环上的线性代数/主理想域上的模

命题（主理想域上的无扭模是自由模）:

设 $R$ 为一个主理想域，并设 $M$ 为一个 $R$ 上的无扭模。则 $M$ 是自由模。

（关于选择公理条件。）

Proof: We consider the set of sets $S\subseteq M$ such that $S$ is linearly independent and $M/\langle S\rangle$ is torsion-free. This set may be equipped with the partial order that is given by inclusion. Suppose then that $A$ is a totally ordered set, and $(S_{\alpha })_{\alpha \in A}$ is a family such that $\alpha \leq \beta \Leftrightarrow S_{\alpha }\subseteq S_{\beta }$ . We claim that an upper bound for this chain is given by the union of the sets $S_{\alpha }$ , which we shall denote by $S_{\infty }$ . Indeed, $S_{\infty }$ is linearly independent, since any linear relation within $S_{\infty }$ involves only finitely many elements of $S_{\infty }$ , and we may find a sufficiently large (w.r.t. the order of $A$ ) $\alpha \in A$ such that all these elements are contained within $S_{\alpha }$ , so that by the linear independence of $S_{\alpha }$ the given linear relation must be trivial. Moreover, $S_{\infty }$ has the property that $M/\langle S_{\infty }\rangle$ is torsion-free, since if $m\in M$ and $r\in R$ are given such that $m\notin S_{\infty }$ , but $rm\in S_{\infty }$ (ie. the equivalence class of $m$ in $M/\langle S_{\infty }\rangle$ is torsion), then $rm$ is a linear combination of finitely many elements of $S_{\infty }$ , so that once more we find a sufficiently large $\alpha \in A$ such that $rm\in S_{\alpha }$ , and then the equivalence class of $m$ in $R/\langle S_{\alpha }\rangle$ is torsion, a contradiction.

因此，可以应用佐恩引理，它得到一个最大线性无关的 $S\subseteq M$ 使得 $M/\langle S\rangle$ 是无扭的。我们将假设 $\langle S\rangle \neq M$ 引导到一个矛盾。实际上，如果我们有 $\langle S\rangle \neq M$ ，则将存在一个元素 $m\in M\setminus \langle S\rangle$ 。则集合 $S\cup \{m\}$ 将是线性无关的，因为如果存在线性关系