Let $R$ be a commutative ring. Is the polynomial ring $R[x]$ a free module? For example, $\mathbb{Z}_{n}$ is not a free module over $\mathbb{Z}_{n}$ because $\forall a \in\mathbb{Z}_{n}$ $na=0$. It seems that we can do the same with $\mathbb{Z}_{n}[x]$. If $\lbrace 1,x,x^{2},...,x^{k}\rbrace$ is a basis then $\lambda_{0}=\lambda_{1}=…=\lambda_{k}=n \Rightarrow \lambda_{0}+\lambda_{1}x+...+\lambda_{k}x^{k}=0 \Rightarrow \mathbb{Z}_{n}[x] $ is not a free module..?
Thank you in advance!!
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Given a commutative ring $R,$ the polynomial ring $R[x]$ is a free $R$-module. 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 $R.$
Actually, the polynomial ring $R[X]$ over a commutative ring $R$ is defined as the free $R$-module $R^{(\mathbf N)}$ ($R$-sequences with finite support), endowed with termwise addition and scalar multiplication, and a multiplication.
In this context, the special sequence $(0,1, 0,\dots,0,\dots)$ is denoted $X$, and one checks that $$X^2=(0,0, 1,0\dots), \qquad X^3=(0,0,0,1, 0,\dots)$$ and so on.