$R$ commutative ring and $R_{n}[x]$ the set of polynomials of degree less or equal to $n$ with coefficients on $R$. Show that $R_{n}[x]$ is $R-$module and $R_{n-1}[x] \cong R_{n}[x]/R$, for $n\geq 1$.
Answer:
$R_{n}[x]$ is $R-$module iff $R\times R_{n}[x] \rightarrow R_{n}[x]$ (the image of $(r,f(x))$ being denoted by $rf(x)$) .
$f(x) = a_{n}x^{n} + . . . + a_{0}$ so
$rf(x)=ra_{n}x^{n} + . . . + ra_{0} \in R_{n}[x]$
For all $r,s \in R$ and $f(x), g(x) \in R_{n}[x]$:
$r(f(x) + g(x)) = rf(x) + rg(x)$
$(r+s)f(x) = rf(x) + sf(x)$
$r(sf(x)) = (rs)f(x)$
$1_{R}f(x) = f(x)$ for all $f(x) \in R_{n}[x]$
Then $R_{n}[x]$ is $R-$module.
Is that right? What about the second part? Any solution, any hint?
Your solution to the first part is correct.
For the second part, instead of working with $R_n[x]$, note that $R_n[x]$ is isomorphic to $R^n$, the set of $n$-tuples of $R$. (This should be pretty obvious, but if not, take a homomorphism $\phi: R_n[x] \to R^n$ mapping $a_nx^n + \cdots + a_0$ to $(a_n, \cdots, a_0)$, then invoke the isomorphism theorem.)
Once you have this isomorphism, you're done, as $R^{n-1}$ is clearly isomorphic to $R^n/R$.