I was told by the author of the answer here Showing that the rank of $M$ is exactly $1.$ that: Free submodules of an integral domain $R$ are exactly the principal ideals of $R.$
I am wondering which theorem, proposition or example in Dummit & Foote, third edition, say this. Could someone clarify this to me please?
Submodules of $R$ are precisely the ideals of $R$ (this is by definition of an ideal). Free submodules of $R$ are thus ideals which are isomorphic (as modules) to $R^n$ for some $n$ (the rank of the free module). Say the ideal $I$ is generated freely by $a_1,…,a_n \in R$
We show that if $R$ is integral then $n\leq 1$. Indeed otherwise $a_1$ and $a_2$ are distinct, so the submodules they generate inside $I$ intersect only in zero (here we use that the generating set was free). But $a_1a_2=a_2a_1$ is in both $\langle a_1\rangle$ and $\langle a_2\rangle$. This is possible only if $a_1a_2=0$, showing $R$ isn’t integral.