Let $R$ be a ring and $f\in R[x]$ a monic polynomial. Consider the quotient map $R[x]\to R[x]/(f)$. Endow the image $R[x]/(f)$ with the structure of an $R[x]$-module by defining multiplication by elements of $R[x]$ as follows: if $g\in R[x]$ and $\overline h \in R[x]/(f)$, then $g\times \overline f=\overline g\overline h$ where the multiplication on the RHS is the multiplication in the quotient ring, namely $\overline g\overline h= \overline gh$. In particular, $R[x]/(f)$ is now an $R$-module. How do I show that the ordered set $(\overline 1,\overline x,\dots, \overline x^{n-1})$ forms an $R$-basis for the $R$-module $R[x]/(f)$?
There are two things to show: $(\overline 1,\overline x,\dots, \overline x^{n-1})$ generates $R[x]/(f)$ and $(\overline 1,\overline x,\dots, \overline x^{n-1})$ is linearly independent over $R$. For the latter:
Assume $\alpha_0\times \overline 1+\dots+\alpha_{n-1}\times \overline x^{n-1}=0$. This means by definitions that $$\overline {\alpha_0}+\dots+\overline{\alpha_{n-1}x^{n-1}}=\overline{\alpha_0+\dots+\alpha_{n-1}x^{n-1}}=0,$$ but how do I conclude from this that the $\alpha_i$s are zero? It's also not clear how to see that the set spans $R[x]/(f)$.
If the polynomial $g(X) = \alpha_0 + \cdots + \alpha_{n-1}X^{n-1}$ is the zero class in $R[X]/(f)$, then this means that $f$ divides $g$ over $R[X]$. But this is impossible unless $\alpha_0 = \ldots = \alpha_{n-1} = 0$; indeed, since $f$ is monic, any nonzero multiple of $f$ in $R[X]$ has degree at least $\deg(f) > \deg(g)$.
To show this set spans, let $g(X) \in R[X]$ be any polynomial. Since $f$ is monic, we can perform Euclidean division on $g(X)$ by $f(X)$ to write $g(X)$ as $g(X) = f(X)q(X) + r(X)$ for some $q, r \in R[X]$ with $\deg(r) < \deg(f)$ (or $r = 0$). Then $\overline{r} = \overline{g}$ in $R[X]/(f)$, and $\overline{r}$ is in the span of $\overline{1}, \overline{X}, \ldots, \overline{X^{n-1}}$.