When considering the commutative ring $R$ over itself, if its submodule is free, then the submodule equals to the module?

Question

When considering the commutative ring $R$ over itself, then this $R$-module is isomorphic to $R$, but if $I$ is an ideal of $R$, then it is a submodule, if this submodule is free too, then it is isomorphic to $R^n$, as its in $R$, thus it is isomorphic to $R$, it is a contradiction.Any mistake I made in it?

Answer 1

By definition, your ideal is isomorphic to $R^n$ but this isomorphism is quite abstract and your proof shows nothing. Counter-example : the ideal $2 \mathbb{Z}$ in $\mathbb{Z}$.

Exercise : show that your property is true if and only if $R$ is a field.

Answer 2

There are rings for which every right ideal is free, and it's even possible for them to have rank more than 1. If such a ring is also commutative, it is a principal ideal domain, and all the ideals are isomorphic (as modules) to the ring.

Answer 3

@Joseph: Isomorphisms are not same as equality. If the natural inclusion map is an isomorphism, then you can say that it is actually an equality. Otherwise not in general. For example, take the ring $A := k[x_1, x_2, x_3, \cdots],$ k is a field and let $B := k[x_2, x_3, x_4, \cdots].$ Then $B$ is a subring of $A$, and it is also isomorphic to $A$ (why?). But clearly $B \neq A.$ Note that in this case, the natural inclusion is not an isomorphism. You have to produce an isomorphism abstractly.