Kernel of a polynomial p is invariant under a function: how is this possible?

Question

If $q(t) \in F[t],$ then $\operatorname{Ker}(q(f))$ is invariant under $f.$

How is this possible? If I understand the definition of invariance properly, for all elements of $$w \in \operatorname{Ker}(q(f)) \implies f(w) \in \operatorname{Ker}(q(f)).$$

However, the elements of $\operatorname{Ker}(q(f))$ are functions, no? What is the meaning of $f(f)?$ Or am I mistaken about $\operatorname{Ker}(q(f))$ in that it is composed of the functions $f$ such that for $p$ in $F[t],$ $$p(f) = a_0 + a_1f + a_2f^2 + \cdots a_nf^n = 0?$$

Answer 1

I'm going to assume that $q : T \to T$ and $ f : T \to T$.

The way I'm interpreting your notation: $$ \text{Ker}(q(f)) = \{ t \in T \, | \, \, q(f(t)) = 0 \} \subseteq T$$

But in this case, the statement doesn't hold. For example, let $q(t) = t+1$, let $f(t) = 2t$. Then $q \circ f(t) = 2t + 1$. So $$\text{Ker}(q(f)) = \{ -\frac{1}{2} \}$$ but $f(-\frac{1}{2}) = -1 \notin \text{Ker}(q(f))$; so the statement doesn't hold.

That being said, I'm not entirely clear that this is what you meant but hope that it at least goes towards making the question clearer!

Answer 2

I think that this makes sense (only?) in this situation: we have an $F$-vector space $V$, a linear mapping $f:V\to V$, and a polynomial $q\in F[t]$.

In that case $q(f):V\to V$ in the usual way.

Moreover, $$ u\in\ker q(f) \implies q(f)u=0 \implies f q(f)u=0 \implies q(f)f u= 0 \implies fu\in\ker q(f) $$ since $f$ and $q(f)$ commute.

This shows $\ker q(f)$ is $f$-invaraint.