example of free module with bases of diffrent cardinality

Let $k$ be a field and $V$ be an infinite dimensional vector space over $k$. Let ${\{e_i\}}_{i\in I}$ be its basis. Denote by $R=\mathrm{End}(V)$ the ring of endomorphisms of $V$ with standard addition and composition as a multiplication.

Let $J$ be any set such that $|J|\le |I|$.

Proposition. $R$ and ${\prod }_{j\in J}R$ are isomorphic as a $R$-modules.

Proof. Let $\alpha :I\to J\times I$ be a bijection (it exists since $|I|\ge |J|$ and $I$ is infinite) and denote by ${\pi }_1:J\times I\to J$ and ${\pi }_2:J\times I\to I$ the projections. Moreover let ${\delta }_1={\pi }_1\circ \alpha$ and ${\delta }_2={\pi }_2\circ \alpha$.

Recall that ${\prod }_{j\in J}R=\{f:J\to R\}$ (with obvious $R$-module structure) and define a map $\varphi :{\prod }_{j\in J}R\to R$ by defining the endomorphism $\varphi (f)\in R$ for $f\in {\prod }_{j\in J}R$ as follows:

$$\varphi (f)(e_i)=f({\delta }_1(i))(e_{{\delta }_2(i)}).$$

We will show that $\varphi$ is an isomorphism. It is easy to see that $\varphi$ is a $R$-module homomorphism. Therefore it is enough to show that $\varphi$ is injective and surjective.

$1)$ Recall that $\varphi$ is injective if and only if $\ker (\varphi )=0$. So assume that $\varphi (f)=0$ for $f\in {\prod }_{j\in J}R$. Note that $f=0$ if and only if $f(j)=0$ for all $j\in J$ and this is if and only if $f(j)(e_i)=0$ for all $j\in J$ and $i\in I$. So take any $(j,i)\in J\times I$. Then (since $\alpha$ is bijective) there exists $i_0\in I$ such that $\alpha (i_0)=(j,i)$. It follows that ${\delta }_1(i_0)=j$ and ${\delta }_2(i_0)=i$. Thus we have

$$0=\varphi (f)(e_{i_0})=f({\delta }_1(i_0))(e_{{\delta }_2(i_0)})=f(j)(e_i).$$

Since $j$ and $i$ were arbitrary, then $f=0$ which completes this part.

$2)$ We wish to show that $\varphi$ is onto, so take any $h\in R$. Define $f\in {\prod }_{j\in J}R$ by the following formula:

$$f(j)(e_i)=h(e_{{\alpha }^{-1}(j,i)}).$$

It is easy to see that $\varphi (f)=h$. $\mathrm{\square }$

Corollary. For any two numbers $n,m\in \mathbb{N}$ there exists a ring $R$ and a free module $M$ such that $M$ has two bases with cardinality $n,m$ respectively.

Proof. It follows from the proposition, that for $R=\mathrm{End}(V)$ we have

$$R^n\simeq R\simeq R^m.$$

For finite set $J$ module ${\prod }_{j\in J}R$ is free with basis consisting $|J|$ elements (product is the same as direct sum). Therefore (due to existence of previous isomorphisms) $R$-module $R$ has two bases, one of cardinality $n$ and second of cardinality $m$. $\mathrm{\square }$