Solution
= Solution
Apply <polynomial division> in $F[x]$:
$$
f=q m_T+r,\qquad r=0\text{ or }\deg r<\deg m_T.
$$
Evaluation at $T$ gives $r(T)=f(T)-q(T)m_T(T)=0$. A nonzero remainder could be rescaled to a monic <polynomial> annihilating $T$ of smaller <degree>, contradicting the defining minimality. Hence $r=0$ and
$$
\boxed{m_T\mid f.}
$$
In particular, the annihilating <polynomials> form the <principal ideal> $(m_T)$.