I'm reading Artin, and he writes:
Is the fact that $(w_0,tw_0,\dots,t^{n-1}w_0)$ is a basis for $W$ as a vector space obvious? It didn't seem obvious to me so I wrote a formal proof of this fact, which despite of being elementary is not a two-line proof. So I was wondering (1) whether my proof is fine and (2) whether this fact is obvious from some other considerations which I didn't use.
Proof of the fact that $(w_0,tw_0,\dots,t^{n-1}w_0)$ generates $W$.
Let $w\in W$, so $w=g(t)w_0$ for some $g(t)\in F[t]$. Its preimage under the map $F[t]\to W$ given by $h(t)\mapsto h(t)w_0$ is $g(t)$. The image of $g(t)$ under the quotient map $F[t]\to F[t]/(f)$ is $\overline {g(t)}$. This last one can be written as $$\overline {g(t)}=\alpha_0\times \overline 1+\dots+\alpha_{n-1}\times \overline{t^{n-1}}=\alpha_0\overline 1+\dots+\alpha_{n-1}\overline{t^{n-1}},$$ where I have used notation ($\times$) from this question. Further we can write $$\overline{g(t)}=\overline {1\alpha_0 }+\dots+\overline {1 \alpha_{n-1}t^{n-1}}=\overline{1\alpha_0+\dots+1\alpha_{n-1}t^{n-1}}.$$ Consider the image of this element under the isomorphism $F[t]/(f)\to W$. On the one hand, the image is $w$. On the other hand, since $\overline 1$ maps to $w_0$, the image is $\alpha_0 w_0+\dots+\alpha_{n-1}t^{n-1}w_0$. So $w$ is the linear combination of $(w_0,tw_0,\dots,t^{n-1}w_0)$ with coeffieients $(\alpha_0,\alpha_1,\dots,\alpha_{n-1})$.
Proof of independence:
Let $\alpha_0w_0+\alpha_1tw_0+\dots+\alpha_{n-1}t^{n-1}w_0=0$ in $W$. Then $p(t)=\alpha_0+\alpha_1t+\dots+\alpha_{n-1}t^{n-1}$ lies in the kernel of the map $F[t]\to W$ described above. So $f$ divides $p$ in $F[t]$: there is $q\in F[t]$ such that $p(t)=f(t)q(t)$. This is impossible unless all $\alpha_i$ are zero since $f$ is monic and hence $\deg p = n-1 < n = \deg f$. (See the answer here.)