If ideal $I$ of domain $R$ is free $R$-module, how to prove $I$ is principal ideal?
Is this right if $R$ is just a commutative ring?
My thought:
$(1)$ for non-zero ideal $I$ and $J$, let $0\not=a \in I, 0\not=b \in J$, since $R$ is domain, $0\not=ab\in I \cap J$, so $I \cap J\not =\ \varnothing$.
$(2)$ Suppose $\{e_i\}_{i\in \Lambda}$ is basis of free $R$-module $I$, then $I=\bigoplus_{i\in \Lambda}Re_i$, and every $R{e_i}$ is principal ideal generated by $e_i$. Inaccurately speaking, $I$ is generated by elements in $\bigcap_{i\in \Lambda}Re_i$. If $\Lambda$ is a finite set, then that's right. What if $\Lambda$ is an infinite set?
And what if $R$ is just a commutative ring istead of a domain? Thanks in advance!
Suppose $I$ is an ideal and free with basis $\{e_i\}_{i\in\Lambda}$. As you noted, $Re_i\cap Re_j$ is not trivial. If $i,j\in\Lambda$ and $i\ne j$, this contradicts free-ness. Hence $\Lambda$ has at most one element, so $I=Re_1$ or $I=0$.
Interestingly, the Claim remains true for commutative rings: If $I$ is an ideal and free with basis $\{e_i\}_{i\in\Lambda}$, then the $e_i$ cannot be zero divisors (because the basis property postulates that $ce_i=0$ implies $c=0$). Hence the above Argument is still valid, i.e., $0\ne e_ie_j\in Re_i\cap Re_j$ for $i\ne j$.
Or with a slightly different twist: Consider $$ a:=\sum_{i\in\Lambda}c_ie_i$$ where for some $j,k\in\Lambda$ with $j\ne k$ we set $c_i=\begin{cases}e_k&i=j\\-e_j&i=k\\0&\text{otherwise}\end{cases}$. Then $a=0$, hence by the Basis property, $c_i=0$ for all $i\in\Lambda$ in particular $e_j=e_k=0$, which is absurd.