Showing that the rank of $M$ is exactly $1.$

Question

Here is the question I am trying to solve:

Let $R = \mathbb Z[x]$ and let $M = (2,x)$ be the ideal generated by $2$ and $x,$ considered as a submodule of $R.$ Show that $\{2,x\}$ is not a basis of $M.$ Show that the rank of $M$ is $1$ but $M$ is not free of rank $1.$

My thoughts

I was able to prove the first part and proving the first part showed that the rank of $M$ is at most one, how can I show that it is exactly one in a simple and clear way?

I think we need to find a single element in $M$ that is not a multiple of any other element in $M.$ how about considering the element $x \in M$ and saying that since $x$ is not divisible by 2 (which I am not very sure if the wording is correct), it is not a multiple of any other element in $M$ (is this statement correct?) Thus, the rank of $M$ is exactly 1.

I saw a discussion about this here Prove $(2, x)$ is not a free $R$-module. but I still did not get the solution.

I saw also online the following idea " R is an integral domain, hence every set is linearly independent, so rank of M is 1" but I do not quite well understand why $R$ being an integral domain leads to that rank of M is 1, could anyone explain to me this idea please if it is the correct idea of solution?

Any help will be greatly appreciated!

Edit:

My definition of rank is: For any integral domain $R,$ the rank of an $R$-module $M$ is the maximal number of $R$-linearly independent elements of $M.$

Answer 1

Any nonzero ideal $M$ of an integral domain $R$ has rank $1$ as an $R$-module.

Indeed, it has rank $\leq 1$ since two elements $a,b\in M$ are always $R$-dependent: indeed, we have $ba-ab=0$, which is a nonzero dependence relation if $(a,b)\neq (0,0)$ (if $a=b=0$, just consider $a-b=0$).

Now pick any non zero element $x$ in $M$. An dependence relation writes $ax=0$, where $a\in R$.But since $x$ is non zero and $R$ is an integral domain, then $a=0$ , so there is non non trivial dependence relation. Done.

Note in your case that $M$ is not free, since free submodules of an integral domain $R$ are exactly the principal ideals of $R$.