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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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    $\begingroup$ 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. $\endgroup$ Jun 22, 2020 at 22:13
  • $\begingroup$ @Carlo Aha! Thanks! $\endgroup$
    – Ingrid
    Jun 22, 2020 at 22:14

2 Answers 2

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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.

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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 $R.$

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