Maximal linear independent set in free module of finite rank over commutative ring with unity Let $R$ be a commutative ring with unity. Consider the usual $R$-module $R^n$. Let $S$ be a maximal linearly independent subset of $R^n$ over $R$, i.e. $S$ is linearly independent and if $S \subseteq S' $ with $S'$ linearly independent in $R^n$ over $R$ then $S=S'$. We know that $|S| \le n$. Is it true that $|S|=n$ ? 
I know this is true if $R$ is an integral domain, but I don't know what happens otherwise. Please help. Thanks in advance. 
 A: Let me basically translate parts of Les familles libres maximales d'un module ont-elles le meme cardinal by M. Lazarus which are relevant to you.
Let $A$ be a commutative unital ring, $M$ an $A$-module. If $S\subset M$ is a linear independent subset then denote by $L(S)$ the submodule generated by $S$. By $\Lambda^mM$ we mean the $m$th exterior power.
(1.2) Any linearly independent subset $S\subset A^n$ has at most $n$ elements.
Proof: Let $S$ be an independent set of $p$ elements of $A^n$. We have [an injection] $L(S)\hookrightarrow A^n$, from which we find [an injection]
$\Lambda^p L(S)\hookrightarrow \Lambda^p A^n$. Since $\Lambda^p L(S)\simeq A$, we conclude that $\Lambda^p A^n\neq 0$, hence $p\leq n$.
(1.1), minors: Let $(x_1, \dots, x_p)$ be elements of $A^n$. We call the minors of order $p$ of $n\times p$ matrix with the columns the coordinates of $x_i$, in the canonical basis of $A^n$, the minors of order $p$ of the system $(x_1, \dots, x_p)$. If $1\leq i_1<i_2<\cdots< i_p\leq n$ we will denote the corresponding minor by $\Delta^{i_1<\cdots<i_p}$.
From now on assume $A$ is Noetherian. The notation $\operatorname{Ass}A$ means the associated primes of $A$: the set of prime ideals of $A$ which are also annihilators of some element of $A$. Since $A$ is Noetherian, $\operatorname{Ass}A$ is finite and contains the minimal ideals of $A$. Also, the set of zero-divisors of $A$ is the union of associated primes.
(3.1) Let $I$ be an ideal of $A$. $\operatorname{Ass}I\neq 0 \Longleftrightarrow \exists \mathfrak{p}\in\operatorname{Ass}A: I\subset \mathfrak{p}$.
Proof: If $\operatorname{Ann}I\neq 0$, then every element of $I$ is a zero-divisor; therefore $I\subset \bigcup_{\mathfrak{p}\subset\operatorname{Ass}A}\mathfrak{p}$, and according to [prime] avoidance lemma, $I\subset \mathfrak{p}$ for some $\mathfrak{p}\in\operatorname{Ass}A$. Conversely, if $I\subset \mathfrak{p}$, with $\mathfrak{p}\in\operatorname{Ass}A$, then $\mathfrak{p}=\operatorname{Ann}x$ for some nonzero $x$. Therefore $xI=0$, meaning $\operatorname{Ann}I\neq 0$.
(3.2) Let $M\subset A^n$. Suppose that $\forall x\in M$, $\operatorname{Ann}x\neq 0$ [$\operatorname{Ann}=$ annihilator ideal]. Then $\operatorname{Ann}M\neq 0$.
Proof: If $x=(a_1, \dots, a_n)\in M$ then there exists $\lambda\neq 0$ such that $\lambda a_i=0$ for all $i$. As a result $\operatorname{Ann}\left(\sum_i \langle a_i\rangle\right)\neq 0$, and by (3.1) there exists $\mathfrak{p}\in \operatorname{Ass}A$ such that $\sum_i \langle a_i\rangle\subset\mathfrak{p}$, meaning $x\in\mathfrak{p}A^n$. So $M\subset\bigcup \mathfrak{p}A^n$, and by avoidance lemma, $M\subset \mathfrak{p}A^n$ for some $\mathfrak{p}$. $\mathfrak{p}=\operatorname{Ann}x$ for some non-zero $x\in M$ which [meaning $\mathfrak{p}$] belongs to $\operatorname{Ann}M$.
(3.3) Let $(x_1, \dots, x_p)$ be a linear independent subset of $A^n$. If $p<n$, then there exists $x$ such that $(x_1, \dots, x_p, x)$ is an independent set.
Proof: Let $\Delta^{i_1<\cdots<i_p}$ be a minor of order $p$ of the system $(x_1, \dots, x_p)$. If $x=(a_1, \dots, a_n)$, the minors of order $p+1\leq n$ of the system $(x_1, \dots, x_p, x)$ are, up to a sign,
$$
\Delta_x^{i_1<\cdots<i_{p+1}}=\sum_{k=1}^{p+1}(-1)^k a_{i_k}\Delta^{i_1<\cdots< \hat{i}_k <\cdots<i_{p+1}}
$$
($\hat{i}_k$ means that the index $i_k$ is omitted). Let $N={n\choose p+1}$. Let $\varphi: A^n\to A^N$ be an $A$-linear map given by
$$
\varphi(x)=\Big(\Delta_{x}^{i_1<\cdots<i_{p+1}}\Big)_{i_1<\cdots<i_{p+1}}
$$
and let $M=\operatorname{Im}\varphi$.

*

*[Claim:] $\operatorname{Ann}M = 0$. Suppose not, let $\lambda \in\operatorname{Ann}M$ and $\lambda \neq 0$. For any sequence $i_1<\cdots<i_p$, there exists $i$ such that $i\neq i_k$ for all $k$. So
$$
\lambda \varphi(e_i)=0\Longrightarrow 0=\lambda \Delta^{i_1<\cdots <i<\cdots <i_p}=\pm \lambda \Delta_{e_i}^{i_1<\cdots<\hat{i}<\cdots<i_p}=
\pm \lambda \Delta^{i_1<\cdots<i_p}
$$
which means $(x_1, \dots, x_p)$ is not linearly independent.

*By (3.2), there exists $x\in A^n$ such that $\operatorname{Ann}\varphi(x)=0$, which by design means $(x_1, \dots, x_p, x)$ is a linearly independent subset.


(3.4) Let $A$ be a Noetherian ring and $L$ a free $A$-module. Then any maximal independent subset has cardinality $\dim L$.

The proof is just the combination of (1.2) + (3.3). This is however only for "Noetherian" rings.
Edit: the property you are actually using is that the set of zero-divisors is the union of associated primes. This is true tautologically for any integral domain, Noetherian or not.
P.S. Just because a math text is in French does not mean you can't read it! Worst case scenario use google translate, that's how I learned how to read math literature in French.
