Finding the $f$-conductor of of basis vectors

Let $f$ be a linear operator on $\Bbb{K}^4$ whose matrix is $$\begin{bmatrix} c & 0 & 0 & 0 \\ 1 & c & 0 &0 \\ 0 &1 &c & 0 \\0& 0 & 1 & c \end{bmatrix}$$

Let $W=\ker(f-cI_4)$

Find the generators of $S(e_4,W) ,S(e_3,W), S(e_2,W),S(e_1,W)$. Where $S(e_i,W)=\{P(x)\in \Bbb{K}[x] \mid P(f)(e_i)\in W\}$ and $e_i$ are the standard basis vectors.

What I've done: I showed that $W$ is spanned by $e_4$. So for $S(e_4,W)$ , its the set of consant times $e_4$, is it right to say $S(e_4,W)=<c>$?

As for $S(e_3,W)$ : $(f-cI_n)(e_3)=e_4$ hence $S(e_3,W)$ is the set of all polynomials in $\Bbb{K}[x]$ of the form $a_1x+\ldots+a_nx^n$ . I still don't know how to write or find the generator.

As for $S(e_3,W)$ and $S(e_4,W)$ we have $(f-cI_n)(e_2)=e_3$ and $(f-cI_n)(e_1)=e_2$ so upon another application of $(f-cI_n)$ we get $e_4$ . My problem is that I don't know how to write the generators of these sets, especially for the last three sets.

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$f-cI$ has last two columns zero. Its kernel cannot be spanned by $\mathbf{e}_4$ since the kernel is two dimensional. What is true, however, is that $\mathbf{e}_4$ (and $\mathbf{e}_3$) is in $W$. This is enough to conclude that every polynomial is in $S(\mathbf{e}_4,\ W)$ because $P(f)$ commutes with $f-cI$ for all $P\in \Bbb{K}[x]$. Therefore the monic generator of $S(\mathbf{e}_4,\ W)$ (and likewise for $S(\mathbf{e}_3,\ W)$) is $1$. –  EuYu Apr 4 '13 at 23:28
@EuYu I am very sorry I got no answers on this so I did not notice the typo . I hope I didn't take much from your time –  user10444 Apr 4 '13 at 23:38

Since $f-cI$ has kernel spanned by $\mathbf{e}_4$ it follows that every polynomial $P\in \mathbb{K}[x]$ is an element of $S$. This is because $P(f)$ commutes with $f-cI$ for all $P$ and therefore $$(f-cI)\left[P(f)\mathbf{e}_4\right] = P(f)\left[(f-cI)\mathbf{e}_4\right] = P(f)\mathbf{0} = \mathbf{0}$$ Therefore the monic generator for $S(\mathbf{e}_4,\ W)$ is $1$ (indeed any non-zero constant can act as a suitable generator as you correctly concluded. We will talk about monic generators for the remainder of the question).

By definition, the generator must divide each element of the ideal. You have actually already done the majority of the work by noticing the shift property of $(f-cI)$.

We have $(x-c)\in S(\mathbf{e}_3,\ W)$ so the generator is either $1$, which is impossible since $\mathbf{e}_3\notin W$, or it is $(x-c)$ which is the case here.

Likewise, we have $(x-c)^2 \in S(\mathbf{e}_2,\ W)$ so the generator is either $1$ (no good since $\mathbf{e}_2\notin W$) or $(x-c)$ (again, no good since $(f-cI)\mathbf{e}_2 = \mathbf{e}_3\notin W$) or $(x-c)^2$, which is the case here.

The case for $S(\mathbf{e}_1,\ W)$ is very similar. Try your hand at that.

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Thank you very much that cleared a lot up for me, again sorry for the typo. Concerning $S(e_1,W)$ we have $(f-cI)^3e_1=0$ then $(x-c)^3\in S(e_1,W)$ hence the generator is either $1, x-c, (x-c)^2, (x-c)^3$. $1$ does not work since $e_1 \notin W$ neither is it $x-c$ since $(f-cI)e_1=e_2$ neither is it $(x-c)^2$ since $(f-cI)^2e_1=e_3$ hence it is $(x-c)^3$ –  user10444 Apr 5 '13 at 0:37
@user10444 That's exactly right! –  EuYu Apr 5 '13 at 1:38