# rank of a free submodule of a free module of infinite rank

I am currently studying the free modules and I am stuck in the following question. Please help me.

I know that if $$M$$ is a free module on an infinite subset $$A$$ over a ring $$R$$ (not necessarily commutative) with unity and if $$B$$ is another subset of $$M$$ such that $$M$$ is also free on $$B$$, then $$card(A) = card(B)$$. What I questioned myself is the following:

Let $$M$$ be a free module on an infinite subset $$A$$ over a ring $$R$$ (not necessarily commutative) with unity. If $$N$$ is a submodule of $$M$$ such that $$N$$ is free on a subset $$B$$, then is this true that $$card(B) \leq card(A)$$?

I know that since $$R$$ is non-commutative, therefore, the cardinalities of the free bases for $$N$$ may differ. But, my question is, if $$B$$ is any one of the free bases for $$N$$, then is this true that the cardinality of $$B$$ is smaller than or equal to the cardinality of $$A$$? I also know that if $$R$$ is a PID, then the answer is affirmative. But I am not able to solve the above-mentioned question. Please help me.

Let $$V$$ be an uncountable-dimensional vector space over some field $$k$$ and let $$R=\operatorname{End}_k(V)$$. Note that then for any $$k$$-vector space $$W$$, there is a natural left $$R$$-module structure on $$\operatorname{Hom}_k(W,V)$$ which coincides with the usual left $$R$$-module structure on $$R$$ when $$W=V$$. So as a left $$R$$-module, $$R=\operatorname{Hom}_k(V,V)\cong \operatorname{Hom}_k(V^{\oplus{\dim V}},V)\cong \operatorname{Hom}_k(V,V)^{\dim V}=R^{\dim V},$$ since $$V\cong V^{\oplus{\dim V}}$$. In particular, this means there is a submodule of $$R$$ which is free of rank $$\dim V$$. So, if you take a free $$R$$-module on a countably infinite set, it has a submodule which is free on an uncountable set.