Direct sum of free module?

Question

Let $R$ be a PID, and let $F$ be a free module with basis $e_1, e_2$. Let $u\in F$ such that $u = me_1 + me_2$ where $m \in R$. How can I show that $Ru$ is a direct summand of $F$ if and only if $m$ is a unit?

Answer 1

If $Ru$ is a direct summand then $F = Ru \oplus Rw$ for some $w = ae_1 + be_2$ (I'm using $R$ a PID to conclude that the other summand must be free of rank $1$). Write $e_1 + e_2 = xu + yw$. We cannot have $u, w \in R(e_1 + e_2)$ otherwise they wouldn't generate $F$, so we must have $y = 0$. Then $x = \frac1m$.

On the other hand if $m$ is invertible let $w = e_1$. Show that any $xe_1 + ye_2 \in F$ can be written in the form $au + bw$ for some $a, b \in R$.