Prove $(2, x)$ is not a free $R$-module.

Solution 1:

Your first part is correct.

As for the second part, think bigger: show that any two elements of an ideal $I$ in a commutative ring are linearly dependent. (Hint: this is just as easy as the first part!)

As for the third part: in view of the second part, in any commutative ring the ideals which are free as $R$-modules are necessarily principal ideals. (In a domain the converse is also true: any ideal of the form $\langle a \rangle$ is isomorphic to $R$ as an $R$-module. In a general ring this is true iff $a$ is not a zero-divisor: if $a$ is a zero-divisor, $\langle a \rangle$ has a nonzero annihilator ideal $I$ and $\langle a \rangle \cong R/I$ is not a free module.) So you want to show that $I = \langle 2,x \rangle$ is not principal, which is fairly straightforward (and in fact has been asked on this site before). One possible starting point: notice that $I$ is a prime ideal...