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Homecardinalities of bases for modules

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# 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 $A=\{a_{1},\ldots,a_{n}\}$ 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 $C=\{b_{1},\ldots,b_{m}\}$ be this finite subset (of $B$). $C$ is linearly independent because $B$ is. If $C\neq B$, 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 $b=r_{1}b_{1}+\cdots+r_{m}b_{m}$, or $0=-b+r_{1}b_{1}+\cdots r_{m}b_{m}$, 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 $|A|\geq\aleph_{0}$, the smallest infinite cardinal, and $B$ is another basis for $M$. We want to show that $|B|=|A|$. First, notice that $|B|\geq\aleph_{0}$ 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 $B^{{\prime}}\neq B$, 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

$|B|=\left|\bigcup_{{a\in A}}B_{a}\right|\leq\aleph_{0}|A|=|A|.$ |

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

$|A|\leq\aleph_{0}|B|\leq|B|.$ |

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

## Mathematics Subject Classification

16D40*no label found*13C05

*no label found*15A03

*no label found*

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