Let M be an R-module. We say that M to be Co-generator. If for any family $(M_{i})_{i\in I}$, I is non-empty set, $M_{i}$ is submodule of M such that ${O_{M}}=\cap_{i\in I} M_{i}$ then there is $B\subset I$ is finite that ${O_{M}}=\cap_{i \in B} M_{i}$.
But we have some modules such that any family $(M_{i})_{i\in I}$ if they satisfy ${O_{M}}=\cap_{i \in I} M_{i}$, then there must be $i\in I$ that $M_{i}={0}$:
Example: V is K-vector space with 1-dimensional.
Now, we only consider the modules such that there is a family $(M_{i})_{i\in I}$, I is non-empty set, $M_{i}$ is submodule and $M_{i} \neq {0}$ for all $i\in I$ such that ${O_{M}}=\cap_{i\in I} M_{i}$.
Then my question is :
Whether the following assertion is right?
for any family $(M_{i})_{i\in I}$, I is non-empty set, $M_{i}$ is submodule of M such that ${O_{M}}=\cap_{i\in I} M_{i}$ then there is $B\subset I$ is finite that ${O_{M}}=\cap_{i\in B} M_{i}$ $\Longleftrightarrow$ there is $n\in N$, $x_{1},...,x_{n}\in M$ and $x_{1},...,x_{n} \neq 0$ such that ${O_{M}}=(x_{1}) \cap ... \cap (x_{n})$.
the direction $\Longrightarrow$ is easy by hypothesis:
there is a family $(M_{i})_{i\in I}$, I is non-empty set, $M_{i}$ is submodule and $M_{i}\neq {0}$ for all $i\in I$ such that ${O_{M}}=\cap_{i\in I} M_{i}$
so we choose $x_{i}\in M_{i}$, $x_{i} \neq 0$ then ${O_{M}}=\cap_{i\in I} (x_{i})$, so there is $n\in N$, $x_{1},...,x_{n} \in M$ and $x_{1},...,x_{n} \neq 0$ such that ${O_{M}}=(x_{1})\cap ... \cap (x_{n})$.
Can you help me investigating the inverse direction?.
