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 $\lambda=\mu=0$).

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 $(A \oplus A \oplus \cdots) \hookrightarrow B$.