I was checking my notes from Hoffman-Kunze book on linear algebra where I came across this result.
Assume $W,V,Z$ are vector spaces where $W \subseteq V \subseteq Z$ are subspaces of $Z$. The linear operator $T:Z\rightarrow Z$ is defined and $W,V$ are invariant under $T$ (i.e. if $v\in V , w\in W$ then $m(T)v\in V,m(T)w\in W$ for any polynomial $m$ over the scalar field). If $\beta\in Z$ is a vector, then the $T-$conductor of $\beta$ into $V$, $f(t)\in S_T(\beta;V)$, must divide the $T-$conductor of $\beta$ into $W$, $g(t)\in S_T(\beta;W)$, i.e. $f(t)\mid g(t)$ or equivalently $g(t)=k(t)f(t)$ for some polynomial $k(t)$ ($f(t)$ and $g(t)$ are monic generators of the conductor ideals, see definition below please).
First the definition
Let W be an invariant subspace of $T$ and let a be $\alpha\in V$. The $T-$conductor of $\alpha$ into $W$ is the set $S_T(\alpha ; W)$ , which consists of all polynomials $g$ (over the scalar field $F$) such that $g(T)\alpha\in W$. $S_T(\alpha ; W)$ is an ideal in the polynomial algebra $F[t]$ and $\color{red}{\text{the unique monic generator of the ideal } S_T(a ; W) \text{ is also called the } T-\text{conductor of }\alpha \\ \text{ into } W}$
Now my question here are two-fold:
$\qquad 1$-Why degree of $g$ is greater than $f$?
$\qquad 2$-Why $f(t)\mid g(t)$?
If it helps, I encountered this in the chapter $7$, "cyclic decomposition and sub-spaces", in the proof of theorem $3$.
Thanks in advance
If $\phi \in S_T(\beta;W)$, then $\phi(T)\beta \in W \subseteq V$ implies $\phi \in S_T(\beta;V).$ Therefore $S_T(\beta;W) \subseteq S_T(\beta;V).$
In particular, as long as you accept that $S_T(\beta;V) = (f)$ and $S_T(\beta;W)=(g)$ are nonzero principal ideals of $F[t]$, it follows that $g \in S_T(\beta;W) \subseteq S_T(\beta;V) = (f).$ This means that $g(t) = f(t) \cdot q(t)$ for some $q(t) \in F[t]$ so that $f(t) \mid g(t)$ and therefore the degree of $f$ is less than or equal to the degree of $g$.