I've been looking for proofs of the following well-known theorem that don't appeal to linear-algebraic techniques (matrices/determinants): For $R$ commutative with $1$, any two bases for a free module $M$ over $R$ have equal cardinality.
I found one in Robert Ash's Basic Abstract Algebra, which I quote almost in full.
If $I$ is a maximal ideal of $R$,then $k=R/I$ is a field,and $V =M/IM$
is a vector space over k. [By $IM$ we mean all finite sums $a_ix_i$ with
$ai \in I$ and $xi \in M$; thus $IM$ is a submodule of $M$. If $r+I
> \in k$ and $x+IM \in M/IM$, we take $(r+I)(x+IM)$ to be $rx + IM$]Now if $(x_i)$ is a basis for $M$, let $\bar{x_i} = xi+IM$. Since the
$x_i$ span $M$, the $\bar{x_i}$ span $M/IM$. If
$\sum\bar{a_i}\bar{x_i} = 0$, then $a_ix_i \in IM$. Thus $\sum
a_ix_i = \sum b_jx_j$ with $b_j \in I$. Since the $x_i$ form a basis,
we must have $a_i = b_j$ for some $j$. Consequently $a_i \in I$, so
that $\bar{a_i} =0$ in $k$. We conclude that the $\bar{x_i}$ form a
basis for $V$ over $k$, and since the dimension of $V$ over $k$
depends only on $M, R$ and $I$, and not on a particular basis for $M$,
the result follows.
The part in bold is confusing me. How exactly does the result follow from this?
Best Answer
The definition of $V$ was given before introducing a basis $(x_i)$ of $M$. Therefore the dimension $d$ of $V$ is a well-defined number. Now if $(x_i)$ is any basis of $M$, then $(\overline{x_i})$ is a basis of $V$, hence has cardinality $d$. Therefore all bases $(x_i)$ of $M$ have the same cardinality $d$.