Let $M$ be a finitely generated $F$-module where $F$ is a field. Let $\{m_1,....m_n\}$ be the generating set for $M$.

Then, given $x \in M$, $\exists r_1,r_2,...r_n \in F$ such that $r_1m_1+....+r_nm_n=x$.

If $M$ is a completely reducible module, then given any submodule $N \leq M$, $\exists Z \leq M$ such that $M = Z \oplus N$.

So, back to the original statement. If $M$ is a module over a field, then is $M$ a direct product of copies of $F$? Why is this? If this is so, then the statement is simple, since any submodule of $M$ is just a direct product of some of the summands that compose $M$, and the summands that you don't use form a submodule of $M$ that is the complement of this first one.

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    $\begingroup$ Do you know what vector spaces are? And that modules over a field $F$ are $F$-vector spaces? $\endgroup$ – Paul K Jul 15 '19 at 19:32
  • $\begingroup$ Ahhh, right, hah $\endgroup$ – MSV Jul 15 '19 at 19:36

$F$ is a simple $F$ module, and by basic linear algebra, every $F$ module is isomorphic to one of the form $\oplus_{i\in I}F$ for some index set $I$.


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