Finitely generated module over a field is always completely reducible

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.

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