# cardinalities of bases for modules

Let $R$ be a ring and $M$ a left module over $R$.

###### Proposition 1.

If $M$ has a finite basis, then all bases for $M$ are finite.

###### Proof.

Suppose $A=\{{a}_{1},\mathrm{\dots},{a}_{n}\}$ is a finite basis for $M$, and $B$ is another basis for $M$. Each element in $A$ can be expressed as a finite linear combination^{} of elements in $B$. Since $A$ is finite, only a finite number of elements in $B$ are needed to express elements of $A$. Let $C=\{{b}_{1},\mathrm{\dots},{b}_{m}\}$ be this finite subset (of $B$). $C$ is linearly independent^{} because $B$ is. If $C\ne B$, pick $b\in B-C$. Then $b$ is expressible as a linear combination of elements of $A$, and subsequently a linear combination of elements of $C$. This means that $b={r}_{1}{b}_{1}+\mathrm{\cdots}+{r}_{m}{b}_{m}$, or $0=-b+{r}_{1}{b}_{1}+\mathrm{\cdots}{r}_{m}{b}_{m}$, contradicting the linear independence of $C$.
∎

###### Proposition 2.

If $M$ has an infinite^{} basis, then all bases for $M$ have the same cardinality.

###### Proof.

Suppose $A$ be a basis for $M$ with $|A|\ge {\mathrm{\aleph}}_{0}$, the smallest infinite cardinal, and $B$ is another basis for $M$. We want to show that $|B|=|A|$. First, notice that $|B|\ge {\mathrm{\aleph}}_{0}$ by the previous proposition^{}. Each element $a\in A$ can be expressed as a *finite* linear combination of elements of $B$, so let ${B}_{a}$ be the collection^{} of these elements. Now, ${B}_{a}$ is uniquely determined by $a$, as $B$ is a basis. Also, ${B}_{a}$ is finite. Let

$${B}^{\prime}=\bigcup _{a\in A}{B}_{a}.$$ |

Since $A$ spans $M$, so does ${B}^{\prime}$. If ${B}^{\prime}\ne B$, pick $b\in B-{B}^{\prime}$, so that $b$ is a linear combination of elements of ${B}^{\prime}$. Moving $b$ to the other side of the expression and we have expressed $0$ as a non-trivial linear combination of elements of $B$, contradicting the linear independence of $B$. Therefore ${B}^{\prime}=B$. This means

$$|B|=\left|\bigcup _{a\in A}{B}_{a}\right|\le {\mathrm{\aleph}}_{0}|A|=|A|.$$ |

Similarly, every element in $B$ is expressible as a finite linear combination of elements in $A$, and using the same argument^{} as above,

$$|A|\le {\mathrm{\aleph}}_{0}|B|\le |B|.$$ |

By Schroeder-Bernstein theorem, the two inequalities can be combined to form the equality $|A|=|B|$. ∎

Title | cardinalities of bases for modules |
---|---|

Canonical name | CardinalitiesOfBasesForModules |

Date of creation | 2013-03-22 18:06:33 |

Last modified on | 2013-03-22 18:06:33 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 9 |

Author | CWoo (3771) |

Entry type | Theorem |

Classification | msc 16D40 |

Classification | msc 13C05 |

Classification | msc 15A03 |