kernel of the evaluation homomorphism I have a question about the kernel of the evaluation homomorphism: Let V be a vector space over a field K, $\text{dim}_KV=n$ and $\phi\in End_K(V)$ which means $\phi:V\to V$ is an endomorphism. Consider the evaluation homomorphism:
$$ev_{\phi}:K[x]\to End_K(V),\;\;   \sum\limits_{i=0}^na_ix^i\mapsto \sum\limits_{i=0}^na_i\phi^i$$ 
$ev_{\phi}$ is a homomorphism of rings, therefore $\text{ker}(ev_{\phi})$ is an ideal of $K[x]$. My question: Why there exists a normed polynomial of degree$\le n^2$ in $\text{ker}(ev_{\phi})$ ?  My first try: $ev_{\phi}$ is a homomorphism of vector spaces, therefore $\text{dim}_K K[x]=\text{dim}_K\text{ker}(ev_{\phi})+\text{dim}_K\text{im}(ev_{\phi})$. Now it is $\text{dim}_K K[x]=\infty$, $\text{dim}_K\text{im}(ev_{\phi})\le \text{dim}_KEnd_k(V)=n^2$,so there must exist a nontrivial polynomial in the kernel of $ev_{\phi}$. But this is not a complete answer of this question, I don't see why there is a normed polynomial of degree $\le n^2$ in the kernel of the evaluation homomorphism. Could you explain me that? 
Edit: a fact which could be important too: $K[x]$ is a principal ideal domain, so $\text{ker}(ev_{\phi})$ is generated by one polynomial.
 A: You correctly argued that the kernel cannot be just $\{0\}$. It then has (like any nonzero ideal of $K[X]$) a monic generator; if $P$ is such a generator, then $\dim K[X]/(P)=\deg P$ (since every class in $\dim K[X]/(P)$ has precisely one representative of degree${}<\deg P$, as follows from the characterisation of Euclidean division by$~P$). Arguing as you did, you get $\deg P\leq n^2=\dim_K(\operatorname{End}_K V)$.
But you can do better than this. The Cayley-Hamilton theorem gives you a monic polynomial of degree$~n$ (namely the characteristic polynomial) in the kernel of the evaluation map, so in fact $\deg P\leq n$. This is the sharpest bound you get in general. Note that $P$ is called the minimal polynomial of$~\phi$, and that its degree can be strictly less than$~n$ (though it is at least$~1$ if $n\neq0$).
A: Note that you've shows that the dimension of $\textrm{Im}(ev_\phi)$ is less than or equal to $n^2$.  This means that the quotient $K[x]/\textrm{ker}(ev_\phi)$ is also less than or equal to $n^2$ dimensional, by the first isomorphism theorem.  You've already mentioned that $K[x]$ is a PID, so write $f(x)$ for the generator of the ideal $\textrm{ker}(ev_\phi)$.  The dimension of $K[x]/\langle f(x) \rangle$ as a vector space over $F$ will then be $\deg(f)$, as every equivalence class has a unique representative $g(x)$ satisfying $\deg(g) < \deg(f)$.  This shows that $\deg(f) \le n^2$.
To ensure that $f$ is normed (which is also referred to as monic), you can rescale it to force the leading term to have $1$ as its coefficient.  Note that taking scalar multiples does not alter the ideal generated by a polynomial.
