Addendum A to 373

An analogous true statement to Prp 373.4 can be formulated for minimal generating subsets. The proof again proceeds by Zorn’s lemma, although less straightforward.

Lemma 429.1(Lemma).

Let $M$ be an $R$-module, and let $S\subseteq M$ be a generating subset. Then there exists a minimal generating subset of $M$ contained in $S$.

Proof.

Call a subset $X\subseteq M$ independent if $⟨X⟩\ne ⟨X\setminus \{x\}⟩$ for all $x\in X$ (see Rmk 2). Let $\mathcal{I}$ be the family of independent subsets of $S$. Partially order $\mathcal{I}$ by inclusion. Let $\mathcal{C}\subseteq \mathcal{I}$ be a chain. Let $X={\bigcup }_{A\in \mathcal{C}}A$. We contend that $X$ is independent. Indeed, if $x\in X$ is such that $⟨X⟩=⟨X\setminus \{x\}⟩$, then $x$ can be expressed as a finite linear combination ${\sum }_{i=1}^n{\alpha }_i{x}_i$, where $x_i\in X$. There must be an $A\in \mathcal{C}$ which contains the finite set $\{x,x_1,x_2,\dots ,x_n\}$, which results in a contradiction since $⟨A⟩=⟨A\setminus \{x\}⟩$.

By Zorn’s lemma, $\mathcal{I}$ has a maximal element, say $Y$. We contend that $Y$ generates $M$. If this is not the case, there must exist $s\in S$ such that $s\notin ⟨Y⟩$; then $Y\cup \{s\}$ is independent, contradicting the maximality of $Y$. Thus, $Y$ generates $M$. No proper subset of $Y$ can generate $M$, since $Y$ is independent.□

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Remark 429.2(Remark).

We’ve encountered several different definitions of independence that are not to be confused:

1. linear independence, as defined in Def 373.1;
2. term-wise independence, as defined in Def 398.8;
3. non-redundancy independence, as defined in the proof of Lem 1.

Clearly, $(1)⟹(2)⟹(3)$; the reverse implications are generally false. For example, $\{2,3\}\subseteq \mathbb{Z}$ are non-redundant, but not term-wise independent, since $3(2)+(-2)(3)=0$.

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Using Lem 1, we can provide another proof of the fact that modules over fields are free. Let $M$ be a $k$-module. Let $A$ be a minimal generating set of $M$; we will show that that $A$ is a basis. We have a surjection $\phi :k^{\oplus A}\twoheadrightarrow M$. Suppose $\ker \phi$ is nontrivial, i.e, there exists a relation ${\sum }_{i=1}^n{c}_i{a}_i=0$ in $M$, where $a_1,\dots ,a_n\in A$. Since $k$ is a field, we can write

$$\begin{array}{r}{a}_1=\sum _{i=2}^n\frac{c_i}{c_1}{a}_i,\end{array}$$

so $M$ is generated by $A\setminus \{a_1\}$, contradicting the minimality of $A$. Thus, $\phi$ is an isomorphism, and $M$ is free.