Any field element $x \in L$ satisfies its own characteristic polynomial.

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I am reading Janusz's Algebraic Number Fields:

Let $L/K$ be a finite field extension. Each element $x \in L$ gives rise to a linear transformation $r_x: L \to L$ defined by $y \mapsto yx$.

The characteristic polynomial of the element $x \in L$ is defined to be the determinant $$ f(t) = \det (tI - r_x) $$ Then $f(r_x)$ is the linear transformation that multiplies an element of $L$ by $f(x)$, so $f(x) = 0$.

I am not seeing why $f(r_x)$ is multiplication by $f(x)$. Actually, perhaps more urgently, I don't even know how to interpret $f(r_x)$, as $f$ seems to be a polynomial in $t$.

Alternatively, is the Cayley-Hamilton, which states that the minimal polynomial divides the characteristic polynomial, not carry over from linear algebra? In which case, since I know $x$ satisfies the minimal polynomial, it would therefore satisfy the characteristic polynomial too.

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For example take the poly $X^2+1$ then $$(r^2+1)(y)=r^2y+y=rxy+y=x^2y+y=(x^2+1)y.$$ The linear transform $f(r)$ is the zero transform, by Cayley-Hamilton, but it is also multiplication by $f(x)$.