Background. I noted that there were some conflicting
notations in Nantel's notes for the course as well as
John's Campbell's solutions
for some of the exercises that were assigned in 2017
so I wrote Nantel and asked him the following question:
There is a question (see exercise 1.9):
Show that if $( {\mathcal C}, {\mathcal U} )$ is a concrete category
and $F$ is an functor from $F : $ Set $\rightarrow {\mathcal C}$
mapping a set $X$ in Set to the free objects $F[X]$ in ${\mathcal C}$,
prove that if $X \simeq X'$, then $F[X] \simeq F[X']$.
but then in your notes there is a notation that "this doesn't mean that
$F[X] \simeq\!\!\!\!\!\!/~~ F[Y]$" for R-Mod.
Can you explain with a little more detail what you had
in mind with that remark?
His response:
The category of R-module is concrete (build on top of set)
but the statement is false.
There are some ring R such that the free R-module do not satisfy the statement
[The statement is true if R is commutative, but not true in general]
Here is an example of a (non-commutative) ring $R$ and a free-module $M$ such that
$M=F(X)=F(Y)$ but
$X$ and Y do not have the same cardinality (in fact we can make $Y$ be any cardinality)
Consider $\mathbb{Z}^\omega=\mathbb{Z} \times \mathbb{Z} \times ...$ as a $\mathbb{Z}$ module.
Let $R = End_ {\mathbb{Z}}(\mathbb{Z}^\omega).$
Think of $M=R$ as a left $R$-module.
Let $\phi_1 (a_1, a_2, ...) = (a_1, a_3, a_5, ...)$.
Let $\phi_2 (a_1, a_2, ...) = (a_2, a_4, a_6, ...)$.
Let $\psi_1 (a_1,a_2,...) = (a_1,0,a_2,0,...)$.
Let $\psi_2 (a_1,a_2,...) = (0,a_1,0,a_2,...)$.
Then $(\psi_1 \phi_1+ \psi_2 \phi_2)(a_1,a_2,...) = (a_1,a_2,...)$
so $\psi_1 \phi_1 + \psi_2 \phi_2 = 1$.
So if $\alpha_1 \phi_1 + \alpha_2 \phi_2 = 0$, then
$0 = \alpha_1 \phi_1 \psi_1 + \alpha_2 \phi_2 \psi_1 = \alpha_1$.
Similarly, $0 = \alpha_1 \phi_1 \psi_2 + \alpha_2 \phi_2 \psi_2 = \alpha_2$.
So $Y=\{\phi_1, \phi_2\}$ is a basis for $R$.
Remark that $R$ also has the basis $X=\{Id\}$:
As a left $R$-module, $R=F[X]=F[Y]$ but $1=|X|\ne |Y|=2$
So Nantel stated that this example proves that the problem
stated on the web page is wrong, but as far as I can tell, it
doesn't seem to contradict it.
It is however a very nice example of a module where the notion of rank
isn't properly defined. This is exercise 27 in [DF section 10.3]