Let $R$ be commutative ring with 1. Prove that if $R[x]$ finitely generated as $R$-module, then $R=\{0\}$.
Proof.
$R[x]$ finitely generated as $R$-module imply for all $f(x)\in R[x]$ there exist $r_1,r_2,\ldots,r_n$ such that for some $f_1(x),f_2(x),\ldots,f_n(x)\in R[x]$, $$f(x)=r_1f_1(x)+r_2f_2(x)+\ldots+r_nf_n(x).$$
But how to conclude $R=\{0\}$? I don't know to prove it. Any hint to prove it?
I try proving contrapositive as below.
If $R\neq\{0\}$, then $R[x]$ is not finitely generated as $R$-module.
Take any $f(x)=r_0+r_1 x+r_2x^2+\ldots+r_nx^n\in R[x]$. Then there exist $r_0,r_1,r_2,\ldots,r_n\in R$ and $1,x,x^2,\ldots,x^n\in R[x]$ (since $R\neq \{0\}$) such that $$f(x)=r_0(1)+r_1 > (x)+r_2(x^2)+\ldots+r_n(x^n).$$ Then we have the generating set for $R[x]$ is $$\{1,x,x^2,\ldots,x^n\}.$$ I confused why the generating set is finite.
Suppose that $R$ is not the zero ring and $R[x]$ is finitely generated by $f_1,f_2,\dots,f_n$. We can clearly assume that none of the polynomials $f_i$ is zero, so all of them have a degree.
Let $k$ be the maximum degree of the generating polynomials. Now try to make $$ x^{k+1}=r_1f_1+r_2f_2+\dots+r_nf_n \qquad (r_1,r_2,\dots,r_n\in R) $$