# 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 $\mathrm{ker}(\varphi )=0$. 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 $m$. $\mathrm{\square}$

Title | example of free module with bases of diffrent cardinality |
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Canonical name | ExampleOfFreeModuleWithBasesOfDiffrentCardinality |

Date of creation | 2013-03-22 18:07:18 |

Last modified on | 2013-03-22 18:07:18 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 13 |

Author | joking (16130) |

Entry type | Example |

Classification | msc 16D40 |

Related topic | IBN |