1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The answer to your question is no. There exist rings A such that the free module A is isomorphic to the free module A^3. (Yes, it is weird) $\endgroup$ Aug 29, 2023 at 0:19
  • 1
    $\begingroup$ You should look into Lam's book on modules, which has a chapter dedicated to this sort of thing. $\endgroup$ Aug 29, 2023 at 0:19

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .