# Free module $R[x]$

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..?

• In fact, $R$ is always a free $R$-module. I think you mean that $\mathbb Z / n \mathbb Z$ is not a free $\mathbb Z$-module. Indeed, every element $m$ of $\mathbb Z / n \mathbb Z$ is torsion (since $nm = 0,$ as you correctly observed), hence it is a torsion module. Jun 22, 2020 at 22:13
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.
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 $$R.$$