The null-ideal of a companion matrix I am stuck with the following exercise (all required definitions and theorems are provided below):

Let $R$ be a commutative ring, $f \in R[x]$ monic of degree $n$ and let $C_f \in M_n(R)$ be the companion matrix of $f$. Show that the characteristic polynomial $\chi_{C_f}$ of $C_f \in M_n(R)$ is $f$. Then use McCoy's Theorem to show that $N_R(C_f)$ is principal and $N_R(C_f) = f(x)R[x]$.

I have already shown that $\chi_{C_f} = f$, but I have no idea how I should do the second part, for McCoy's Theorem does not say anything about principality. I further recognize that McCoy's Theorem directly implies
$$N_R(C_f) = f(x)R[x] =(\chi_{C_f}(x)R[x] :_{R[x]} J_{n-1}(xI-C)) = (f(x)R[x] :_{R[x]} J_{n-1}(xI-C)),$$
which already contains the desired expression $f(x)R[x]$, but I do not see how to derive $N_R(C_f) = f(x)R[x]$ from that (the definition of ideal quotient is new to me).

Definition: Let $R$ be a commutative ring and $C \in M_n(R)$. The null ideal of $C$ in $R[x]$ is defined as $$N_R(C) = \{g \in R[x] \mid g(C) = 0 \}$$
McCoy's Theorem: Let $C$ be an $n \times n$ matrix with entries in a commutative ring $R$. For $1 \le k \le n$ let $J_k(xI-C)$ be the ideal of $R[x]$ generated by the $k \times k$ minors of $xI-C$. Then
$$ N_R(C) = (J_n (xI-C) :_{R[x]} J_{n-1}(xI-C)) = (\chi_{C}(x)R[x] :_{R[x]} J_{n-1}(xI-C)) $$
(The two latter expressions are ideal quotients.)

Could you please give me a hint?
 A: As you know (recently perhaps), an ideal quotient $(A:B)$ is defined as
$$
  (A:B) = \{ r \colon rB \subseteq A \}.
$$
Since the ideal $A$ has the property of absorbing products, for any $r \in A$ we have $rB \subseteq A$. So $A \subseteq (A:B)$ automatically.
Often it happens that $(A:B)$ is strictly larger than $A$.
But look at what you are supposed to show in this exercise.
You know that $N_R(C) = ((\chi_C(x)) : J_{n-1}(xI-C))$ for any (square, $n \times n$) matrix $C$,
and specifically for a companion matrix $C = C_f$ you're supposed to show that
$N_R(C_f) = (\chi_C(x)) = (f)$.
That means your goal is to show, for this case, where $C = C_f$, you have the equality
$((\chi_C(x)) : J_{n-1}(xI-C)) = (\chi_C(x))$.
In the earlier notation, you want to show that $(A:B) = A$.
This is not an automatic equality, it has to be shown in this particular case.
What could make it true?
Well, one inclusion is automatic, the inclusion $A \subseteq (A:B)$.
What about the other inclusion?
Suppose $r \in (A:B)$, meaning $rB \subseteq A$.
Why would this imply $r \in A$?
Well, $rB \subseteq A$ would indeed imply $r \in A$, if we had $1 \in B$.
So, the hint is: for a companion matrix $C = C_f$,
show that $1 \in J_{n-1}(xI-C)$.
Note that the matrix $xI-C$ has a whole bunch of $(n-1) \times (n-1)$ minors ($n^2$ of them).
It will not be interesting to list them out or to work out all of them.
You want one thing only, to find if $1$ is in that ideal.
The fewer minors you can deal with, the better (simpler).
Best of all would be if you can get it down to just one single minor...
