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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...

Erik
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3 Answers3

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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 $$\frac{(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})}{n!}\text{ unordered bases.}$$

Arturo Magidin
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As you have $p^2-1$ choices for the first and $p$ multiples to avoid for the second, you get $(p^2-1)(p^2-p)$.

  • I worked a bit further and I stomped in the same conclusion but I think it is not it.. is $(m_{1},m_{2})$ a base, than i count it 2 times with this formula...or not? – Erik Apr 19 '20 at 14:55
  • Oh. Bases are usually ordered. So no. –  Apr 19 '20 at 14:59
  • Not in this case, consider them as families of Vectors that are equal if they contain the same vectors – Erik Apr 19 '20 at 15:08
  • Why is that? Normally ${m_1,m_2}\ne {m_2,m_1}$ The order matters. Change the order and you have a different basis.. –  Apr 19 '20 at 15:24
  • It is a consequence of how my course built bases,they are a set then not a tupple. then is ${m_1,m_2}= {m_2,m_1}$ – Erik Apr 19 '20 at 15:32
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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 : $$(p^{2}-1)\cdotp(p-1)=p(p-1)^{2}(p+1)$$

$\square$

Erik
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