Number of bases in $\mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$ I have to calculate how many bases I have in $\mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$  as a $\mathbb{Z}\mathit{/p}\mathbb{Z}$-vector space.
I assumed that there have to be $\frac{p^{2}(p^{2}-1)}{2}$ since $\dim(\mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z})=2$ and I can remove from the standard basis $(e_{1}, e_{2})$ one vector and put any inside for it to still be a base.
That gave me the intuition that for every $(m_{1},m_{2})$ with $m_{1},m_{2}$ in $\mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$ is:
$$m_{1}\neq m_{2} \Rightarrow (m_{1},m_{2})\ \ \mathbf{ is\space a \space \space base \space  of \space  } \mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$$
I struggle to give a satisfying demonstration of that.
Can somebody help me out?
EDIT
I got further like that:
$K := \mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$ 
Let $m_{i}\neq 0_{k}$ be a vector in $K$ 
Then is $M_{i}:=K$\ $span(m_{i})$  the set of linear indipendet vectors to $m_{i}$.
Then: $$\forall m_{i}\space in \space K:|M_{i}|=p^2-p=p(p-1)$$
Now I would be tempted to say that then the number of bases is $$(p^{2}-1)*p(p-1)=p(p-1)^{2}(p+1)$$
But I am pretty sure I need to avoid repetition someway...
 A: It depends on whether you want ordered bases (the standard when dealing with vector spaces) or unordered bases.
For ordered bases, pick the first vector; all you need to do is avoid the zero vector, so you have $p^2-1$ possibilities. For the second vector, you can pick anything that is not a scalar multiple of the first vector (there are $p$ of them), so that gives you $p^2-p$ possibilities. This gives $(p^2-1)(p^2-p)$ possible ordered bases.
For unordered bases, you are counting each basis twice, so all you need to do is divide by $2$, giving $\frac{1}{2}(p^2-1)(p^2-p)$ possibilities. This makes sense for any prime; if $p$ is odd then both factors are even; if $p=2$, then the second factor is even.
This generalizes to $(\mathbb{Z}/p\mathbb{Z})^n$. For ordered bases, after the first two choices you want to pick a third vector that is not a linear combination of the first two (there are $p^2$ of them); etc. So you have
$$(p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1})\text{ ordered bases.}$$
If you want unordered bases, then each base has been counted $n!$ times, so you get
$$\frac{(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})}{n!}\text{ unordered bases.}$$
A: As you have $p^2-1$ choices for the first and $p$ multiples to avoid for the second, you get $(p^2-1)(p^2-p)$.
A: Out of Why number of bases of $\mathbb{F}_p^2$ equals order of $GL_2(\mathbb{F}_p)$?  I can be sure that the answer is $$p(p+1)(p-1)^{2}$$
Proof:
Let $K := \mathbb{Z}\mathit{/p}\mathbb{Z} \times \mathbb{Z}\mathit{/p}\mathbb{Z}$ and $m_{i}\in K:m_{i}\neq0_{K}$.
We define $M_{i}:= K$ \ $\mathbf{span}(m_{i})$, the set of all vectors in $K$ that are linear indipendent to $m_{i}$.
Then:
$$\forall m_{i} \in K : |M_{i}|=p^2 - p  =p(p-1)$$ wich is the number of bases that can be obtained from a single vector different than $0_{K}$
I have then $p^2 -1$ such vectors and the total number of bases is then :
$$(p^{2}-1)\cdotp(p-1)=p(p-1)^{2}(p+1)$$
$\square$
