The Stacks project

Proof. This follows quite easily once we prove that the characteristic polynomial $\bar P(T) \in \kappa (\mathfrak p)[T]$ of the multiplication map $m_{\bar f}: A \otimes _ R \kappa (\mathfrak p) \to A \otimes _ R \kappa (\mathfrak p)$ which multiplies elements of $A \otimes _ R \kappa (\mathfrak p)$ by $\bar f$, the image of $f$ viewed in $\kappa (\mathfrak p)$, is just the image of $P(T)$ in $\kappa (\mathfrak p)[T]$. Let $(a_{ij})$ be the matrix of the map $m_ f$ with entries in $R$, using a basis $e_1, \ldots , e_ n$ of $A$ as an $R$-module. Then, $A \otimes _ R \kappa (\mathfrak p) \cong (R \otimes _ R \kappa (\mathfrak p))^{\oplus n} = \kappa (\mathfrak p)^ n$, which is an $n$-dimensional vector space over $\kappa (\mathfrak p)$ with basis $e_1 \otimes 1, \ldots , e_ n \otimes 1$. The image $\bar f = f \otimes 1$, and so the multiplication map $m_{\bar f}$ has matrix $(a_{ij} \otimes 1)$. Thus, the characteristic polynomial is precisely the image of $P(T)$.

From linear algebra, we know that a linear transformation acts nilpotently on an $n$-dimensional vector space if and only if the characteristic polynomial is $T^ n$ (since the characteristic polynomial divides some power of the minimal polynomial). Hence, $f$ acts nilpotently on $A \otimes _ R \kappa (\mathfrak p)$ if and only if $\bar P(T) = T^ n$. This occurs if and only if $r_ i \in \mathfrak p$ for all $0 \leq i \leq n - 1$, that is when $\mathfrak p \in V(r_0, \ldots , r_{n - 1}).$ $\square$