characteristic vs minimal polynomial

Question

Let $L$ be a finite field extension of $K$. For every element $\theta$ in $L$ define the characteristic polynomial of $\theta$ as follows $$\operatorname{char}_{\theta}(X):=\det(X\cdot id_L-r_\theta)$$ where $r_{\theta}$ is the $K$-linear map given by $$r_{\theta}:L\longrightarrow L\qquad;\qquad x\longmapsto x\theta$$

Let finally $f(X)$ be the minimal polynomial of $\theta$ over $K$.

My question is: $\operatorname{char}_{\theta}=f?$

Answer 1

This is true iff $L = K[\theta]$ by examining degrees.

Answer 2

Suppose $d = [L : K(\theta)]$. Then $\chi_{char(\theta)}(x) = \chi_{min(\theta)}(x)^d$ assuming we've normalized both polynomials to be monic.