= Solution
Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:
$$
\boxed{\bigl(f(B)\bigr)_{k,k+1}=f'(A),}
$$
where $f'$ is the <formal derivative>. Thus the candidate is $g=f'$. If the <irreducible polynomial> $f$ has $f'\ne0$, its <polynomial degree> makes $f$ and $f'$ coprime. Equivalently, $f$ is a <separable polynomial>. This includes every irreducible <polynomial> over a <field> of characteristic zero, and every irreducible <polynomial> over a <perfect field>.
Under this separability hypothesis, <Bézout's identity for polynomials> gives $a(x)f(x)+b(x)f'(x)=1$. Evaluating at $A$ proves that $C=f'(A)$ is an <invertible matrix>. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a <nilpotent matrix> with $f(B)^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, so
$$
\bigl(f(B)^{N-1}\bigr)_{1,N}=C^{N-1}\ne0.
$$
For $N=1$, the corresponding assertion is simply $f(B)^0=I\ne0$. The <block upper triangular matrix> $B$ has
$$
\chi_U(x)=\det(B-xI)=\det(A-xI)^N=(-1)^{nN}f(x)^N.
$$
The <minimal polynomial> divides this <characteristic polynomial>. Since $f$ is an <irreducible polynomial>, it must be a power of $f$; the exact <nilpotency index> just found forces
$$
\boxed{m_U=f^N,\qquad \chi_U=(-1)^{nN}f^N\quad\text{if }f\text{ is separable}.}
$$
\b[The arbitrary-field qualification is necessary.] In the <rational function field> $F=\mathbb F_2(t)$, take $f(x)=x^2-t$. This is an <irreducible polynomial>: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its <formal derivative> is zero. With $N=2$,
$$
A=\begin{pmatrix}0&1\\t&0\end{pmatrix},\qquad
B=\begin{pmatrix}A&I\\0&A\end{pmatrix},\qquad
B^2=tI_4.
$$
The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative <polynomial> $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general <field>.
There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h'\ne0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an <irreducible polynomial>, so $h$ and $h'$ are coprime. The <binomial theorem> in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the <Hasse derivatives>, gives
$$
f(B)=K^qH,
$$
where $H$ commutes with $K$ and is a <block upper triangular matrix> whose diagonal blocks are $h'(A^q)$. These blocks are invertible by <Bézout's identity for polynomials>, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, and
$$
f(B)^s=K^{qs}H^s=0\quad\Longleftrightarrow\quad qs\geq N.
$$
The same <characteristic polynomial> and divisibility argument now prove the corrected arbitrary-field answer:
$$
\boxed{m_U(x)=f(x)^{\lceil N/q\rceil},\qquad
\chi_U(x)=(-1)^{nN}f(x)^N.}
$$
This is the <minimal polynomial of a companion block chain>; $q=1$ recovers the intended separable case.
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