I am new to module theory and as of now am not very comfortable with the subject. So can somebody please check whether my claim and its proof is okay?
Consider the following statement:
Let $M$ be a finitely generated module over a nonzero ring (commutative with identity) $R$. Say $M$ can be generated by the set $\{x_1,\ldots, x_n\}\subseteq M$. Then given $n+1$ elements $y_1,\ldots, y_{n+1}\in M$, there exist $a_1, \ldots, a_{n+1}\in R$ not all zero such that $$ a_1y_1+\cdots+a_{n+1}y_{n+1}=0 $$
Purported Proof: Let $f:R^n\to M$ be the unique module homomorphism which sends $e_i=(0, \ldots,0, 1, 0,\ldots, 0)$ to $x_i$. Then $f$ is surjective and we can thus find $t_1, \ldots, t_{n+1}\in R^n$ such that $f(t_j)=y_j$ for all $1\leq j\leq n+1$. Since rank of the free module $R^n$ is $n$, the members $t_1,\ldots, t_{n+1}$ must be linearly dependent. So we have $a_1, \ldots, a_{n+1}\in R$ not all zero such that $a_1t_1+\cdots+a_{n+1}t_{n+1}=0$. Operating $f$ on both sides of the last equation, we get the desired result.