 Solution

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ 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 $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ 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, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe 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

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 $N=2$,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 $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ 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, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.