Consider the morphism of schemes $\operatorname{Spec} k[t] \rightarrow \operatorname{Spec} k[u]$ given by the corresponding homomorphism of rings $\phi: k[u]\rightarrow k[t]$, $u \rightarrow p(t)$ for some degree $n$ polynomial in $t$. Then the claim is that the morphism is finite because $\phi$ makes $k[t]$ into a finitely generated $k[u]$-module (generated by $1,t,t^2,...,t^{n-1}$).
I don't understand how $k[t]$ is generated here as $k[u]$ module. I believe $k[u]$ acts on $k[t]$ via $u\cdot a_{(t)}=p_{(t)}*a_{(t)}$, for any element $a_{(t)}\in k[t]$, and in general $q_{(u)}\cdot a_{(t)}=q(p_{(t)})*a_{(t)}$ for some polynomial $q_{(u)}\in k[u]$ ("*" is multiplication of polynomials). So I don't see how can I reproduce any polynomial in $t$ by linear combinations of first $n$ degrees monomials $t^k$, $k=0,1,...,n-1$, and those very special polynomials of form $q(p_{(t)})$ multiplying polynomials of degree $<n$ that can be obtained from the monomials. Probably, there is a mistake in what I said so far or I just don't see how this works out. Any help is appreciated.
Thank you!
Following @red_trumpet comment, I can solve this by induction on the leading power $m$ of a general polynomial $r_{(t)}=a_mt^m+...\in k[t]$. If $m<n$, it`s immediate. Let $m=nk+a$. Then $q_{(t)}=u^{(m-a)/n}\cdot t^a$ is a polynomial with leading power $m$, say $q_{(t)}=b_mt^m+...$. So $r_{(t)}-\frac{a_m}{b_m}q_{(t)}$ is of degree m-1. By induction on $m$, I can produce the entire $r_{(t)}$ this way, so $k[t]$ really is generated as a $k[u]$ module by first $n$ powers of $t$.