= Solution
{c}
The zero <polynomial> belongs to $I_v$. If $f,g\in I_v$, linearity gives $(f-g)(T)v=0$. If $a\in K[X]$ and $f\in I_v$, <polynomial> evaluation obeys $(af)(T)=a(T)f(T)$, so $(af)(T)v=0$. Thus $I_v$ is an <ideal> of the <polynomial ring> $K[X]$.
The $n+1$ vectors $v,Tv,\ldots,T^nv$ are linearly dependent in the $n$-dimensional <vector space>. Their dependence supplies a nonzero <polynomial> of <degree> at most $n$ in $I_v$. Hence this <ideal> is nonzero. A <polynomial> ring over a <field> is a <Euclidean domain> under <degree>, and every <ideal> in a Euclidean domain is principal. It follows that $I_v=(m_v)$ for a nonzero <polynomial> $m_v$, which can be scaled uniquely to be monic.
For completeness, uniqueness follows because any two generators divide each other and hence differ by a nonzero constant; the monic normalization forces that constant to be one. Since $m_v$ divides the <polynomial> of <degree> at most $n$ constructed above, its <degree> is at most $n$. The generator characterization gives
$$
\boxed{m_v\mid f\quad\Longleftrightarrow\quad f\in I_v\quad\Longleftrightarrow\quad f(T)v=0.}
$$
This is the <minimal polynomial of a vector under a linear operator>. If $v=0$, the <ideal> is the whole ring and the unique monic generator is the constant <polynomial> one.
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