cdots\beta^{d-1}a_0.$$ Now this doesn't really look like the above $p_\beta$, so either I think I made a mistake (but I double checked) or somehow the $\beta$'s would have to vanish - but I don't know how/why they would do that. Any ideas please ?
 A: What does the linear map $p_\beta(f)$ look like? You will use the (obvious) fact that if $p_\beta(x)=x^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0$, then $p_\beta(\beta)=0$.
Edit: Denote the minimal and characteristic polynomials of $f$ by $m_f(x)$ and $c_f(x)$ respectively. $p_\beta(f)$ refers to the linear operator $f^d+a_{d-1}f^{d-1}+\cdots+a_1f+a_0\iota$ where $\iota$ is the identity on $F(\beta)$. Since $f(v)=\beta v$ we have $f^k(v)=\beta^k v$ and therefore
$$p_\beta(f)(v)=(\beta^d+a_{d-1}\beta^{d-1}+\cdots+a_1\beta+a_0)v=0.$$
This shows that $m_f(x)$ divides $p_\beta(x)$. But $p_\beta(x)$ is irreducible, so $m_f(x)=p_\beta(x)$. Furthermore, $\deg(m_f(x))=d=\deg(c_f(x))$ and $m_f(x)$ divides $c_f(x)$, so $c_f(x)=m_f(x)=p_\beta(x)$.
My other comment is this: If you write down the matrix of $f$ with respect to your basis $B$, you will get the companion matrix of $p_\beta(x)$. It is well-known that the characteristic polynomial of the companion matrix of some $p(x)$ is just $p(x)$, so that's another way of doing the question. Your result is even clearer/"obvious" if you look at the $F[x]$-module associated with $f$.
