Any two basis of a free abelian group $F$ have the same cardinal

Question

The proof goes as follows:

Let $A$, $B$ be basis for $F$. For a fixed prime $p$, it is easy to see that the quotient group $F/pF$ is a vector space over $\Bbb Z /p \Bbb Z$ and that the cosets $\{a + pF: a \in A\}$ form a basis. And $\dim F/pF =\text{card } A = \text{card } B$.

I don't see how $F/pF$ is a vector space over $\Bbb Z / p\Bbb Z$. I understand that $\Bbb Z / p \Bbb Z = \{ c + p \Bbb Z : c \in \Bbb Z\}$, where $p \Bbb Z = \{ p*z : z \in \Bbb Z\}$ (multiplication operation), but how is $pF$ defined since $F/pF = \{ h + pF: h \in F\}$ and free abelian groups are defined in terms of addition?

Answer 1

The sum in $\;V_p:=F/pF\;:\;\;\left[u+(pF)\right](+)\left[v+(pF)\right]:=(u+v) +(pF)\;$

Multiplication of element by scalar $\;t\in\Bbb F_p\;:\;\;t\cdot\left[u+(pF)\right]:=(tu)+(pF)\;$

The above makes sense since $\;p\left[u+(pF)\right]=(pu)+(pF)=pF=\overline 0\;\text{ in}\;\;V_p$ .

In the above, $\;u+v\;$ and etc. means the usual sum of elements in the free abelian group $\;F\;$

You can change the above to multiplicative notation if you want (that is what I did first but I changed it right away to additive notation since I think it is more used this way...)

Answer 2

$pF$ is the subgroup of $F$ consisting of all elements of the form $pa : a \in F$. As usual, $pa$ means $a$ added to itself $p$ times.

Already $V = F/pF$ is an abelian group. To give it the structure of a vector space over $k = \mathbb{Z}/p\mathbb{Z}$ is to define scalar multiplication on $V$. That is, you need to define a map

$$k \times V \rightarrow V, (k,v) \mapsto k \cdot v$$

satisfying the usual properties. The natural thing to do is to define

$$(x+ p \mathbb{Z}) \cdot (v + pF) = xv + pF$$

where $xv$ is the usual meaning ($v$ added to itself $x$ times). It isn't clear that this operation is well defined, so you need to show this. Specifically, you need to show that if $x + p\mathbb{Z} = x' + p\mathbb{Z}$, and $v + pF = v' + pF$, then $xv + pF = x'v' + pF$.

After you do this, it should be straightforward to verify that $V$ is a vector space over $k$ in this way.