The annihilator ideal is nonzero and has a unique monic generator; $p(T)=0$ if and only if ${\mu }_T\mid p$

The zero polynomial lies in $\mathrm{Ann}(T)$. If $p(T)=q(T)=0$, then $(p-q)(T)=0$; and if $h\in F[x]$ and $p(T)=0$, distributivity of the finite polynomial sums and associativity of composition give $(hp)(T)=h(T)p(T)=0$. Hence $\mathrm{Ann}(T)$ is an ideal.

By [L2], ${\chi }_T\in \mathrm{Ann}(T)$. Fix an ordered basis $\mathcal{B}$ of the $n$-dimensional $V$; by [L3], ${\chi }_T={\chi }_{[T{]}_{\mathcal{B}}^{\mathcal{B}}}$ with $[T{]}_{\mathcal{B}}^{\mathcal{B}}\in M_n(F)$, so ${\chi }_T$ is monic of degree $n$ and in particular nonzero. Thus the ideal is nonzero, including when $V=0$, where ${\chi }_T=1$.

If $u$ and $v$ are monic generators of this nonzero ideal, then $u=av$ and $v=bu$ for some polynomials $a,b$. Degrees force $a,b$ to be nonzero constants, and monicity forces both constants to be $1$; hence $u=v$.

By [L1], write $\mathrm{Ann}(T)=(g)$ with $g\mathrm{\ne }0$. Multiplying $g$ by the inverse of its leading coefficient gives a monic generator ${\mu }_T$ by [L4], and step 1.3 shows that this monic generator is unique.

Finally, $p(T)=0$ means $p\in \mathrm{Ann}(T)=({\mu }_T)$, which is equivalent to $p={\mu }_{T}q$ for some $q\in F[x]$, that is, ${\mu }_T\mid p$. When $V=0$, step 1.2 gives $(1)=F[x]$, so its monic generator is $1$.