MHB How can we conclude from that that I is a principal ideal?

[mathmari]

Hey!

Let $R$ be a commutative ring with unit.
I want to show that if $I$ is an ideal of $R$ then $I$ is a free $R$-module iff it is a principal ideal that is generated by an element $a$ that is not a zero-divisor in $R$.

Suppose that $I$ is an ideal of $R$ and it is a free $R$-module.
Then it has a basis, i.e., a generating set consisting of linearly independent elements.
How can we conclude from that that $I$ is a principal ideal? (Wondering)

 

[mathmari]

Suppose that $I$ is a free ideal, so it has a basis, i.e., a generating set consisting of linearly independent elements. Suppose that $I$ is not a principal domain.
Does this mean that the basis has more than one element? (Wondering)
Suppose that it has two elements in the basis, say $x_1,x_2$.
Then $I=\{r_1x_1+r_2x_2\}$.
Do we take $r_1=x_2$ and $r_2=−x_1$ ? (Wondering)
Then we would have $x_2x_1−x_1x_2=0$. But since the set is linearly independent, and $x_1,x_2$ are non-zero, we have a contradiction.
Therefore, it is generated by one element $a$. Can we say also that $a$ is not a zero-divisor in $R$ ? (Wondering)
Is this correct? (Wondering)

For the other direction, suppose that $I$ is a principal ideal, that is generated by an element $a$ that is not a zero-divisor in $R$.
How could we conclude that $I$ is a free $R$-module? (Wondering)