Vtyjkyuk

I've once asked a similar question only about groups, but I am interested whether the logic is still sound:

$(1)$Let $S$ be a generating set of a ring $R$, and denote $kappa=vert Svert$. Then $vert Rvert leq kappa$, and if $kappa<aleph_0$ then $vert R vert leq aleph_0$.

This is because any element is written as a finite sum of finite products of elements in $S$. Thus if we denote:

$R_n,m:=Big underseti=1oversetnsum undersetj=1oversetm_iprod a_i,j Bigvert m_ileq m, ; a_i,j subseteq S Big$

We can see that $R=undersetn,m=1oversetinftycupR_n,m$. Hence:

$vert Rvertleq undersetn=1oversetinftysumundersetm=1oversetinftysumvert R_n,mvert$. Where if $kappageq aleph_0$ then $vert R_n,mvert=kappa$, and if $kappa<aleph_0$ then $vert R_n,mvert leq aleph_0$.

$(2)$The natural follow up question (using similar logic) is for a Module $M$ over a ring $R$, if $S$ generates the module $M$ such that $vert Svert= kappageq aleph_0$ is it true that:

$vert M vert=kappa$ if $kappa geq vert Rvert$, and $vert Mvert=vert Rvert$ if $vert Rvert>kappa$.

Let's work under the assumption that both $|R|$ and $|S|$ are infinite. You can work out the other cases.

There is a surjective map from the set $Sigma_<omega(Rtimes S)$ of finite sequences from the set $Rtimes S$ onto the set $M$. By standard cardinality arguments,
$$
|Sigma_<omega(Rtimes S)|=|Rtimes S|
$$
so we can conclude that
$$
|S|le|M|le|Rtimes S|
$$

It is known that
$$
|Rtimes S|=begincases
|R| & text$ \[4px]
|S| & textS
endcases
$$

Therefore, if $|R|le|S|$ you can argue that $|M|=|S|$.

If $|S|<|R|$, one can have $|M|<|R|$. Let $R=J_p$ be the ring of $p$-adic integers; then $R/pR$ is finite (isomorphic to $mathbbZ/pmathbbZ$) and we can consider the module $M=(R/pR)^(aleph_0)$ (countable direct sum of copies of $R/pR$). Then we can take $S=M$ and $|S|=|M|=aleph_0$, but $|R|=2^aleph_0$.

What's the obstruction? There is no injective homomorphism (no injective map, actually) $Rto M$, in this case.