Module generated by $n$ elements, containing $n+1$ independent elements Let $A$ be a ring and $M$ a $A$-module such that $M$ is generated by some system of $n$ elements, but contains some other system of $n+1$ linearly independent elements. I want to show that $M$ must contain an infinite system of linearly independent elements. In fact I've been able to prove it under the additional assumption that there is no zero divisor of $A$. Indeed it is easy then to prove by induction that there are two ideals in A, $I$ and $J$, both nonzeros and such that their intersection is trivial ; and then we can construct the infinite system by another induction.
But without this assumption I'm stuck. Could you help?
(Edit. I put the edited part in slanted font)
Here's my proof in case there is no zero divisor. 
First, $A$ contains two nonzero ideals $I$ and $J$ such that $I \cap J=\{0\}$. Indeed we shall argue by contradiction; let $n$ be minimal such that $A^n$ contains some independent family of $n+1$ elements (it is easy to lift the hypothesis to $A^n$). If $n=1$ then the claim is obvious (just pick $Ax$ and $Ay$ where $x,y$ are independent). Otherwise, let $(x_1,\cdots,x_{n+1})$ be some independent family of $A^n$. Let us project $A^n$ onto $A^{n-1}=A^{n-1} \times \{0\}$. It gives us a family $(y_1,\ldots,y_{n+1})$. Since $n$ is minimal, we know $(y_1,\ldots,y_n)$ is not independent. So we can find some nontrivial $\lambda_1,\ldots,\lambda_n$ such that
$$ \sum_{i=1}^n \lambda_i x_i = (0,\ldots,0,\lambda) $$
where $\lambda \in A$ (but $\lambda \neq 0$). We can assume for example that $\lambda_1 \neq 0$. Then by the same token there is some nontrivial sequence $\mu_2,\ldots,\mu_{n+1}$ and some $\mu \neq 0$ such that
$$ \sum_{i=2}^{n+1} \mu_i x_i = (0,\ldots,0,\mu) $$
We assumed that any two nonzeros ideals of $A$ intersect nontrivially. Consequently there must exist $\alpha,\beta$ such that $\alpha \lambda = \beta \mu \neq 0$. Then we can prove that $(x_1,\ldots,x_{n+1})$ is not independent, a contradiction. 
So let $I,J$ be nonzero ideals of $A$ such that $I \cap J=\{0\}$. We can build the infinite independent family by induction. We just put $x_{n+1}'=\alpha x_{n+1}$ and $x_{n+2}'=\beta x_{n+1}$ ($\alpha \in I,\beta \in J$ and $\alpha,\beta \neq 0$). Then the family $(x_1,\ldots,x_n,x_{n+1}',x_{n+2}')$ is independent if the family $(x_1,\ldots,x_{n+1})$ is independent (because $\lambda \alpha + \mu \beta=0$ can happen only if $\lambda=\mu=0$).
 A: To say that $M$ contains $k$ linearly independent elements $\{x_1, \cdots, x_k\}$ is to say that the homomorphism $R^k \hookrightarrow M$, defined by sending the $i$th basis element to $x_i$, is injective.  As you mention above, if $M$ is generated by $n$ elements and contains $n+1$ linearly independent elements, it follows that we may "lift" to $R^n$ so that we have an injective homomorphisms $R^{n+1} \hookrightarrow R^n$.
I claim that if there is such an embedding $R^{n+1} \hookrightarrow R^n$, then $R^n$ contains an infinite linearly independent set.  Once this is proved, it will follow that $M$ contains an infinite linearly independent set, because $R^n \hookrightarrow R^{n+1} \hookrightarrow M$.
So suppose we have an embedding $R^{n+1} \hookrightarrow R^n$. The image of this injection can be written as $X_1 \oplus Y_1$, where $X_1 \cong R$ and $Y_1 \cong R^n$. Now we may proceed inductively to find submodules $(X_{k+1} \oplus Y_{k+1}) \subseteq Y_k$ where $X_k \cong R$ and $Y_k \cong R^n$. We obtain a chain of inclusions $$R^n \supseteq X_1 \oplus Y_1 \supseteq X_1 \oplus (X_2 \oplus Y_2)  \supseteq X_1 \oplus X_2 \oplus (X_3 \oplus Y_3) \cdots $$ As a result, there is a submodule that is an infinite direct sum $$X_1 \oplus X_2 \oplus X_3 \oplus \cdots \subseteq R^n$$ where all $X_i \cong R$. Clearly it follows that $R^n$ has an infinite linearly independent subset.
Notes: (1) A relevant search term might be "strong rank condition" for rings. (2) The proof above generalizes to say that if $A$ and $B$ are modules such that $A \oplus B \hookrightarrow B$, then there is an embedding $(A \oplus A \oplus \cdots) \hookrightarrow B$.
A: It depends on what you mean by "linearly independent"; if what you mean is "minimal set of generators for the module they generate", then the statement is incorrect, even under the assumption that $A$ is an integral domain.
For instance, let $k$ be a field, let $A = k[X,Y]$ (where $X$ and $Y$ are indeterminates), and let $M$ be $A$, as an $A$-module.  Then the set $\{1\}$ of course generates $M$ as an $A$-module, but so does the set $\{X, 1-XY\}$.  Moreover, both are minimal generating sets for $M$.  That is, if you remove either $X$ or $1-XY$ from the latter set, you generate a proper ideal of $A$.
For what it's worth, the set $\{X, 1-XY\}$ is also linearly independent over $k$ if you think of $A$ as a $k$-vector space, in case that was what you meant..
A: Assume that there is a linearly independent set of cardinal $n+1$ in $A^n$. In this case we get a monomorphism $\iota:A^{n+1}\to A^n$. Now, the morphism $\iota\oplus Id:A^{n+2}\to A^{n+1}$ is also a monomorphism, so that $\iota \circ (\iota\oplus Id):A^{n+2}\to A^n$ gives us a linearly independent set of cardinal $n+2$ in $A^n$. The end of the proof follows from an easy induction and the comment of timofei on his question. As pointed out in the comments this only shows that there is a linearly independent set of size $m$ for all $m>n$.
A: I think I have a proof. 
I showed supra that there must exist some independent family $(\alpha,\beta)$ in $A$. Then we check by induction on $n$ that the family 
$$ (\alpha,\alpha \beta,\ldots,\alpha \beta^n) $$
(where $n \geq 1$) is independent. It is easy to finish the proof.
Maybe it is possible do adapt this reasoning and construct an independent family without resorting to the case $n=1$?
