# 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 $|J|\leq|I|$.

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

Proof. Let $\alpha:I\rightarrow J\times I$ be a bijection (it exists since $|I|\geq|J|$ and $I$ is infinite) and denote by $\pi_{1}:J\times I\rightarrow J$ and $\pi_{2}:J\times I\rightarrow I$ 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 $\phi:\prod_{j\in J}R\rightarrow R$ by defining the endomorphism $\phi(f)\in R$ for $f\in\prod_{j\in J}R$ as follows:

 $\phi(f)(e_{i})=f(\delta_{1}(i))(e_{\delta_{2}(i)}).$

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

$1)$ Recall that $\phi$ is injective if and only if $\mathrm{ker}(\phi)=0$. So assume that $\phi(f)=0$ 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

 $0=\phi(f)(e_{i_{0}})=f(\delta_{1}(i_{0}))(e_{\delta_{2}(i_{0})})=f(j)(e_{i}).$

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

$2)$ We wish to show that $\phi$ 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 $\phi(f)=h$. $\square$

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 $m$. $\square$

Title example of free module with bases of diffrent cardinality ExampleOfFreeModuleWithBasesOfDiffrentCardinality 2013-03-22 18:07:18 2013-03-22 18:07:18 joking (16130) joking (16130) 13 joking (16130) Example msc 16D40 IBN