Here is the question I am trying to solve:
Prove that the polynomial ring $R[x]$ in the indeterminate $x$ over the commutative ring $R$ is a flat $R$-module.
My thoughts:
We know by corollary 42 in D&F, third edition, that free modules are flat. I know that any free module has a basis but is the converse true? Is any module that had a basis is a free module. Here is my proof depending on that the converse is true (if not please tell me the correct solution):
Given a commutative ring $R,$ the polynomial ring $R[x]$ is a free $R$-module because for instance, every element in $R[x]$ can be written as $a_0 + a_1 x + \cdots + a_n x^n$ for some non-negative integer $n$ and some elements $a_i$ in $R,$ hence the linearly independent elements $1, x, x^2, \dots$ generate $R[x]$ over $R.$