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Denote by $\Bbb Z_5^2$ the direct product $\Bbb Z_5 \times \Bbb Z_5$. This is a vector space with coefficients from $\Bbb Z_5$. Determine the points on the line formed by $([1]_5, [2]_5)$ and find out how many lines there are in $\Bbb Z_5^2$.

The points on the line spanned by $([1]_5, [2]_5)$ seem to be of the form $$r(t)=([1]_5, [2]_5)+([1]_5, [2]_5)t$$ where $t \in \Bbb Z_5$? How can I start to figure out how many lines there are in total? It seems that I have to figure out how many different points $([x]_5,[y]_5)$ I can consider? Or can I just start somehow shifting the line formed by $([1]_5,[2]_5)$ until I get back to it?

Jen
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  • Since everything is small and finite here, I think that "determine the points" in this exercise expects you to concretely write down a list of all the points, rather than just a recipe for generating them. – Troposphere Dec 03 '21 at 12:13
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    The line can be written plainly $r(t)=([1]_5, [2]_5)t$... – Jean Marie Dec 03 '21 at 12:25
  • https://math.stackexchange.com/q/315762 – Jean Marie Dec 03 '21 at 12:32
  • Let me use $a_5$ instead of $[a]_5$ for the convenience.

    Two lines formed by nonzero points $(a_5, b_5)$ and $(c_5, d_5)$ are the same when $\exists k \in \mathbb{Z}_5$ s.t. $(a_5, b_5) = k(c_5, d_5)$. Note that for any line $L$ formed by a nonzero point $(a_5, b_5)$, $k(a_5, b_5)$ generates $L$ for $k = 1_5, 2_5, 3_5, 4_5$. There are $5\times 5 -1 = 24$ nonzero points, so there are $24/4 = 6$ lines. They are generated by $(1_5, 0_5)$, $(1_5, 1_5)$, $(1_5, 2_5)$, $(1_5, 3_5)$, $(1_5, 4_5)$, $(0_5, 1_5)$ respectively.

     – dust05 Dec 03 '21 at 12:40
  • This holds for any prime $p$, i.e. there are $(p^2-1)/(p-1)=p+1$ lines (on the origin) in $\mathbb{Z}_p^2$. – dust05 Dec 03 '21 at 12:40

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