Let $R$ be a ring and $B$ an $R$-algebra that is free as an $R$-module of finite rank $N$. For any element $f\in B$, we may consider the $R$-linear endomorphism of $B$ defined by multiplication by $f$. We then define the characteristic polynomial of $f$, denoted $\chi _f(T)$, by $$\chi _{B/R}^f(T)=\det(T-f)\in R[T].$$ In particular, we define the norm of $f$ as $$\operatorname{N}_{B/R}(f)=(-1)^N\chi _{B/R}^f(0)\in R.$$
Now, under the same context, we may consider the $R[T]$-algebra $B[T]$, which is free of finite rank $N$ as an $R[T]$-module. My question is the following:
Is the following identity in $R[T]$ true?
$$N_{B[T]/R[T]}(T-f)=\chi _{B/R}^f(T)$$
In other words, is the characteristic polynomial of $f\in B$ the norm of $T-f$ relative to $B[T]/R[T]$?
If it is true, could someone please sketch a proof for it or give me a reference?
Thank you very much for your help.