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 $|A|=|B|$.∎